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#### Page No 2.33:

#### Question 1:

Find the zeros of each of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:

(i) *f*(*x*) = *x*^{2} − 2*x* − 8

(ii) *g*(*s*) = 4*s*^{2} − 4*s* + 1

(iii) *h*(*t*) = *t*^{2} − 15

(iv) 6*x*^{2} − 3 − 7*x*

(v) $p\left(x\right)={x}^{2}+2\sqrt{2}x-6$

(vi) $q\left(x\right)=\sqrt{3}{x}^{2}+10x+7\sqrt{3}$

(vii)$f\left(x\right)={x}^{2}-\left(\sqrt{3}+1\right)x+\sqrt{3}$

(viii) *g*(*x*) = *a*(*x*^{2} + 1) − *x*(*a*^{2} + 1)

(ix) $h\left(s\right)=2{s}^{2}-\left(1+2\sqrt{2}\right)s+\sqrt{2}$

(x) $f\left(v\right)={v}^{2}+4\sqrt{3}v-15$

(xi) $p\left(y\right)={y}^{2}+\frac{3\sqrt{5}}{2}y-5$

(xii) $q\left(y\right)=7{y}^{2}-\frac{11}{3}y-\frac{2}{3}$

#### Answer:

(i) We have,

*f*(*x*) = *x*^{2} − 2*x* − 8

*f*(*x*) = *x*^{2} + 2*x* − 4*x* − 8

*f*(*x*) = *x* (*x* + 2) − 4(*x* + 2)

*f*(*x*) = (*x* + 2) (*x* − 4)

The zeros of *f*(*x*) are given by

*f*(*x*) = 0

*x*^{2} − 2*x* − 8 = 0

(*x* + 2) (*x* − 4) = 0

*x* + 2 = 0

*x* = −2

Or

*x* − 4 = 0

*x* = 4

Thus, the zeros of *f*(*x*) = *x*^{2} − 2*x* − 8 are *α* = −2 and *β* = 4

Now,

and

Therefore, sum of the zeros =

Product of the zeros

_{= − 2 × 4}

_{= −8}

and

Therefore,

Product of the zeros =

Hence, the relation-ship between the zeros and coefficient are verified.

(ii) Given

When have,

*g*(*s*) = 4*s*^{2} − 4*s* + 1

*g*(*s*) = 4*s*^{2} − 2*s* − 2*s* + 1

*g*(*s*) = 2*s* (2*s* − 1) − 1(2*s* − 1)

*g*(*s*) = (2*s* − 1) (2*s* − 1)

The zeros of *g*(*s*) are given by

Or

Thus, the zeros ofare

and

Now, sum of the zeros

and

Therefore, sum of the zeros =

Product of the zeros

and =

Therefore, the product of the zeros =

Hence, the relation-ship between the zeros and coefficient are verified.

(iii) Given

We have,

$h\left(t\right)={t}^{2}-15\phantom{\rule{0ex}{0ex}}h\left(t\right)={\left(t\right)}^{2}-{\left(\sqrt{15}\right)}^{2}\phantom{\rule{0ex}{0ex}}h\left(t\right)=\left(t+\sqrt{15}\right)\left(t-\sqrt{15}\right)$

The zeros of are given by

$h\left(t\right)=0\phantom{\rule{0ex}{0ex}}\left(t-\sqrt{15}\right)\left(t+\sqrt{15}\right)=0\phantom{\rule{0ex}{0ex}}\left(t-\sqrt{15}\right)=0\phantom{\rule{0ex}{0ex}}t=\sqrt{15}\phantom{\rule{0ex}{0ex}}\mathrm{or}\phantom{\rule{0ex}{0ex}}\left(t+\sqrt{15}\right)=0\phantom{\rule{0ex}{0ex}}t=-\sqrt{15}$

$\mathrm{Hence},\mathrm{the}\mathrm{zeros}\mathrm{of}h\left(t\right)\mathrm{are}\alpha =\sqrt{15}\mathrm{and}\beta =-\sqrt{15}.$

Now,

Sum of the zeros

and =

Therefore, sum of the zeros =

also,

Product of the zeros = *αβ*

and,

Therefore, the product of the zeros =

Hence, The relationship between the zeros and coefficient are verified.

(iv) Given

We have,

The zeros of are given by

Or

Thus, the zeros of are and.

Now,

Sum of the zeros = *α* + *β*

and, =

Therefore, sum of the zeros =

Product of the zeros = *α* × *β*

and, =

Product of zeros =

Hence, the relation between the zeros and its coefficient are verified.

(v) Given

We have,

The zeros of are given by

Or

Thus, The zeros of areand

Now,

Sum of the zeros = *α* + *β*

and,

Therefore, Sum of the zeros =

Product of the zeros

and

Therefore, The product of the zeros =

Hence, the relation-ship between the zeros and coefficient are verified.

(vi) Given

We have,

The zeros of g(*x*) are given by

Or

Thus, the zeros of are and.

Now,

Sum of the zeros = *α* + *β*

and =

Therefore, sum of the zeros =

Product of zeros = *α* × *β*

and =

Therefore, the product of the zeros =

Hence, the relation-ship between the zeros and coefficient are verified.

(vii) Given

The zeros of ƒ(*x*) are given by

Or

Thus, the zeros of are *α* = 1 and

Now,

Sum of zeros = *α* + *β*

And,

Therefore, sum of the zeros =

Product of the zeros = *αβ*

And

=

Product of zeros =

Hence, the relation-ship between the zeros and coefficient are verified.

(viii) Given

The zeros of *g*(*x*) are given by

or

Thus, the zeros of are

and

Sum of the zeros = *α* + *β*

and, =

Product of the zeros

And, =

Therefore,

Product of the zeros =

Hence, the relation-ship between the zeros and coefficient are verified.

(ix) $h\left(s\right)=2{s}^{2}-\left(1+2\sqrt{2}\right)s+\sqrt{2}$

$h\left(s\right)=2{s}^{2}-s-2\sqrt{2}s+\sqrt{2}\phantom{\rule{0ex}{0ex}}h\left(s\right)=s\left(2s-1\right)-\sqrt{2}\left(2s-1\right)\phantom{\rule{0ex}{0ex}}h\left(s\right)=\left(2s-1\right)\left(s-\sqrt{2}\right)$

The zeros of *h*(*s*) are given by

*h*(*s*) = 0

$2{s}^{2}-\left(1+2\sqrt{2}\right)s+\sqrt{2}=0\phantom{\rule{0ex}{0ex}}\left(2s-1\right)\left(s-\sqrt{2}\right)=0\phantom{\rule{0ex}{0ex}}\left(2s-1\right)=0\mathrm{or}\left(s-\sqrt{2}\right)=0\phantom{\rule{0ex}{0ex}}s=\frac{1}{2}\mathrm{or}s=\sqrt{2}$

Thus, the zeros of $h\left(s\right)=\left(2s-1\right)\left(s-\sqrt{2}\right)\mathrm{are}\alpha =\frac{1}{2}\mathrm{and}\beta =\sqrt{2}$.

Now,

Sum of the zeros = $\alpha +\beta $

$=\frac{1}{2}+\sqrt{2}$

and

$\frac{-\mathrm{Coefficient}\mathrm{of}s}{\mathrm{Coefficient}\mathrm{of}{s}^{2}}\phantom{\rule{0ex}{0ex}}=-\left(\frac{-\left(1+2\sqrt{2}\right)}{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{1+2\sqrt{2}}{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}+\sqrt{2}$

Therefore, sum of the zeros =

Product of the zeros

$=\frac{1}{2}\times \sqrt{2}=\frac{1}{\sqrt{2}}$

and

$\frac{\mathrm{Constant}\mathrm{term}}{\mathrm{Coefficient}\mathrm{of}{s}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

Therefore,

Product of the zeros =

Hence, the relation-ship between the zeros and coefficient are verified.

(x) $f\left(v\right)={v}^{2}+4\sqrt{3}v-15$

$f\left(v\right)={v}^{2}+5\sqrt{3}v-\sqrt{3}v-15\phantom{\rule{0ex}{0ex}}={v}^{2}-\sqrt{3}v+5\sqrt{3}v-15\phantom{\rule{0ex}{0ex}}=v\left(v-\sqrt{3}\right)+5\sqrt{3}\left(v-\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=\left(v-\sqrt{3}\right)\left(v+5\sqrt{3}\right)$

The zeros of *f*(*v*) are given by

*f*(*v*) = 0

${v}^{2}+4\sqrt{3}v-15=0\phantom{\rule{0ex}{0ex}}\left(v+5\sqrt{3}\right)\left(v-\sqrt{3}\right)=0\phantom{\rule{0ex}{0ex}}\left(v-\sqrt{3}\right)=0\mathrm{or}\left(v+5\sqrt{3}\right)=0\phantom{\rule{0ex}{0ex}}v=\sqrt{3}\mathrm{or}v=-5\sqrt{3}$

Thus, the zeros of $f\left(v\right)=\left(v-\sqrt{3}\right)\left(v+5\sqrt{3}\right)\mathrm{are}\alpha =\sqrt{3}\mathrm{and}\beta =-5\sqrt{3}$.

Now,

Sum of the zeros = $\alpha +\beta $

$=\sqrt{3}-5\sqrt{3}=-4\sqrt{3}$

and

$\frac{-\mathrm{Coefficient}\mathrm{of}v}{\mathrm{Coefficient}\mathrm{of}{v}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{-4\sqrt{3}}{1}=-4\sqrt{3}$

Therefore, sum of the zeros =

Product of the zeros

$=\sqrt{3}\times \left(-5\sqrt{3}\right)=-15$

and

$\frac{\mathrm{Constant}\mathrm{term}}{\mathrm{Coefficient}\mathrm{of}{v}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{-15}{1}=-15$

Therefore,

Product of the zeros =

Hence, the relation-ship between the zeros and coefficient are verified.

(xi) $p\left(y\right)={y}^{2}+\frac{3\sqrt{5}}{2}y-5$

$p\left(y\right)=\frac{1}{2}\left(2{y}^{2}+4\sqrt{5}y-\sqrt{5}y-10\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[2y\left(y+2\sqrt{5}\right)-\sqrt{5}\left(y+2\sqrt{5}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\left(2y-\sqrt{5}\right)\left(y+2\sqrt{5}\right)\right]$

The zeros are given by *p*(*y*) = 0.

Thus, the zeros of $p\left(y\right)=\frac{1}{2}\left(2y-\sqrt{5}\right)\left(y+2\sqrt{5}\right)\mathrm{are}\alpha =\frac{\sqrt{5}}{2}\mathrm{and}\beta =-2\sqrt{5}$.

Now,

Sum of the zeros = $\alpha +\beta $

$=\frac{\sqrt{5}}{2}-2\sqrt{5}=\frac{\sqrt{5}-4\sqrt{5}}{2}=\frac{-3\sqrt{5}}{2}$

and

$\frac{-\mathrm{Coefficient}\mathrm{of}y}{\mathrm{Coefficient}\mathrm{of}{y}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{-3\sqrt{5}}{2}}}{1}=\frac{-3\sqrt{5}}{2}$

Therefore, sum of the zeros =

Product of the zeros

$=\frac{\sqrt{5}}{2}\times -2\sqrt{5}\phantom{\rule{0ex}{0ex}}=-5$

and

$\frac{\mathrm{Constant}\mathrm{term}}{\mathrm{Coefficient}\mathrm{of}{y}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{-5}{1}=-5$

Therefore,

Product of the zeros =

Hence, the relation-ship between the zeros and coefficient are verified.

(xii) $q\left(y\right)=7{y}^{2}-\frac{11}{3}y-\frac{2}{3}$

$q\left(y\right)=\frac{1}{3}\left(21{y}^{2}-11y-2\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(21{y}^{2}-14y+3y-2\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[7y\left(3y-2\right)+1\left(3y-2\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[\left(7y+1\right)\left(3y-2\right)\right]$

The zeros are given by *q*(*y*) = 0.

Thus, the zeros of $q\left(y\right)=\frac{1}{3}\left(7y+1\right)\left(3y-2\right)\mathrm{are}\alpha =\frac{-1}{7}\mathrm{and}\beta =\frac{2}{3}$.

Now,

Sum of the zeros = $\alpha +\beta $

$=\frac{-1}{7}+\frac{2}{3}\phantom{\rule{0ex}{0ex}}=\frac{11}{21}$

and

$\frac{-\mathrm{Coefficient}\mathrm{of}y}{\mathrm{Coefficient}\mathrm{of}{y}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{-\left(-{\displaystyle \frac{11}{3}}\right)}{7}=\frac{11}{21}$

Therefore, sum of the zeros =

Product of the zeros

$=\frac{-1}{7}\times \frac{2}{3}\phantom{\rule{0ex}{0ex}}=\frac{-2}{21}$

and

$\frac{\mathrm{Constant}\mathrm{term}}{\mathrm{Coefficient}\mathrm{of}{y}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{-2}{3}}}{7}\phantom{\rule{0ex}{0ex}}=\frac{-2}{21}$

Therefore,

Product of the zeros =

Hence, the relation-ship between the zeros and coefficient are verified.

#### Page No 2.34:

#### Question 2:

For each of the following , find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find the zeroes of these polynomials by factorization.

(i) $-\frac{8}{3},\frac{4}{3}$ (ii) $\frac{21}{8},\frac{5}{16}$ (iii) $-2\sqrt{3},-9$ (iv) $\frac{-3}{2\sqrt{5}},-\frac{1}{2}$Question

#### Answer:

We know that a quadratic polynomial whose sum and product of zeroes are given is

$f\left(x\right)=k\left\{{x}^{2}-\left(\mathrm{Sum}\mathrm{of}\mathrm{zeroes}\right)x+\mathrm{Product}\mathrm{of}\mathrm{zeroes}\right\}$

(i) We have, sum of zeroes = $\frac{-8}{3}$ and product of zeroes = $\frac{4}{3}$

So, the required quadratic polynomial will be $f\left(x\right)=k\left({x}^{2}+\frac{8}{3}x+\frac{4}{3}\right)$

$f\left(x\right)=k\left({x}^{2}+\frac{8}{3}x+\frac{4}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{3}\left(3{x}^{2}+8x+4\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{3}\left(3{x}^{2}+6x+2x+4\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{3}\left(3x\left(x+2\right)+2\left(x+2\right)\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{3}\left(3x+2\right)\left(x+2\right)$

Now, the zeroes are given by *f*(*x*) = 0.

Thus, $x=-\frac{2}{3}\mathrm{and}x=-2$

(ii) We have, sum of zeroes = $\frac{21}{8}$ and product of zeroes = $\frac{5}{16}$

So, the required quadratic polynomial will be $f\left(x\right)=k\left({x}^{2}-\frac{21}{8}x+\frac{5}{16}\right)$

$f\left(x\right)=k\left({x}^{2}-\frac{21}{8}x+\frac{5}{16}\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{16}\left(16{x}^{2}-42x+5\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{16}\left(16{x}^{2}-40x-2x+5\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{16}\left(16{x}^{2}-2x-40x+5\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{3}\left(2x\left(8x-1\right)-5\left(8x-1\right)\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{3}\left(8x-1\right)\left(2x-5\right)$

Now, the zeroes are given by *f*(*x*) = 0.

Thus, $x=\frac{1}{8}\mathrm{and}x=\frac{5}{2}$

(iii) We have, sum of zeroes = $-2\sqrt{3}$ and product of zeroes = −9.

So, the required quadratic polynomial will be $f\left(x\right)=k\left({x}^{2}+2\sqrt{3}x-9\right)$.

$f\left(x\right)=k\left({x}^{2}+2\sqrt{3}x-9\right)\phantom{\rule{0ex}{0ex}}=k\left({x}^{2}+3\sqrt{3}x-\sqrt{3}x-9\right)\phantom{\rule{0ex}{0ex}}=k\left(x+3\sqrt{3}\right)\left(x-\sqrt{3}\right)$

Now, the zeroes are given by *f*(*x*) = 0.

Thus, $x=-3\sqrt{3}\mathrm{and}x=\sqrt{3}$.

(iv) We have, sum of zeroes = $\frac{-3}{2\sqrt{5}}$ and product of zeroes = $\frac{-1}{2}$

So, the required quadratic polynomial will be $f\left(x\right)=k\left({x}^{2}+\frac{3}{2\sqrt{5}}x-\frac{1}{2}\right)$.

$f\left(x\right)=\frac{k}{2\sqrt{5}}\left(2\sqrt{5}{x}^{2}+3x-\sqrt{5}\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{2\sqrt{5}}\left(2\sqrt{5}{x}^{2}+5x-2x-\sqrt{5}\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{2\sqrt{5}}\left[\sqrt{5}x\left(2x+\sqrt{5}\right)-1\left(2x+\sqrt{5}x\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{k}{2\sqrt{5}}\left(2x+\sqrt{5}\right)\left(\sqrt{5}x-1\right)$

Now, the zeroes are given by *f*(*x*) = 0.

Thus, $x=\frac{-\sqrt{5}}{2}\mathrm{and}x=\frac{1}{\sqrt{5}}$.

#### Page No 2.34:

#### Question 3:

If α and β are the zeros of the quadratic polynomial *f*(*x*) = *x*^{2} − 5*x* + 4, find the value of $\frac{1}{\mathrm{\alpha}}+\frac{1}{\mathrm{\beta}}-2\mathrm{\alpha \beta}$.

#### Answer:

Sinceand are the zeros of the quadratic polynomial

Therefore

=

= 5

We have,

By substituting and we get ,

Taking least common factor we get ,

Hence, the value of is.

#### Page No 2.34:

#### Question 4:

If α and β are the zeros of the quadratic polynomial *p*(*y*) = 5*y*^{2} − 7*y* + 1, find the value of $\frac{1}{\mathrm{\alpha}}+\frac{1}{\mathrm{\beta}}$.

#### Answer:

Since α and β are the zeros of the quadratic polynomial

We have,

By substituting and we get ,

Hence, the value of is.

#### Page No 2.34:

#### Question 5:

If α and β are the zeros of the quadratic polynomial *f*(*x*) = *x*^{2} − *x* − 4, find the value of $\frac{1}{\mathrm{\alpha}}+\frac{1}{\mathrm{\beta}}-\mathrm{\alpha \beta}$.

#### Answer:

Since and are the zeros of the quadratic polynomials

sum of the zeros =

Product if zeros =

We have,

By substituting and we get ,

Hence, the value ofis.

#### Page No 2.34:

#### Question 6:

If α and β are the zeros of the quadratic polynomial *f*(*x*) = *x*^{2} + *x* − 2, find the value of $\frac{1}{\mathrm{\alpha}}-\frac{1}{\mathrm{\beta}}$.

#### Answer:

Since and are the zeros of the quadratic polynomials

Sum of the zeros =

Product if zeros =

We have,

By substituting and we get ,

By substituting in we get ,

Taking square root on both sides we get

.

Hence, the value of is.

#### Page No 2.34:

#### Question 7:

If one zero of the quadratic polynomial *f*(*x*) = 4*x*^{2} − 8*kx* − 9 is negative of the other, find the value of *k*.

#### Answer:

Since and are the zeros of the quadratic polynomial

= 0

Hence, the Value of is .

#### Page No 2.34:

#### Question 8:

If the sum of the zeros of the quadratic polynomial *f*(*t*) = *kt*^{2} + 2*t* + 3*k* is equal to their product, find the value of *k*.

#### Answer:

Let be the zeros of the polynomial.Then,

It is given that the sum of the zero of the quadratic polynomial is equal to their product then, we have

Hence, the value of k is

#### Page No 2.34:

#### Question 9:

If α and β are the zeros of the quadratic polynomial *p*(*x*) = 4*x*^{2} − 5*x* − 1, find the value of α^{2}β + αβ^{2}.

#### Answer:

Since and are the zeros of the quadratic polynomials

Sum of the zeros =

Product of zeros =

We have,

By substituting and in, we get

Hence, the value of is.

#### Page No 2.34:

#### Question 10:

If α and β are the zeros of the quadratic polynomial *f*(*t*) = *t*^{2} − 4*t* + 3, find the value of α^{4}β^{3} + α^{3}β^{4}.

#### Answer:

Since and are the zeros of the quadratic polynomial

We have

Hence, the value of is .

#### Page No 2.34:

#### Question 11:

If α and β are the zeros of the quadratic polynomial *f*(*x*) = 6*x*^{2} + *x* − 2, find the value of $\frac{\mathrm{\alpha}}{\mathrm{\beta}}+\frac{\mathrm{\beta}}{\mathrm{\alpha}}$.

#### Answer:

Since and are the zeros of the quadratics polynomial

*f* (*x*)=

sum of zeros =

Product of the zeros =

We have,

By substituting and we get ,

Hence, the value of is .

#### Page No 2.34:

#### Question 12:

If α and β are the zeros of the quadratic polynomial *p*(*s*) = 3*s*^{2} − 6*s* + 4, find the value of $\frac{\mathrm{\alpha}}{\mathrm{\beta}}+\frac{\mathrm{\beta}}{\mathrm{a}}+2\left(\frac{1}{\mathrm{\alpha}}+\frac{1}{\mathrm{\beta}}\right)+3\mathrm{\alpha \beta}$.

#### Answer:

Since α and β are the zeros of the quadratic polynomial

We have,

By substituting and we get ,

Hence, the value of is

#### Page No 2.35:

#### Question 13:

If the squared difference of the zeros of the quadratic polynomial *f*(*x*) = *x*^{2} + *px* + 45 is equal to 144, find the value of *p*.

#### Answer:

Given and are the zeros of the quadratic polynomial

We have,

Substituting and then we get,

Hence, the value of is .

#### Page No 2.35:

#### Question 14:

If α and β are the zeros of the quadratic polynomial *f*(*x*) = *x*^{2} − *px* + *q*, prove that $\frac{{\mathrm{\alpha}}^{2}}{{\mathrm{\beta}}^{2}}+\frac{{\mathrm{\beta}}^{2}}{{\mathrm{\alpha}}^{2}}=\frac{{p}^{2}}{{q}^{2}}-\frac{4{p}^{2}}{q}+2$.

#### Answer:

Since and are the zeros of the quadratic polynomial

= *p*

We have,

Hence, it is proved that is equal to .

#### Page No 2.35:

#### Question 15:

If α and β are the zeros of the quadratic polynomial *f*(*x*) = *x*^{2} − *p*(*x* + 1) − *c*, show that (α + 1) (β + 1) = 1 − *c*.

#### Answer:

Since and are the zeros of the quadratic polynomial

Then

=

=

We have to prove that

Substituting and we get,

Hence, it is shown that.

#### Page No 2.35:

#### Question 16:

If α and β are the zeros of a quadratic polynomial such that α + β = 24 and α − β = 8, find a quadratic polynomial have α and β as its zeros.

#### Answer:

Given

……(i)

……(ii)

By subtracting equation from we get

Substituting in equation we get,

Let S and P denote respectively the sum and product of zeros of the required polynomial. then,

Hence, the required polynomial if is given by

Hence, required equation is where *k* is any non-zeros real number.

#### Page No 2.35:

#### Question 17:

If α and β are the zeros of the quadratic polynomial *f*(*x*) = *x*^{2} − 1, find the quadratic polynomial whose zeros are $\frac{2\mathrm{\alpha}}{\mathrm{\beta}}\mathrm{and}\frac{2\mathrm{\beta}}{\mathrm{\alpha}}$.

#### Answer:

Since α and β are the zeros of the quadratic polynomial

The roots are

Let S and P denote respectively the sum and product of zeros of the required polynomial. Then,

Taking least common factor we get,

Hence, the required polynomial is given by,

Hence, required equation is Where *k* is any non zero real number.

#### Page No 2.35:

#### Question 18:

If α and β are the zeros of the quadratic polynomial *f*(*x*) = *x*^{2} − 3*x* − 2, find a quadratic polynomial whose zeros are $\frac{1}{2\mathrm{\alpha}+\mathrm{\beta}}\mathrm{and}\frac{1}{2\mathrm{\beta}+\mathrm{\alpha}}$.

#### Answer:

Since are the zeros of the quadratic polynomial

The roots are and

Let S and P denote respectively the sum and the product of zero of the required polynomial . Then,

Taking least common factor then we have ,

By substituting and we get ,

By substituting and we get ,

Hence ,the required polynomial is given by

Hence, the required equation is Where *k* is any non zero real number.

#### Page No 2.35:

#### Question 19:

If α and β are the zeros of the polynomial* f*(*x*) = *x*^{2} + *px* + *q*, from a polynomial whose zeros are (α + β)^{2} and (α − β)^{2}.

#### Answer:

If* * and are the zeros of the quadratic polynomial

Let S and P denote respectively the sums and product of the zeros of the polynomial whose zeros are and. Then,

The required polynomial of is given by

, where *k* is any non-zero real number.

#### Page No 2.35:

#### Question 20:

If α and β are the zeros of the quadratic polynomial *f*(*x*) = *x*^{2} − 2*x* + 3, find a polynomial whose roots are (i) α + 2, β + 2 (ii) $\frac{\mathrm{\alpha}-1}{\mathrm{\alpha}+1},\frac{\mathrm{\beta}-1}{\mathrm{\beta}+1}$.

#### Answer:

(i) Since and are the zeros of the quadratic polynomial

Product of the zeros =

Let S and P denote respectively the sums and product of the polynomial whose zeros

Therefore the required polynomial *f (x)* is given by

Hence, the required equation is.

(ii) Since and are the zeros of the quadratic polynomial

Product of the zeros =

Let S and P denote respectively the sums and product of the polynomial whose zeros

By substituting and we get ,

The required polynomial *f (x)* is given by,

Hence, the required equation is , where *k* is any non zero real number .

#### Page No 2.35:

#### Question 21:

If α and β are the zeroes of the quadratic polynomial *f*(*x*) = *ax*^{2} + *bx* + *c*, then evaluate :

(i) α − β

(ii) $\frac{1}{\mathrm{\alpha}}-\frac{1}{\mathrm{\beta}}$

(iii) $\frac{1}{\mathrm{\alpha}}+\frac{1}{\mathrm{\beta}}-2\mathrm{\alpha \beta}$

(iv) α^{2}β − αβ^{2}

(v) α^{4} + β^{4}

(vi) $\frac{1}{a\mathrm{\alpha}+b}+\frac{1}{a\mathrm{\beta}+b}$

(vii) $\frac{\mathrm{\beta}}{a\mathrm{\alpha}+b}+\frac{\mathrm{\alpha}}{a\mathrm{\beta}+b}$

(viii) $a\left(\frac{{\mathrm{\alpha}}^{2}}{\mathrm{\beta}}+\frac{{\mathrm{\beta}}^{2}}{\mathrm{\alpha}}\right)+b\left(\frac{\mathrm{\alpha}}{\mathrm{\beta}}+\frac{\mathrm{\beta}}{\mathrm{\alpha}}\right)$

#### Answer:

(i) Given *α* and are the zeros of the quadratic polynomial *f* (*x*)

=

We have,

Substituting and then we get,

Hence, the value of is.

(ii) Given *α* and are the zeros of the quadratic polynomial *f* (*x*)

=

We have,

Substituting and then we get,

By taking least common factor we get,

Hence the value of is .

Given *α* and are the zeros of the quadratic polynomial *f* (*x*)

=

We have,

By cross multiplication we get,

By substituting and we get ,

Hence the value of is.

Given *α* and are the zeros of the quadratic polynomial *f* (*x*)

=

We have,

By taking common factor we get,

By substituting and we get ,

Hence the value of is.

Given *α* and are the zeros of the quadratic polynomial *f* (*x*)

=

We have,

By substituting and we get ,

By taking least common factor we get

${\alpha}^{4}+{\beta}^{4}={\left[{\left(\frac{-b}{a}\right)}^{2}-2\times \left(\frac{c}{a}\right)\right]}^{2}-2\times {\left(\frac{c}{a}\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}={\left[\frac{{b}^{2}}{{a}^{2}}-\frac{2c}{a}\right]}^{2}-2\times {\left(\frac{c}{a}\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}={\left[\frac{{b}^{2}-2ac}{{a}^{2}}\right]}^{2}-2\times \frac{{c}^{2}}{{a}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{{\left({b}^{2}-2ac\right)}^{2}}{{a}^{4}}-2\times \frac{{c}^{2}}{{a}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{{\left({b}^{2}-2ac\right)}^{2}-2{c}^{2}{a}^{2}}{{a}^{4}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{the}\mathrm{value}\mathrm{of}{\alpha}^{4}+{\beta}^{4}\mathrm{is}\frac{{\left({b}^{2}-2ac\right)}^{2}-2{c}^{2}{a}^{2}}{{a}^{4}}.$

(*vi*) Since

*α*and are the zeros of the quadratic polynomial

=

We have,

By substituting and we get ,

Hence, the value of is .

(*vii*) Since *α* and are the zeros of the quadratic polynomial

=

We have,

By substituting and we get ,

Hence, the value of is .

(*viii*) Since *α* and are the zeros of the quadratic polynomial

=

We have,

By substituting and we get ,

Hence, the value of is .

#### Page No 2.43:

#### Question 1:

Verify that the numbers given along side of the cubic polynomials below are their zeros. Also, verify the relationship between the zeros and coefficients in each case:

(i) $f\left(x\right)=2{x}^{3}+{x}^{2}-5x+2;\frac{1}{2},1,-2$

(ii) *g*(*x*) = *x*^{3} − 4*x*^{2} + 5*x* − 2; 2, 1, 1

#### Answer:

We have,

So, and are the zeros of polynomial p(x)

Let and . Then

From

Taking least common factor we get,

From

From

Hence, it is verified that the numbers given along side of the cubic polynomials are their zeros and also verified the relationship between the zeros and coefficients

(ii) We have,

So 2,1and 1 are the zeros of the polynomial g(x)

Let and. Then,

From

From

From

Hence, it is verified that the numbers given along side of the cubic polynomials are their

zeros and also verified the relationship between the zeros and coefficients.

#### Page No 2.43:

#### Question 2:

Find the cubic polynomial with the sum, sum of the product of its zeros taken two at a time, and product of its zeros as 3, −1 and − 3 respectively.

#### Answer:

If and are the zeros of a cubic polynomial *f* (*x*), then

where *k* is any non-zero real number.

Here,

Therefore

Hence, cubic polynomial is, where *k* is any non-zero real number.

#### Page No 2.43:

#### Question 3:

Find all zeroes of the polynomial 3*x*^{3} + 10*x*^{2} – 9*x* – 4, if one of its zeroes is 1.

#### Answer:

Let *f*(*x*) = 3*x*^{3} + 10*x*^{2} – 9*x* – 4

It is given that one of its zeroes is 1

Therefore, one factor of *f*(*x*) is (*x* – 1).

We get another factor of *f*(*x*) by dividing it with (*x* – 1).

On division, we get the quotient 3*x*^{2 }+ 13*x* + 4.

$\Rightarrow f\left(x\right)=\left(x-1\right)\left(3{x}^{2}+13x+4\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(3{x}^{2}+12x+x+4\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(3x\left(x+4\right)+1\left(x+4\right)\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(3x+1\right)\left(x+4\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{To}\mathrm{find}\mathrm{the}\mathrm{zeroes},\mathrm{we}\mathrm{put}f\left(x\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-1\right)\left(3x+1\right)\left(x+4\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-1\right)=0\mathrm{or}\left(3x+1\right)=0\mathrm{or}\left(x+4\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=1,-\frac{1}{3},-4$

Hence, all the zeroes of the polynomial *f*(*x*) are $1,-\frac{1}{3}\mathrm{and}-4.$

#### Page No 2.43:

#### Question 4:

If 4 is a zero of the cubic polynomial *x*^{3} − 3*x*^{2} − 10*x* + 24, find its other two zeroes.

#### Answer:

Given 4 is a zero of a cubic polynomial ${x}^{3}-3{x}^{2}-10x+24$

$\Rightarrow \left(x-4\right)$is the factor of polynomial ${x}^{3}-3{x}^{2}-10x+24$

Therefore, we have

To find the other two zeroes of the given polynomial, we need to find the zeroes of the quotient ${x}^{2}+x-6$.

$\mathrm{i}.\mathrm{e}.{x}^{2}+x-6=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+3x-2x-6=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(x+3\right)-2\left(x+3\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x+3\right)\left(x-2\right)=0$

$\Rightarrow x+3=0\mathrm{or}x-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=-3\mathrm{or}x=2$

Hence, the other two zeroes of the given polynomial are 2 and $-3$.

#### Page No 2.43:

#### Question 5:

If the zeros of the polynomial *f*(*x*) = 2*x*^{3} − 15*x*^{2} + 37*x* − 30 are in A.P., find them.

#### Answer:

Let and be the zeros of the polynomial

Therefore

Sum of the zeros =

Product of the zeros =

Substituting we get

Therefore, substituting and in ,and

Hence, the zeros of the polynomial are .

#### Page No 2.43:

#### Question 6:

Find the condition that the zeros of the polynomial *f*(*x*) = *x*^{3} + 3*px*^{2} + 3*qx* + *r* may be in A.P.

#### Answer:

Let and be the zeros of the polynomials .Then,

Sum of the zeros =

Since is a zero of the polynomial .Therefore,

Substituting we get,

Hence, the condition for the given polynomial is .

#### Page No 2.43:

#### Question 7:

If the zeros of the polynomial *f*(*x*) = *ax*^{3} + 3*bx*^{2} + 3*cx* + *d* are in A.P., prove that 2*b*^{3} − 3*abc* + *a*^{2}*d* = 0.

#### Answer:

Let and be the zeros of the polynomial *f*(*x*). Then,

Sum of the zeros =

Since *a* is a zero of the polynomial *f*(*x*).

Therefore,

Hence, it is proved that _{.}

#### Page No 2.43:

#### Question 8:

If the zeros of the polynomial *f*(*x*) = *x*^{3} − 12*x*^{2} + 39*x* + *k* are in A.P., find the value of *k*.

#### Answer:

Let and be the zeros of the polynomial _{.}

Then,

Sum of the zeros =

Since is a zero of the polynomial

Hence, the value of *k* is.

#### Page No 2.57:

#### Question 1:

Apply division algorithm to find the quotient *q*(*x*) and remainder *r*(*x*) in dividing *f*(*x*) by *g*(*x*) in each of the following :

(i)* f*(*x*) = *x*^{3} − 6*x*^{2} + 11*x* − 6, *g*(*x*) = *x*^{2} + *x* + 1

(ii) *f*(*x*) = 10*x*^{4} + 17*x*^{3} − 62*x*^{2} + 30*x* − 3, *g*(*x*) = 2*x*^{2} + 7*x* + 1

(iii) *f*(*x*) = 4*x*^{3}^{ }+ 8*x* + 8*x*^{2} + 7, *g*(*x*) = 2*x*^{2} − *x* + 1

(iv) *f*(*x*) = 15*x*^{3} − 20*x*^{2} + 13*x* − 12, *g*(*x*) = 2 − 2*x* + *x*^{2}

#### Answer:

We have

Here, degree and

Degree

Therefore, quotient is of degree and the remainder is of degree less than 2

Let and

Using division algorithm, we have

Equating the co-efficients of various powers of on both sides, we get

On equating the co-efficient of

On equating the co-efficient of

Substituting

On equating the co-efficient of

Substituting and we get,

On equating the constant terms

Substituting we get,

Therefore,

Quotient

And remainder

Hence, the quotient and remainder is given by,

.

We have

Here, Degree and

Degree

Therefore, quotient is of degree and remainder is of degree less than 2

Let and

Using division algorithm, we have

Equating the co-efficients of various powers on both sides, we get

On equating the co-efficient of

On equating the co-efficient of

Substituting we get

On equating the co-efficient of

Substituting and, we get

On equating the co-efficient of

Substituting and,we get

On equating constant term, we get

Substituting c=-2, we get

Therefore, quotient

Remainder

Hence, the quotient and remainder are and .

we have

Here, Degree and

Degree

Therefore, quotient is of degree and

Remainder is of degree less than

Let and

Using division algorithm, we have

Equating the co-efficient of various Powers of on both sides, we get

On equating the co-efficient of

On equating the co-efficient of

Substituting we get

On equating the co-efficient of

Substituting and we get

On equating the constant term, we get

Substituting, we get

Therefore, quotient

Remainder

Hence, the quotient and remainder are and.

Given,

Here, Degree and

Degree

Therefore, quotient is of degree and

Remainder is of degree less than

Let and

Using division algorithm, we have

Equating the co-efficients of various powers of on both sides, we get

On equating the co-efficient of

On equating the co-efficient of

Substituting , we get

On equating the co-efficient of

Substituting and, we get

On equating constant term

Substituting , we get

Therefore, quotient

Remainder

Hence, the quotient and remainder are and.

#### Page No 2.57:

#### Question 2:

Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm :

(i) $g\left(t\right)={t}^{2}-3,f\left(t\right)=2{t}^{4}+3{t}^{3}-2{t}^{2}-9t-12$

(ii) $g\left(x\right)={x}^{3}-3x+1,f\left(x\right)={x}^{5}-4{x}^{3}+{x}^{2}+3x+1$

(iii) $g\left(x\right)=2{x}^{2}-x+3,f\left(x\right)=6{x}^{5}-{x}^{4}+4{x}^{3}-5{x}^{2}-x-15$

#### Answer:

. Given

Here, degree and

Degree

Therefore, quotient is of degree

Remainder is of degree or less

Let and

Using division algorithm, we have

Equating co-efficient of various powers of t, we get

On equating the co-efficient of

On equating the co-efficient of

On equating the co-efficient of

Substituting, we get

On equating the co-efficient of

Substituting, we get

On equating constant term

Substituting, we get

Quotient

=

Remainder

Clearly,

Hence, is a factor of

(ii) Given

Here, Degree and

Degree

Therefore, quotient is of degree

Remainder is of degree1

Let and

Using division algorithm, we have

Equating the co-efficient of various powers of on both sides, we get

On equating the co-efficient of

On equating the co-efficient of

On equating the co-efficient of

Substituting we get

On equating the co-efficient of

Substituting and, we get

On equating constant term, we get

Substituting, we get

Therefore, quotient

Remainder

Clearly,

Hence, is not a factor of

(iii) Given,

Here, Degree and

Degree

Therefore, quotient is of degree and

Remainder is of degree less than

Let and

Using division algorithm, we have

Equating the co-efficient of various powers of on both sides, we get

On equating the co-efficient of

On equating the co-efficient of

Substituting, we get

On equating the co-efficient of

Substituting and, we get

On equating the co-efficient of

Substituting, we get

On equating the co-efficient of

Substituting and, we get

On equating constant term

Substituting, we get

Therefore, Quotient

Remainder

Clearly,

Hence, is a factor of

#### Page No 2.57:

#### Question 3:

Obtain all zeros of the polynomial *f*(*x*) = 2*x*^{4} + *x*^{3} − 14*x*^{2} − 19*x* − 6, if two of its zeros are −2 and −1.

#### Answer:

We know that, if is a zero of a polynomial, and then is a factor of.

Since and are zeros of .

Therefore

is a factor of .Now, We divide by to find the other zeros of .

By using division algorithm we have,

Hence, the zeros of the given polynomials are.

#### Page No 2.57:

#### Question 4:

Obtain all zeros of *f*(*x*) = *x*^{3} + 13*x*^{2} + 32*x* + 20, if one of its zeros is −2.

#### Answer:

Since −2 is one zero of .

Therefore, we know that, if is a zero of a polynomial, then is a factor of is a factor of .

Now, we divide by to find the others zeros of .

By using that division algorithm we have,

Hence, the zeros of the given polynomials are .

#### Page No 2.57:

#### Question 5:

Obtain all zeros of the polynomial *f*(*x*) = *x*^{4} − 3*x*^{3} − *x*^{2} + 9*x* − 6, if two of its zeros are $-\sqrt{3}\mathrm{and}\sqrt{3}$

#### Answer:

We know that if is a zero of a polynomial, then is a factor of .

Since and are zeros of .

Therefore

is a factor of .Now, we divide by to find the other zeros of .

By using division algorithm we have, .

Hence, the zeros of the given polynomials are .

#### Page No 2.57:

#### Question 6:

Find all zeros of the polynomial *f*(*x*) = 2*x*^{4} − 2*x*^{3} − 7*x*^{2} + 3*x* + 6, if its two zeros are $-\sqrt{\frac{3}{2}}\mathrm{and}\sqrt{\frac{3}{2}}$.

#### Answer:

Since and are two zeros of .Therefore,

is a factor of .

Also is a factor of .

Let us now divide by. We have,

By using division algorithm we have,

Hence, The zeros of are .

#### Page No 2.57:

#### Question 7:

Find all the zeros of the polynomial *x*^{4} + *x*^{3} − 34*x*^{2} − 4*x* + 120, if two of its zeros are 2 and −2.

#### Answer:

We know that if is a zero of a polynomial, then is a factor of .

Since, and are zeros of .

Therefore

is a factor of .Now, we divide by to find the other zeros of .

By using division algorithm we have

Hence, the zeros of the given polynomial are .

#### Page No 2.57:

#### Question 8:

Find all zeros of the polynomial 2*x*^{4} + 7*x*^{3} − 19*x*^{2} − 14*x* + 30, if two of its zeros are $\sqrt{2}\mathrm{and}-\sqrt{2}$.

#### Answer:

We know that if is a zero of a polynomial, and then is a factor of .

Since and are zeros of .

Therefore

is a factor of . Now, we divide by to find the zero of .

By using division algorithm we have

Hence, the zeros of the given polynomial are .

#### Page No 2.58:

#### Question 9:

Find all the zeros of the polynomial 2*x*^{3} + *x*^{2} − 6*x* − 3, if two of its zeros are $-\sqrt{3}\mathrm{and}\sqrt{3}$.

#### Answer:

We know that if is a zero of a polynomial, and then is a factor of .

Since and are zeros of .

Therefore

is a factor of .Now, we divide by to find the other zeros of .

By using division algorithm we have

Hence, the zeros of the given polynomial are.

#### Page No 2.58:

#### Question 10:

Find all the zeros of the polynomial *x*^{3} + 3*x*^{2} − 2*x* − 6, if two of its zeros are $-\sqrt{2}\mathrm{and}\sqrt{2}$.

#### Answer:

We know that if is a zero of a polynomial, and then is a factor of .

Since and are zeros of .

Therefore

is a factor of .Now, we divide by to find the other zeros of .

By using division algorithm we have

Hence, the zeros of the given polynomials are .

#### Page No 2.58:

#### Question 11:

Find all zeros of the polynomial 2*x*^{4} – 9*x*^{3} + 5*x*^{2} +3*x* – 1, if two of its zeros are $2+\sqrt{3}\mathrm{and}2-\sqrt{3}$.

#### Answer:

It is given that $2+\sqrt{3}$ and $2-\sqrt{3}$ are two zeroes of the polynomial *f*(*x*) = 2*x*^{4} – 9*x*^{3} + 5*x*^{2} + 3*x* – 1.

$\begin{array}{rcl}\therefore \left\{x-\left(2+\sqrt{3}\right)\right\}\left\{x-\left(2-\sqrt{3}\right)\right\}& =& \left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)\\ & =& {\left(x-2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}\\ & =& {x}^{2}-4x+4-3\\ & =& {x}^{2}-4x+1\end{array}$

is a factor of *f*(*x*)

Now, divide *f*(*x*) by *x*^{2} – 4*x* + 1.

∴ *f*(*x*) = (*x*^{2} – 4*x* + 1) (2*x*^{2} – *x* – 1)

Hence, other two zeroes of *f*(*x*) are the zeroes of the polynomial 2*x*^{2} – *x* – 1.

2*x*^{2} – *x* – 1 = 2*x*^{2} – 2*x* + *x* – 1 = 2*x*(*x* – 1) + 1 (*x* – 1) = (2*x* + 1) (*x* – 1)

Hence, the other two zeroes are $-\frac{1}{2}$ and 1.

#### Page No 2.58:

#### Question 12:

For what value of *k*, is the polynomial *f*(*x*) = 3*x*^{4} – 9*x*^{3} + *x*^{2} + 15*x* +* k *completely divisible by 3*x*^{2} – 5?

#### Answer:

Let *f*(*x*) = 3*x*^{4} – 9*x*^{3} + *x*^{2} + 15*x* +* k*

It is given that *f*(*x*) is completely divisible by 3*x*^{2} – 5.

Therefore, one factor of *f*(*x*) is (3*x*^{2} – 5).

We get another factor of *f*(*x*) by dividing it with (3*x*^{2} – 5).

On division, we get the quotient *x*^{2 }– 3*x* + 2 and the remainder *k* + 10.

Since, *f*(*x*) = 3*x*^{4} – 9*x*^{3} + *x*^{2} + 15*x* +* k *completely divisible by 3*x*^{2} – 5

Therefore, remainder must be zero.

$\Rightarrow k+10=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=-10$

Hence, the value of *k* is –10.

#### Page No 2.58:

#### Question 13:

What must be added to the polynomial *f*(*x*) = *x*^{4} + 2*x*^{3} − 2*x*^{2} + *x* − 1 so that the resulting polynomial is exactly divisible by *x*^{2} + 2*x* − 3?

#### Answer:

We know that,

Clearly , Right hand side is divisible by .

Therefore, Left hand side is also divisible by .Thus, if we add to , then the resulting polynomial is divisible by.

Let us now find the remainder when is divided by.

Hence, we should add to so that the resulting polynomial is divisible by .

#### Page No 2.58:

#### Question 14:

What must be subtracted from the polynomial *f*(*x*) = *x*^{4} + 2*x*^{3} − 13*x*^{2} − 12*x* + 21 so that the resulting polynomial is exactly divisible by *x*^{2} − 4*x* + 3?

#### Answer:

We know that DividendQuotientDivisorRemainder.

DividendRemainderQuotientDivisor.

Clearly, Right hand side of the above result is divisible by the divisor.

Therefore, left hand side is also divisible by the divisor.

Thus, if we subtract remainder from the dividend, then it will be exactly divisible by the divisor.

Dividing by

Therefore, quotient and remainder.

Thus, if we subtract the remainder from , it will be divisible by .

#### Page No 2.58:

#### Question 15:

Given that $\sqrt{2}$ is a zero of the cubic polynomial $6{x}^{3}+\sqrt{2}{x}^{2}-10x-4\sqrt{2}$ , find its other two zeroes .

#### Answer:

We know that if is a zero of a polynomial, and then is a factor of .

It is given tha $\sqrt{2}$ is a zero of the cubic polynomial = $6{x}^{3}+\sqrt{2}{x}^{2}-10x-4\sqrt{2}$.

Therefore, $x-\sqrt{2}$ is a factor of. Now, we divide $6{x}^{3}+\sqrt{2}{x}^{2}-10x-4\sqrt{2}$ by $x-\sqrt{2}$ to find the other zeroes of.

∴ Quotient = $6{x}^{2}+7\sqrt{2}x+4$ and remainder = 0.

By using division algorithm, we have $f\left(x\right)=g\left(x\right)\times q\left(x\right)+r\left(x\right)$.

$f\left(x\right)=\left(x-\sqrt{2}\right)\left(6{x}^{2}+7\sqrt{2}x+4\right)+0\phantom{\rule{0ex}{0ex}}=\left(x-\sqrt{2}\right)\left(\sqrt{2}x+1\right)\left(3\sqrt{2}x+4\right)$

Hence, the other two zeroes of the given polynomial are $-\frac{1}{\sqrt{2}}\mathrm{and}-\frac{4}{3\sqrt{2}}$.

#### Page No 2.58:

#### Question 16:

Given that $x-\sqrt{5}$ is a factor of the cubic polynomial ${x}^{3}-3\sqrt{5}{x}^{2}+13x-3\sqrt{5}$ , find all the zeroes of the polynomial .

#### Answer:

We know that if is a zero of a polynomial, and then is a factor of .

It is given that $x-\sqrt{5}$ is a factor of = ${x}^{3}-3\sqrt{5}{x}^{2}+13x-3\sqrt{5}$.

Now, we divide ${x}^{3}-3\sqrt{5}{x}^{2}+13x-3\sqrt{5}$ by $x-\sqrt{5}$ to find the other zeroes of .

â€‹

∴ Quotient = ${x}^{2}-2\sqrt{5}x+3$ and remainder = 0.

By using division algorithm, we have $f\left(x\right)=g\left(x\right)\times q\left(x\right)+r\left(x\right)$.

$f\left(x\right)=\left(x-\sqrt{5}\right)\left({x}^{2}-2\sqrt{5}x+3\right)+0\phantom{\rule{0ex}{0ex}}=\left(x-\sqrt{5}\right)\left[x-\left(\sqrt{5}+\sqrt{2}\right)\right]\left[x-\left(\sqrt{5}-\sqrt{2}\right)\right]$

Hence, the zeroes of the given polynomial are $\sqrt{5},\sqrt{5}+\sqrt{2}\mathrm{and}\sqrt{5}-\sqrt{2}$.

#### Page No 2.58:

#### Question 1:

If α, β are the zeros of the polynomial *f*(*x*) = *x*^{2} + *x* + 1, then $\frac{1}{\mathrm{\alpha}}+\frac{1}{\mathrm{\beta}}=$

(a) 1

(b) −1

(c) 0

(d) None of these

#### Answer:

Since and are the zeros of the quadratic polynomial

We have

The value of is

Hence, the correct choice is _{.}

#### Page No 2.58:

#### Question 2:

If α, β are the zeros of the polynomial *p*(*x*) = 4*x*^{2} + 3*x* + 7, then $\frac{1}{\mathrm{\alpha}}+\frac{1}{\mathrm{\beta}}$ is equal to

(a) $\frac{7}{3}$

(b) $-\frac{7}{3}$

(c) $\frac{3}{7}$

(d) $-\frac{3}{7}$

#### Answer:

Since and are the zeros of the quadratic polynomial

We have

The value of is.

Hence, the correct choice is

#### Page No 2.58:

#### Question 3:

If one zero of the polynomial *f*(*x*) = (*k*^{2} + 4)*x*^{2} + 13*x* + 4*k* is reciprocal of the other, then *k* =

(a) 2

(b) −2

(c) 1

(d) −1

#### Answer:

We are given then

One root of the polynomial is reciprocal of the other. Then, we have

1

Hence the correct choice is

#### Page No 2.59:

#### Question 4:

If the sum of the zeros of the polynomial *f*(x) = 2*x*^{3} − 3*kx*^{2} + 4*x* − 5 is 6, then the value of* k* is

(a) 2

(b) 4

(c) −2

(d) −4

#### Answer:

Let, be the zeros of the polynomial and we are given that

$\alpha +\beta +\gamma $=.

Then,

$\alpha +\beta +\gamma =\frac{-\mathrm{Coefficient}\mathrm{of}x}{\mathrm{Coefficient}\mathrm{of}{x}^{2}}\phantom{\rule{0ex}{0ex}}=-\frac{(-3k)}{2}=\frac{3k}{2}$

It is given that

$\alpha +\beta +\gamma $=

Substituting $\alpha +\beta +\gamma =\frac{3k}{2}$, we get

The value of *k* is.

Hence, the correct alternative is

#### Page No 2.59:

#### Question 5:

If α and β are the zeros of the polynomial *f*(*x*) = *x*^{2} + *px* + *q*, then a polynomial having $\frac{1}{\mathrm{\alpha}}\mathrm{and}\frac{1}{\mathrm{\beta}}$ is its zero is

(a) *x*^{2} + *qx* +* p*

(b) *x*^{2} − *px* + *q*

(c) *qx*^{2} + *px* + 1

(d) *px*^{2} + *qx* + 1

#### Answer:

Let and be the zeros of the polynomial.Then,

And

Let S and R denote respectively the sum and product of the zeros of a polynomial

Whose zeros are and .then

Hence, the required polynomial whose sum and product of zeros are S and R is given by

So

Hence, the correct choice is

#### Page No 2.59:

#### Question 6:

If α, β are the zeros of polynomial *f*(*x*) = *x*^{2} − *p* (*x* + 1) − *c*, then (α + 1) (β + 1) =

(a) *c* − 1

(b) 1 − *c*

(c) *c*

(d) 1 + *c*

#### Answer:

Since and are the zeros of quadratic polynomial

We have

The value of is.

Hence, the correct choice is

#### Page No 2.59:

#### Question 7:

If α, β are the zeros of the polynomial *f*(*x*) = *x*^{2} − *p*(*x* + 1) − c such that (α +1) (β + 1) = 0, then *c* =

(a) 1

(b) 0

(c) −1

(d) 2

#### Answer:

Since and are the zeros of quadratic polynomial

We have

The value of *c* is.

Hence, the correct alternative is

#### Page No 2.59:

#### Question 8:

If *f*(*x*) = *ax*^{2} + *bx* + *c* has no real zeros and *a* + *b* + *c* < 0, then

(a) *c* = 0

(b) *c* > 0

(c) *c *< 0

(d) None of these

#### Answer:

If has no real zeros and then

Hence, the correct choice is

#### Page No 2.59:

#### Question 9:

If the diagram in Fig. 2.22 shows the graph of the polynomial *f*(*x*) = *ax*^{2} + *bx* + *c*, then

(a)

*a*> 0,

*b*< 0 and

*c*> 0

(b)

*a*< 0,

*b*< 0 and

*c*< 0

(c)

*a*< 0,

*b*> 0 and

*c*> 0

(d)

*a*< 0,

*b*> 0 and

*c*< 0

#### Answer:

Clearly, represent a parabola opening upwards.

Therefore, cuts Y axis at P which lies on . Putting *x* = 0 in , we get *y = c*. So the coordinates of P is. Clearly, P lies on. Therefore

Hence, the correct choice is

#### Page No 2.59:

#### Question 10:

Figure 2.23 show the graph of the polynomial *f*(*x*) = *ax*^{2} + *bx* + *c* for which

(a)

*a*< 0,

*b*> 0 and c > 0

(b)

*a*< 0,

*b*< 0 and c > 0

(c)

*a*< 0,

*b*< 0 and c < 0

(d)

*a*> 0,

*b*> 0 and c < 0

#### Answer:

Clearly, represent a parabola opening downwards. Therefore,

cuts *y*-axis at P which lies on. Putting *x* = 0 in, we get *y* = *c*. So the coordinates P are. Clearly, *P* lies on. Therefore

The vertex of the parabola is in the second quadrant. Therefore,

Therefore and

Hence, the correct choice is

#### Page No 2.59:

#### Question 11:

If the product of zeros of the polynomial *f*(*x*) *ax*^{3} − 6*x*^{2} + 11*x* − 6 is 4, then *a* =

(a) $\frac{3}{2}$

(b) $-\frac{3}{2}$

(c) $\frac{2}{3}$

(d) $-\frac{2}{3}$

#### Answer:

Since and are the zeros of quadratic polynomial

So we have

The value of *a* is

Hence, the correct alternative is.

#### Page No 2.59:

#### Question 12:

If zeros of the polynomial *f*(*x*) = *x*^{3} − 3*px*^{2} + *qx* − *r* are in A.P., then

(a) 2*p*^{3} = *pq* − *r*

(b) 2*p*^{3} = *pq* + *r*

(c) *p*^{3} = *pq* − *r*

(d) None of these

#### Answer:

Let be the zeros of the polynomial then

Since *a* is a zero of the polynomial

Therefore,

Substituting .we get

Hence, the correct choice is

#### Page No 2.60:

#### Question 13:

If the product of two zeros of the polynomial *f*(*x*) = 2*x*^{3} + 6*x*^{2} − 4*x* + 9 is 3, then its third zero is

(a) $\frac{3}{2}$

(b) $-\frac{3}{2}$

(c) $\frac{9}{2}$

(d) $-\frac{9}{2}$

#### Answer:

Let be the zeros of polynomial such that

We have,

Putting in, we get

Therefore, the value of third zero is.

Hence, the correct alternative is.

#### Page No 2.60:

#### Question 14:

If the polynomial *f*(*x*) = *ax*^{3} + *bx* − *c* is divisible by the polynomial *g*(*x*) = *x*^{2} + *bx* + *c*, then *ab* =

(a) 1

(b) $\frac{1}{c}$

(c) −1

(d) $-\frac{1}{c}$

#### Answer:

We have to find the value of

Given is divisible by the polynomial

We must have

, for all *x*

So put *x* = 0 in this equation

Since , so

Hence, the correct alternative is _{.}

#### Page No 2.60:

#### Question 15:

If Q.No. 14, *c* =

(a) *b*

(b) 2*b*

(c) 2*b*^{2}

(d) −2*b*

#### Answer:

We have to find the value of *c*

Given is divisible by the polynomial

We must have

for all

…… (1)

Since , so

Now in the equation (1) the condition is true for all *x. *So put *x *= 1 and also we have *ab* = 1

Therefore we have

Substituting and we get,

Hence, the correct alternative is

#### Page No 2.60:

#### Question 16:

If one root of the polynomial *f*(*x*) = 5*x*^{2} + 13*x* + *k* is reciprocal of the other, then the value of *k* is

(a) 0

(b) 5

(c) $\frac{1}{6}$

(d) 6

#### Answer:

If one zero of the polynomial is reciprocal of the other. So

Now we have

Since

Therefore we have

Hence, the correct choice is

#### Page No 2.60:

#### Question 17:

If α, β, γ are the zeros of the polynomial *f*(*x*) = *ax*^{3} + *bx*^{2} +* **cx* + *d*, then $\frac{1}{\mathrm{\alpha}}+\frac{1}{\mathrm{\beta}}+\frac{1}{\mathrm{\gamma}}=$

(a) $-\frac{b}{d}$

(b) $\frac{c}{d}$

(c) $-\frac{c}{d}$

(d) $-\frac{c}{a}$

#### Answer:

We have to find the value of

Given be the zeros of the polynomial

We know that

So

Hence, the correct choice is

#### Page No 2.60:

#### Question 18:

If α, β, γ are the zeros of the polynomial *f*(*x*) = *ax*^{3} + *bx*^{2}^{ }+ *cx* + *d*, then α^{2} + β^{2} + γ^{2} =

(a) $\frac{{b}^{2}-ac}{{a}^{2}}$

(b) $\frac{{b}^{2}-2ac}{a}$

(c) $\frac{{b}^{2}+2ac}{{b}^{2}}$

(d) $\frac{{b}^{2}-2ac}{{a}^{2}}$

#### Answer:

We have to find the value of

Given be the zeros of the polynomial

Now

The value of

Hence, the correct choice is

#### Page No 2.60:

#### Question 19:

If α, β, γ are are the zeros of the polynomial *f*(*x*) = *x*^{3} − *px*^{2} + *qx* − *r*, then $\frac{1}{\mathrm{\alpha \beta}}+\frac{1}{\mathrm{\beta \gamma}}+\frac{1}{\mathrm{\gamma \alpha}}=$

#### Answer:

We have to find the value of

Given be the zeros of the polynomial

Now we calculate the expression

Hence, the correct choice is

#### Page No 2.60:

#### Question 20:

If α, β are the zeros of the polynomial *f*(*x*) = *ax*^{2} + *bx* + *c*, then $\frac{1}{{\mathrm{\alpha}}^{2}}+\frac{1}{{\mathrm{\beta}}^{2}}=$

(a) $\frac{{b}^{2}-2ac}{{a}^{2}}$

(b) $\frac{{b}^{2}-2ac}{{c}^{2}}$

(c) $\frac{{b}^{2}+2ac}{{a}^{2}}$

(d) $\frac{{b}^{2}+2ac}{{c}^{2}}$

#### Answer:

We have to find the value of

Given and are the zeros of the quadratic polynomial f(x)

We have,

Hence, the correct choice is

#### Page No 2.60:

#### Question 21:

If two of the zeros of the cubic polynomial* **ax*^{3} + *bx*^{2} + *cx* + *d* are each equal to zero, then the third zero is

(a) $\frac{-d}{a}$

(b) $\frac{c}{a}$

(c) $\frac{-b}{a}$

(d) $\frac{b}{a}$

#### Answer:

Let and be the zeros of the polynomial

Therefore

The value of

Hence, the correct choice is

#### Page No 2.60:

#### Question 22:

If two zeros *x*^{3} +* **x*^{2} − 5*x* − 5 are $\sqrt{5}\mathrm{and}-\sqrt{5}$, then its third zero is

(a) 1

(b) −1

(c) 2

(d) −2

#### Answer:

Let and be the given zeros and be the third zero of *x*^{3} + *x*^{2} − 5*x* − 5 = 0 then

By substitutingand in

Hence, the correct choice is

#### Page No 2.60:

#### Question 23:

The product of the zeros of *x*^{3} + 4*x*^{2} + *x* − 6 is

(a) −4

(b) 4

(c) 6

(d) −6

#### Answer:

Given be the zeros of the polynomial

Product of the zeros =

The value of Product of the zeros is 6.

Hence, the correct choice is

#### Page No 2.61:

#### Question 24:

What should be added to the polynomial *x*^{2} − 5*x* + 4, so that 3 is the zero of the resulting polynomial?

(a) 1

(b) 2

(c) 4

(d) 5

#### Answer:

If, is a zero of a polynomial thenis a factor of

Since 3 is the zero of the polynomial,

Thereforeis a factor of

Now, we divideby we get

Therefore we should add 2 to the given polynomial

Hence, the correct choice is

#### Page No 2.61:

#### Question 25:

What should be subtracted to the polynomial *x*^{2} − 16*x* + 30, so that 15 is the zero of the resulting polynomial?

(a) 30

(b) 14

(c) 15

(d) 16

#### Answer:

We know that, if, is zero of a polynomial then is a factor of

Since 15 is zero of the polynomial *f *(*x*) = *x*^{2}^{ }− 16*x* + 30, therefore (*x* − 15) is a factor of *f* (*x*)

Now, we divide by we get

Thus we should subtract the remainder from,

Hence, the correct choice is.

#### Page No 2.61:

#### Question 26:

A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is

(a)* **x*^{2} − 9

(b) *x*^{2} + 9

(c) *x*^{2} + 3

(d) *x*^{2} − 3

#### Answer:

Since and are the zeros of the quadratic polynomials such that

If one of zero is 3 then

Substituting in we get

Let S and P denote the sum and product of the zeros of the polynomial respectively then

Hence, the required polynomials is

Hence, the correct choice is

#### Page No 2.61:

#### Question 27:

If two zeroes of the polynomial *x*^{3} + *x*^{2} − 9*x* − 9 are 3 and −3, then its third zero is

(a) −1

(b) 1

(c) −9

(d) 9

#### Answer:

Let and be the given zeros and be the third zero of the polynomial then

Substituting and in, we get

Hence, the correct choice is.

#### Page No 2.61:

#### Question 28:

If $\sqrt{5}\mathrm{and}-\sqrt{5}$ are two zeroes of the polynomial *x*^{3} + 3*x*^{2} − 5*x* − 15, then its third zero is

(a) 3

(b) −3

(c) 5

(d) −5

#### Answer:

Let andbe the given zeros and be the third zero of the polynomial . Then,

Substitutingandin

We get

Hence, the correct choice is

#### Page No 2.61:

#### Question 29:

If *x* + 2 is a factor of *x*^{2} + *ax* + 2*b* and *a* + *b* = 4, then

(a) *a*= 1, *b* = 3

(b) *a* = 3, *b* = 1

(c) *a* = −1, *b* = 5

(d) *a* = 5, *b* = −1

#### Answer:

Given that is a factor of and *a + b*=4

By solving and *a* + *b* = 4 by elimination method we get

Multiply by we get,

. So

By substituting *b* = 1 in *a* + *b* = 4 we get

Then *a* = 3, *b* = 1

Hence, the correct choice is

#### Page No 2.61:

#### Question 30:

The polynomial which when divided by −*x*^{2} + *x* − 1 gives a quotient *x* − 2 and remainder 3, is

(a) *x*^{3} − 3*x*^{2} + 3*x* − 5

(b) −*x*^{3} − 3*x*^{2} − 3*x* − 5

(c) −*x*^{3} + 3*x*^{2} − 3*x* + 5

(d) *x*^{3} − 3*x*^{2} − 3*x* + 5

#### Answer:

We know that

Therefore,

The polynomial which when divided by gives a quotient and remainder 3, is

Hence, the correct choice is.

#### Page No 2.61:

#### Question 31:

The number of polynomials having zeroes −2 and 5 is

(a) 1 (b) 2 (c) 3 (d) more than 3

#### Answer:

Polynomials having zeros −2 and 5 will be of the form

$p\left(x\right)=a{\left(x+2\right)}^{n}{\left(x-5\right)}^{m}$

Here, *n* and *m* can take any value from 1, 2, 3,...

Thus, the number of polynomials will be more than 3.

Hence, the correct answer is option D.

#### Page No 2.61:

#### Question 32:

If one of the zeroes of the quadratic polynomial (*k* – 1) *x*^{2} + *kx* + 1 is –3, then the value of *k* is

(a) $\frac{4}{3}$

(b) $-\frac{4}{3}$

(c) $\frac{2}{3}$

(d) $-\frac{2}{3}$

#### Answer:

The given polynomial is *f*(*x*) = $\left(k-1\right){x}^{2}+kx+1$.

Since $-3$ is one of the zeroes of the given polynomial, so $f\left(-3\right)=0$.

$\left(k-1\right){\left(-3\right)}^{2}+k\left(-3\right)+1=0\phantom{\rule{0ex}{0ex}}\Rightarrow 9\left(k-1\right)-3k+1=0\phantom{\rule{0ex}{0ex}}\Rightarrow 9k-9-3k+1=0\phantom{\rule{0ex}{0ex}}\Rightarrow 6k-8=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{4}{3}$

Hence, the correct answer is option A.

#### Page No 2.61:

#### Question 33:

The zeroes of the quadratic polynomial *x*^{2} + 99*x* + 127 are

(a) both positive (b) both negative

(c) both equal (d) one positive and one negative

#### Answer:

Let *f*(*x*) = *x*^{2} + 99*x* + 127.

Product of the zeroes of *f*(*x*) = $127\times 1=127$ [Product of zeroes = $\frac{c}{a}$ when *f*(*x*) = *ax*^{2} + *bx* + *c*]

Since the product of zeroes is positive, we can say that it is only possible when both zeroes are positive or both zeroes are negative.

Also, sum of the zeroes = –99 [Sum of zeroes = $-\frac{b}{a}$ when *f*(*x*) = *ax*^{2} + *bx* + *c*]

The sum being negative implies that both zeroes are positive is not correct.

So, we conclude that both zeroes are negative.

Hence, the correct answer is option B.

#### Page No 2.61:

#### Question 34:

If the zeroes of the quadratic polynomial ${x}^{2}+\left(a+1\right)x+b$. are 2 and $-3$, then

(a) $a=-7,b=-1\phantom{\rule{0ex}{0ex}}$ (b) $a=5,b=-1$ (c) $a=2,b=-6$ (d) $a=0,b=-6$

#### Answer:

The given quadratic equation is ${x}^{2}+\left(a+1\right)x+b=0$.

Since the zeroes of the given equation are 2 and –3.

So,

$\alpha =2$ and $\beta =-3$

Now,

$\mathrm{Sum}\mathrm{of}\mathrm{zeroes}=-\frac{\mathrm{Coefficient}\mathrm{of}x}{\mathrm{Coefficient}\mathrm{of}{x}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2+\left(-3\right)=-\frac{\left(a+1\right)}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow -1=-a-1\phantom{\rule{0ex}{0ex}}\Rightarrow a=0$

$\mathrm{Product}\mathrm{of}\mathrm{zeroes}=\frac{\mathrm{Constant}\mathrm{term}}{\mathrm{Coefficient}\mathrm{of}{x}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2\times \left(-3\right)=\frac{b}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow b=-6$

So, *a* = 0 and *b* = −6

Hence, the correct answer is option D.

#### Page No 2.61:

#### Question 35:

Given that one of the zeroes of the cubic polynomial $a{x}^{3}+b{x}^{2}+cx+d$ is zero , the product of the other two zeroes is

(a) $\frac{-c}{a}$ (b) $\frac{c}{a}$ (c) 0 (d) $\frac{-b}{a}$

#### Answer:

Let $p\left(x\right)=a{x}^{3}+b{x}^{2}+cx+d$.

Now 0 is the zero of the polynomial.

So, *p*(0) = 0.

$\Rightarrow a{\left(0\right)}^{3}+b{\left(0\right)}^{2}+c\left(0\right)+d=0\phantom{\rule{0ex}{0ex}}\Rightarrow d=0$

So,

$p\left(x\right)=a{x}^{3}+b{x}^{2}+cx=x\left(a{x}^{2}+bx+c\right)$

Putting *p*(*x*) = 0, we get

*x* = 0 or $a{x}^{2}+bx+c=0$ .....(1)

Let *α, β* be the other zeroes of $a{x}^{2}+bx+c=0$.

So, $\alpha \beta =\frac{c}{a}$

Hence, the correct answer is option B.

#### Page No 2.61:

#### Question 36:

The zeroes of the quadratic polynomial *x*^{2} + *ax* + *a, $a\ne 0,$*

*(a) *cannot be both positive (b) cannot both be negative

(c) are always unequal (d) are always equal

#### Answer:

Let *f*(x) = *x*^{2} + *ax* + *a*.

Product of the zeroes of *f*(*x*) = *a* [Product of zeroes = $\frac{c}{a}$ when *f*(x)* = ax*^{2} + *bx* + *c*]

Since the product of zeroes is positive, so the zeroes must be either both positive or both negative.

Also, sum of the zeroes = –*a* [Sum of zeroes = $-\frac{b}{a}$ when *f*(x)* = ax*^{2} + *bx* + *c*]

So, the sum of the zeroes is negative, so the zeroes cannot be both positive.

Hence, the correct answer is option B.

#### Page No 2.61:

#### Question 37:

If one of the zeroes of the cubic polynomial ${x}^{3}+a{x}^{2}+bx+c$is $-1$, then the product of other two zeroes is

(a) $b-a+1$ (b) $b-a-1$ (c) $a-b+1$ (d) $a-b-1$

#### Answer:

Let $p\left(x\right)={x}^{3}+a{x}^{2}+bx+c$.

Now, −1 is a zero of the polynomial.

So, *p*(−1) = 0.

$\Rightarrow {\left(-1\right)}^{3}+a{\left(-1\right)}^{2}+b\left(-1\right)+c=0\phantom{\rule{0ex}{0ex}}\Rightarrow -1+a-b+c=0\phantom{\rule{0ex}{0ex}}\Rightarrow a-b+c=1\phantom{\rule{0ex}{0ex}}\Rightarrow c=1-a+b$

Now, if $\alpha ,\beta ,\gamma $ are the zeroes of the cubic polynomial $a{x}^{3}+b{x}^{2}+cx+d$, then product of zeroes is given by

$\alpha \beta \gamma =-\frac{d}{a}$

So, for the given polynomial, $p\left(x\right)={x}^{3}+a{x}^{2}+bx+c$

$\alpha \beta \left(-1\right)=\frac{-c}{1}=\frac{-\left(1-a+b\right)}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha \beta =1-a+b$

Hence, the correct answer is option A.

#### Page No 2.62:

#### Question 38:

Given that two of the zeroes of the cubic polynomial $a{x}^{3}+b{x}^{2}+cx+d$are 0 , the third zero is

(a) $-\frac{b}{a}$ (b) $\frac{b}{a}$ (c) $\frac{c}{a}$ (d) $-\frac{d}{a}$

#### Answer:

Let the polynomial be $f\left(x\right)=a{x}^{3}+b{x}^{2}+cx+d$.

Suppose the two zeroes of *f* (*x*) are $\alpha =0\mathrm{and}\beta =0$.

We know that,

Sum of the zeros,

$\alpha +\beta +\gamma =\frac{-b}{a}\phantom{\rule{0ex}{0ex}}\Rightarrow 0+0+\gamma =\frac{-b}{a}\phantom{\rule{0ex}{0ex}}\Rightarrow \gamma =\frac{-b}{a}$

Hence, the correct answer is option A.

#### Page No 2.62:

#### Question 39:

If one zero of the quadratic polynomial *x*^{2} + 3*x* + *k * is 2 , then the value of *k* is

(a) 10 (b)$-$10 (c) 5 (d)$-$5

#### Answer:

Let the given polynomial be *f*(*x*) = *x*^{2} + 3*x* + *k*.

Since 2 is one of the zero of the given plynomial, so (*x* − 2) will be a factor of the given polynomial.

Now, *f*(2) = 0

$\Rightarrow {2}^{2}+3\times 2+k=0\phantom{\rule{0ex}{0ex}}\Rightarrow 4+6+k=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=-10$

Hence, the correct answer is option B.

#### Page No 2.62:

#### Question 40:

If the zeroes of the quadratic polynomial $a{x}^{2}+bx+c$, $c\ne 0$ are equal, then

(a) *c* and *a* have opposite signs (b) *c* and* b *have opposite signs

(c)* c * and *a* have the same sign (d) *c* and *b* have the same sign

#### Answer:

Let the given quadratic polynomial be *f*(*x*) = $a{x}^{2}+bx+c$.

Suppose $\alpha $ and $\beta $ be the zeroes of the given polynomial.

Since $\alpha $ and $\beta $ are equal so they will have the same sign i.e. either both are positive or both are negative.

So, $\alpha \beta >0$

But, $\alpha \beta =\frac{c}{a}$

∴ $\frac{c}{a}>0$, which is possible only when both have same sign

Hence, the correct answer is option C.

#### Page No 2.62:

#### Question 41:

If one of the zeroes of a quadratic polynomial of the form ${x}^{2}+ax+b$ is the negative of the other , then it

(a) has no linear term and constant term is negative.

(b) has no linear term and the constant term is positive.

(c) can have a linear term but the constant term is negative .

(d) can have a linear term but the constant term is positive .

#### Answer:

Let the quadratic polynomial be $f\left(x\right)={x}^{2}+ax+b$.

Now, the zeroes are $\alpha \mathrm{and}-\alpha $.

So, the sum of the zeroes is zero.

$\therefore \alpha +\left(-\alpha \right)=\frac{-a}{1}=-a\phantom{\rule{0ex}{0ex}}\Rightarrow a=0$

So, the polynomial becomes $f\left(x\right)={x}^{2}+b$, which is not linear

Also, the product of the zeros,

$\alpha \beta =\frac{b}{1}=b\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha \left(-\alpha \right)=b\phantom{\rule{0ex}{0ex}}\Rightarrow -{\alpha}^{2}=b$

Thus, the constant term is negative.

Hence, the correct answer is option A.

#### Page No 2.62:

#### Question 42:

Which of the following is not the graph of a quadratic polynomial ?

(a)

(b)

(c)

(d)

#### Answer:

For a quadratic polynomial, $a{x}^{2}+bx+c$, the zeros are precisely the *x*-coordinates of the points where the graph representing $y=a{x}^{2}+bx+c$ intersects the *x*-axis.

The graph has one of the two shapes either open upwards like ∪ (parabolic shape) or open downwards like ∩ (parabolic shape) depending on whether *a* > 0 or *a* < 0.

Three cases are thus possible:

a) graph cuts *x*-axis at two distinct points (two zeroes)

b) graph cuts the *x*-axis at exactly one point (one zero)

c) the graph is either completely above the *x*-axis or completely below the *x*-axis (no zeroes)

In option (d), the graph is cutting the *x*-axis at three distinct points and it is not a parabola opening either upwards or downwards.

So, option (d) does not represent the graph of a quadratic polynomial.

Hence, the correct answer is option D.

#### Page No 2.63:

#### Question 1:

Quadratic polynomials, whose zeroes are –4 and 3 are given by ________.

#### Answer:

Quadratic polynomial with given zeroes is

*x*^{2} – (sum of zeroes)*x* + (product of zeroes)

$={x}^{2}-\left(-4+3\right)x+\left(-4\times 3\right)\phantom{\rule{0ex}{0ex}}={x}^{2}-\left(-1\right)x+\left(-12\right)\phantom{\rule{0ex}{0ex}}={x}^{2}+x-12$

Hence, Quadratic polynomials, whose zeroes are –4 and 3 are given by __ x^{2} + x – 12__.

#### Page No 2.63:

#### Question 2:

Quadratic polynomials with rational coefficients having $\sqrt{3}$ as a zero are given by ________.

#### Answer:

Irrational zeroes of a quadratic polynomial always occurs in pairs.

If one zero is $\sqrt{3}$ then, other zero is –$\sqrt{3}$.

Now,

Quadratic polynomial with given zeroes is

*x*^{2} – (sum of zeroes)*x* + (product of zeroes)

$={x}^{2}-\left(\sqrt{3}+\left(-\sqrt{3}\right)\right)x+\left(\sqrt{3}\times -\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}={x}^{2}-\left(\sqrt{3}-\sqrt{3}\right)x+\left(-3\right)\phantom{\rule{0ex}{0ex}}={x}^{2}-3$

Hence, Quadratic polynomials with rational coefficients having $\sqrt{3}$ as a zero are given by __ x^{2} – 3__.

#### Page No 2.63:

#### Question 3:

The number of quadratic polynomials whose zeroes are 2 and – 3, is __________.

#### Answer:

Quadratic polynomial with given zeroes is

*x*^{2} – (sum of zeroes)*x* + (product of zeroes)

$={x}^{2}-\left(-3+2\right)x+\left(-3\times 2\right)\phantom{\rule{0ex}{0ex}}={x}^{2}-\left(-1\right)x+\left(-6\right)\phantom{\rule{0ex}{0ex}}={x}^{2}+x-6$

Hence, the number of quadratic polynomials whose zeroes are 2 and – 3, is __one__.

#### Page No 2.63:

#### Question 4:

The zeroes of the quadratic polynomial *x*^{2} + *x* – 6 are __________.

#### Answer:

Let *f*(*x*) = *x*^{2} + *x* – 6

$\Rightarrow f\left(x\right)={x}^{2}+x-6\phantom{\rule{0ex}{0ex}}={x}^{2}+3x-2x-6\phantom{\rule{0ex}{0ex}}=x\left(x+3\right)-2\left(x+3\right)\phantom{\rule{0ex}{0ex}}=\left(x-2\right)\left(x+3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{To}\mathrm{find}\mathrm{the}\mathrm{zeroes},\mathrm{we}\mathrm{put}f\left(x\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-2\right)\left(x+3\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-2\right)=0\mathrm{or}\left(x+3\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=2,-3$

Hence, the zeroes of the quadratic polynomial *x*^{2} + *x* – 6 are __2 and –3__.

#### Page No 2.63:

#### Question 5:

The quadratic polynomials, the sum and product of whose zeroes are 7 and 12 are given by _________.

#### Answer:

Quadratic polynomial with given zeroes is

*x*^{2} – (sum of zeroes)*x* + (product of zeroes)

$={x}^{2}-\left(7\right)x+\left(12\right)\phantom{\rule{0ex}{0ex}}={x}^{2}-7x+12$

Hence, quadratic polynomials, the sum and product of whose zeroes are 7 and 12 are given by __ x^{2} – 7x + 12__.

#### Page No 2.63:

#### Question 6:

If *x* + 1 is a factor of the polynomial 2*x*^{3} + *ax*^{2} + 4*x* + 1, then *a* = _________.

#### Answer:

Let *f*(*x*) = 2*x*^{3} + *ax*^{2} + 4*x* + 1

It is given that one factor of *f*(*x*) is (*x* + 1).

Therefore, $f\left(x\right)=0\mathrm{when}x=-1$.

On putting *x* = –1 in *f*(*x*) = 0, we get

$2{\left(-1\right)}^{3}+a{\left(-1\right)}^{2}+4\left(-1\right)+1=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2\left(-1\right)+a\left(+1\right)+4\left(-1\right)+1=0\phantom{\rule{0ex}{0ex}}\Rightarrow -2+a-4+1=0\phantom{\rule{0ex}{0ex}}\Rightarrow a-5=0\phantom{\rule{0ex}{0ex}}\Rightarrow a=5$

Hence, *a* = __5__.

#### Page No 2.63:

#### Question 7:

If *a + b + c* = 0, then a zero of the polynomial *ax*^{2} + *bx *+ *c*, is _________.

#### Answer:

Let *f*(*x*) = *ax*^{2} + *bx *+ *c*

It is given that *a + b + c* = 0

Therefore, when *x* = 1, *f*(*x*) = $a{\left(1\right)}^{2}+b\left(1\right)+c=a+b+c=0$

Thus, *x* = 1 is the zero of *f*(*x*).

Hence, if *a + b + c* = 0, then a zero of the polynomial *ax*^{2} + *bx *+ *c*, is __1__.

#### Page No 2.63:

#### Question 8:

If *a + c = b*, then a zero of the polynomial* ax*^{2} + *bx *+* c*, is _________.

#### Answer:

Let *f*(*x*) = *ax*^{2} + *bx *+ *c*

It is given that *a + c = b
⇒ a − b + c = *0

Therefore, when

*x*=

*−*1,

*f*(

*x*) = $a{\left(-1\right)}^{2}+b\left(-1\right)+c=a-b+c=0$

Thus,

*x*=

*−*1 is the zero of

*f*(

*x*).

Hence, if

*a + c = b*, then a zero of the polynomial

*ax*

^{2}+

*bx*+

*c*, is

__.__

*−*1#### Page No 2.63:

#### Question 9:

The graph of a quadratic polynomial intersects the *x-*axis at the most at ________ points.

#### Answer:

A quadratic polynomial has at most 2 zeroes.

Thus, it can intersects the *x-*axis at the most at 2 points.

Hence, the graph of a quadratic polynomial intersects the *x-*axis at the most at __two__ points.

#### Page No 2.63:

#### Question 10:

If two of the zeroes of a cubic polynomial are zero, then it does not have ______ and _______ terms.

#### Answer:

Let *f*(*x*) = *ax*^{3} + *bx*^{2} + *cx* + *d* be a cubic polynomial.

Since, two of the zeroes of a cubic polynomial are zero, then the equation will be *ax*^{3} + *bx*^{2} = 0

Therefore, it does not have the linear term and the constant term.

Hence, If two of the zeroes of a cubic polynomial are zero, then it does not have __linear__ and __constant__ terms.

#### Page No 2.63:

#### Question 11:

If all the zeroes of cubic polynomial *x*^{3} + *ax*^{2 }– *bx *+ *c* are negative then *a, b *and *c* all have _______ sign.

#### Answer:

Let *f*(*x*) = *x*^{3} + *ax*^{2 }– *bx *+ *c*

$\mathrm{Let}\mathrm{the}\mathrm{zeroes}\mathrm{of}f\left(x\right)\mathrm{be}\alpha ,\beta ,\gamma ,\mathrm{where}\mathrm{all}\mathrm{these}\mathrm{zeroes}\mathrm{are}\mathrm{negative}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Then},\phantom{\rule{0ex}{0ex}}\alpha +\beta +\gamma =-a\phantom{\rule{0ex}{0ex}}\mathrm{Thus},a\mathrm{is}\mathrm{positive}\left(\because \alpha ,\beta ,\gamma \mathrm{are}\mathrm{negative}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\alpha \beta +\beta \gamma +\gamma \alpha =-b\phantom{\rule{0ex}{0ex}}\mathrm{Thus},b\mathrm{is}\mathrm{positive}\left(\because \alpha ,\beta ,\gamma \mathrm{are}\mathrm{negative}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\alpha \beta \gamma =-c\phantom{\rule{0ex}{0ex}}\mathrm{Thus},c\mathrm{is}\mathrm{positive}\left(\because \alpha ,\beta ,\gamma \mathrm{are}\mathrm{negative}\right)$

Hence, if all the zeroes of cubic polynomial *x*^{3} + *ax*^{2 }– *bx *+ *c* are negative then *a, b *and *c* all have __positive__ sign.

#### Page No 2.63:

#### Question 12:

If the zeroes of the quadratic polynomial *ax*^{2} + *x* + *a* are equal, then *a* = ________.

#### Answer:

Let *f*(*x*) = *ax*^{2} + *x* + *a*

$\mathrm{Let}\mathrm{the}\mathrm{zeroes}\mathrm{of}f\left(x\right)\mathrm{be}\alpha \mathrm{and}\alpha .\left(\because \mathrm{the}\mathrm{zeroes}\mathrm{are}\mathrm{equal}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Then},\phantom{\rule{0ex}{0ex}}\alpha +\alpha =-\frac{1}{a}\mathrm{and}\alpha \times \alpha =\frac{a}{a}\phantom{\rule{0ex}{0ex}}\Rightarrow 2\alpha =-\frac{1}{a}\mathrm{and}{\alpha}^{2}=1\phantom{\rule{0ex}{0ex}}\Rightarrow a=-\frac{1}{2\alpha}\mathrm{and}\alpha =\pm \sqrt{1}\phantom{\rule{0ex}{0ex}}\Rightarrow a=-\frac{1}{2\alpha}\mathrm{and}\alpha =\pm 1\phantom{\rule{0ex}{0ex}}\mathrm{For}\alpha =1,a=-\frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{For}\alpha =-1,a=\frac{1}{2}$

Hence, if the zeroes of the quadratic polynomial *ax*^{2} + *x* + *a* are equal, then *a* = $\overline{)\pm \frac{1}{2}}.$

#### Page No 2.63:

#### Question 13:

If the zeroes of the quadratic polynomial *ax*^{2} + *bx + c *are both negative, then *a, b *and *c* all have the _______ sign.

#### Answer:

Let *f*(*x*) = *ax*^{2} + *bx + c*

$\mathrm{Let}\mathrm{the}\mathrm{zeroes}\mathrm{of}f\left(x\right)\mathrm{be}\alpha ,\beta ,\mathrm{where}\mathrm{all}\mathrm{these}\mathrm{zeroes}\mathrm{are}\mathrm{negative}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Then},\phantom{\rule{0ex}{0ex}}\alpha +\beta =-\frac{b}{a}\phantom{\rule{0ex}{0ex}}\because \alpha ,\beta \mathrm{are}\mathrm{negative}\phantom{\rule{0ex}{0ex}}\therefore \frac{b}{a}\mathrm{is}\mathrm{positive}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{either}\mathrm{both}a\mathrm{and}b\mathrm{are}\mathrm{positive}\mathrm{or}\mathrm{both}\mathrm{are}\mathrm{negative}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\alpha \beta =\frac{c}{a}\phantom{\rule{0ex}{0ex}}\because \alpha ,\beta \mathrm{are}\mathrm{negative}\phantom{\rule{0ex}{0ex}}\therefore \frac{c}{a}\mathrm{is}\mathrm{positive}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{either}\mathrm{both}a\mathrm{and}c\mathrm{are}\mathrm{positive}\mathrm{or}\mathrm{both}\mathrm{are}\mathrm{negative}.$

Hence, if the zeroes of the quadratic polynomial *ax*^{2} + *bx + c *are both negative, then *a, b *and *c* all have the __same__ sign.

#### Page No 2.63:

#### Question 14:

If *x* + α is a factor of the polynomial 2*x*^{2} + 2α*x *+ 4*x *+ 12, then α = _______.

#### Answer:

Let *f*(*x*) = 2*x*^{2} + 2α*x *+ 4*x *+ 12

It is given that one factor of *f*(*x*) is (*x* + α).

Therefore, $f\left(x\right)=0\mathrm{when}x=-\alpha $.

On putting *x* = –α in *f*(*x*) = 0, we get

$2{\left(-\alpha \right)}^{2}+2\alpha \left(-\alpha \right)+4\left(-\alpha \right)+12=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2{\left(\alpha \right)}^{2}-2{\alpha}^{2}-4\alpha +12=0\phantom{\rule{0ex}{0ex}}\Rightarrow -4\alpha +12=0\phantom{\rule{0ex}{0ex}}\Rightarrow -4\alpha =-12\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha =3$

Hence, if *x* + α is a factor of the polynomial 2*x*^{2} + 2α*x *+ 4*x *+ 12, then α = __3__.

#### Page No 2.63:

#### Question 15:

If –4 is a zero of the polynomial *x*^{2} – *x* – (2*k* + 2), then *k* = _________.

#### Answer:

Let *f*(*x*) = *x*^{2} – *x* – (2*k* + 2)

It is given that –4 is a zero of *f*(*x*).

Therefore, $f\left(x\right)=0\mathrm{when}x=-4$.

On putting *x* = –4 in *f*(*x*) = 0, we get

${\left(-4\right)}^{2}-\left(-4\right)-\left(2k+2\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 16+4-2k-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow 20-2k-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow 18-2k=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2k=18\phantom{\rule{0ex}{0ex}}\Rightarrow k=9$

Hence, if –4 is a zero of the polynomial *x*^{2} – *x* – (2*k* + 2), then *k* = __9__.

#### Page No 2.63:

#### Question 16:

If 4*x*^{2} – 6*x *– *m* is divisible by *x* – 3, then *m* = ________.

#### Answer:

Let *f*(*x*) = 4*x*^{2} – 6*x *– *m*

It is given that *f*(*x*) is divisible by *x* – 3.

Therefore, $f\left(x\right)=0\mathrm{when}x=3$.

On putting *x* = 3 in *f*(*x*) = 0, we get

$4{\left(3\right)}^{2}-6\left(3\right)-m=0\phantom{\rule{0ex}{0ex}}\Rightarrow 4\left(9\right)-6\left(3\right)-m=0\phantom{\rule{0ex}{0ex}}\Rightarrow 36-18-m=0\phantom{\rule{0ex}{0ex}}\Rightarrow 18-m=0\phantom{\rule{0ex}{0ex}}\Rightarrow m=18$

Hence, if 4*x*^{2} – 6*x *– *m* is divisible by *x* – 3, then *m* = __18__.

#### Page No 2.63:

#### Question 17:

If *a, a + b, a *+ 2*b* are zeroes of the cubic polynomial* x*^{3} – 6*x*^{2} + 3*x* + 10, then *a + b* = ________.

#### Answer:

Let *f*(*x*) =* x*^{3} – 6*x*^{2} + 3*x* + 10

$\mathrm{Let}\mathrm{the}\mathrm{zeroes}\mathrm{of}f\left(x\right)\mathrm{be}a,a+b,a+2b.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Then},\phantom{\rule{0ex}{0ex}}\mathrm{Sum}\mathrm{of}\mathrm{zeroes}=6\phantom{\rule{0ex}{0ex}}\Rightarrow a+\left(a+b\right)+\left(a+2b\right)=6\phantom{\rule{0ex}{0ex}}\Rightarrow 3a+3b=6\phantom{\rule{0ex}{0ex}}\Rightarrow 3\left(a+b\right)=6\phantom{\rule{0ex}{0ex}}\Rightarrow a+b=2$

Hence, if *a, a + b, a *+ 2*b* are zeroes of the cubic polynomial* x*^{3} – 6*x*^{2} + 3*x* + 10, then *a + b* = __2__.

#### Page No 2.63:

#### Question 18:

If one zero of the quadratic polynomial 2*x*^{2} – 6*kx* + 6*x* – 7 is negative of the other, then *k* = ________.

#### Answer:

Let *f*(*x*) = 2*x*^{2} – 6*kx* + 6*x* – 7 = 2*x*^{2} + (– 6*k* + 6)*x* – 7

$\mathrm{Let}\mathrm{the}\mathrm{zeroes}\mathrm{of}f\left(x\right)\mathrm{be}\alpha \mathrm{and}-\alpha .\left(\because \mathrm{one}\mathrm{zero}\mathrm{is}\mathrm{negative}\mathrm{of}\mathrm{the}\mathrm{other}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Then},\phantom{\rule{0ex}{0ex}}\left(\alpha \right)+\left(-\alpha \right)=-\frac{-6k+6}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha -\alpha =-(-3k+3)\phantom{\rule{0ex}{0ex}}\Rightarrow 0=3k-3\phantom{\rule{0ex}{0ex}}\Rightarrow 3k=3\phantom{\rule{0ex}{0ex}}\Rightarrow k=1$

Hence, if one zero of the quadratic polynomial 2*x*^{2} – 6*kx* + 6*x* – 7 is negative of the other, then *k* = __1__.

#### Page No 2.63:

#### Question 19:

If the product of the zeroes of the quadratic polynomial *x*^{2} – 3*ax* + 2*a*^{2} – 1 is 7, then *a* = _______.

#### Answer:

Let *f*(*x*) = *x*^{2} – 3*ax* + 2*a*^{2} – 1

$\mathrm{Let}\mathrm{the}\mathrm{zeroes}\mathrm{of}f\left(x\right)\mathrm{be}\alpha \mathrm{and}\beta .\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Then},\phantom{\rule{0ex}{0ex}}\alpha \beta =\frac{2{a}^{2}-1}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha \beta =2{a}^{2}-1\phantom{\rule{0ex}{0ex}}\Rightarrow 2{a}^{2}-1=7\left(\because \mathrm{product}\mathrm{of}\mathrm{zeroes}\mathrm{is}7\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 2{a}^{2}=8\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=4\phantom{\rule{0ex}{0ex}}\Rightarrow a=\pm 2$

Hence, if the product of the zeroes of the quadratic polynomial *x*^{2} – 3*ax* + 2*a*^{2} – 1 is 7, then *a* = __±2__.

#### Page No 2.63:

#### Question 20:

The sum of the zeros of the quadratic polynomial 2*x*^{2} – 3*k* is ________.

#### Answer:

Let *f*(*x*) = 2*x*^{2} – 3*k = *2*x*^{2} + 0*x* – 3*k*

$\mathrm{Let}\mathrm{the}\mathrm{zeroes}\mathrm{of}f\left(x\right)\mathrm{be}\alpha \mathrm{and}\beta .\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Then},\phantom{\rule{0ex}{0ex}}\alpha +\beta =-\frac{0}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha +\beta =0$

Hence, the sum of the zeroes of the quadratic polynomial 2*x*^{2} – 3*k* is __0__.

#### Page No 2.63:

#### Question 21:

The parabola representing a quadratic polynomial *f*(*x*) = *ax*^{2} + *bx *+* c* opens upward when _________.

#### Answer:

Let *f*(*x*) = *ax*^{2} + *bx *+* c*

When a > 0, the parabola representing *f*(*x*) opens upward.

When a < 0, the parabola representing *f*(*x*) opens downward.

Hence, the parabola representing a quadratic polynomial *f*(*x*) = *ax*^{2} + *bx *+* c* opens upward when __ a is positive__.

#### Page No 2.63:

#### Question 22:

The parabola representing a quadratic polynomial *f*(*x*) = *ax*^{2} + *bx* +* c* opens downward when __________.

#### Answer:

Let *f*(*x*) = *ax*^{2} + *bx *+* c*

When a > 0, the parabola representing *f*(*x*) opens upward.

â€‹When a < 0, the parabola representing *f*(*x*) opens downward.

Hence, the parabola representing a quadratic polynomial *f*(*x*) = *ax*^{2} + *bx *+* c* opens downward when __ a is negative__.

#### Page No 2.63:

#### Question 23:

If the parabola represented by *f*(*x*) = *ax*^{2} + *bx + c* cuts *x*-axis at two distinct points, then the polynomial *ax*^{2} + *bx *+ *c* has ________ real zeroes.

#### Answer:

Let *f*(*x*) = *ax*^{2} + *bx *+* c*

If *f*(*x*) has two real and distinct zeroes, the parabola represented by *f*(*x*) cuts *x*-axis at two distinct points.

If *f*(*x*) has two real and equal zeroes, the parabola represented by *f*(*x*) touches *x*-axis at only one distinct point.

Hence, if the parabola represented by *f*(*x*) = *ax*^{2} + *bx + c* cuts *x*-axis at two distinct points, then the polynomial *ax*^{2} + *bx *+ *c* has __2__ real zeroes.

#### Page No 2.64:

#### Question 24:

The maximum number of zeroes which a quadratic polynomial can have is _________.

#### Answer:

Let *f*(*x*) = *ax*^{2} + *bx *+* c*

Maximum number of zeroes of polynomial = Highest power of* x
= *2

Therefore, It has at most 2 zeroes.

Hence, the maximum number of zeroes which a quadratic polynomial can have is

__2__.

#### Page No 2.64:

#### Question 25:

A quadratic polynomial, the sum of whose zeroes is –5 and their product is 6, is

(a) *x*^{2} + 5*x* + 6

(b) *x*^{2} – 5*x* + 6

(c) *x*^{2} – 5*x* – 6

(d) –*x*^{2} + 5*x* + 6

#### Answer:

Let the zeroes be $\alpha $ and $\beta $ respectively.

Therefore, $\alpha +\beta =-5$ and $\alpha \beta =6$.

Hence, the required polynomial is

${x}^{2}-\left(\alpha +\beta \right)x+\alpha \beta \phantom{\rule{0ex}{0ex}}={x}^{2}-\left(-5\right)x+6\phantom{\rule{0ex}{0ex}}={x}^{2}+5x+6$

Hence, the correct answer is option A.

#### Page No 2.64:

#### Question 1:

Define a polynomial with real coefficients.

#### Answer:

In the polynomial,

, and are known as the terms of the polynomial and and are their real coefficients.

For example, is a polynomial and 3 is a real coefficient

#### Page No 2.64:

#### Question 2:

Define degree of a polynomial.

#### Answer:

The exponent of the highest degree term in a polynomial is known as its degree.

In other words, the highest power of *x* in a polynomial is called the degree of the polynomial.

For Example: is a polynomial in the variable *x* of degree 2.

#### Page No 2.64:

#### Question 3:

Write the standard form of a linear polynomial with real coefficients.

#### Answer:

Any linear polynomial in variable with real coefficients is of the form, where are real numbers and

#### Page No 2.64:

#### Question 4:

Write the standard form of a quadratic polynomial with real coefficients.

#### Answer:

Any quadratic polynomial in variable with real coefficients is of the form, where are real numbers and

#### Page No 2.64:

#### Question 5:

Write the standard form of a cubic polynomial with real coefficients.

#### Answer:

The most general form of a cubic polynomial with coefficients as real numbers is of the form, where are real number and

#### Page No 2.64:

#### Question 6:

Define value of polynomial at a point.

#### Answer:

If is a polynomial and is any real number, then the real number obtained by replacing by in, is called the value of at and is denoted by

#### Page No 2.64:

#### Question 7:

Define the zero of a polynomial.

#### Answer:

The zero of a polynomial is defined as any real number such that

#### Page No 2.64:

#### Question 8:

The sum and product of the zeros of a quadratic polynomial are $-\frac{1}{2}$ and −3 respectively. What is the quadratic polynomial.

#### Answer:

Let sum of quadratic polynomial is

Product of the quadratic polynomial is

Let S and P denote the sum and product of the zeros of a polynomial asand.

Then

The required polynomial is given by

Hence, the quadratic polynomial is, where *k* is any non-zero real number

#### Page No 2.64:

#### Question 9:

Write the family of quadratic polynomials having $-\frac{1}{4}$ and 1 as its zeros.

#### Answer:

We know that, if is a zero of a polynomial then is a factor of quadratic polynomials.

Sinceand are zeros of polynomial.

Therefore

Hence, the family of quadratic polynomials is, where *k* is any non-zero real number

#### Page No 2.64:

#### Question 10:

If the product of zeros of the quadratic polynomial *f*(*x*) = *x*^{2} − 4*x* + *k* is 3, find the value of *k*.

#### Answer:

We have to find the value of *k*.

Given,

The product of the zeros of the quadratic polynomial .is

Product of the polynomial

Hence, the value of *k* is.

#### Page No 2.64:

#### Question 11:

If the sum of the zeros of the quadratic polynomial *f*(*x*) = *kx*^{2} − 3*x* + 5 is 1, write the value of *k*.

#### Answer:

We have to find the value of *k*, if the sum of the zeros of the quadratic polynomial is

Given

Sum of the polynomial

Hence, the value of *k* is

#### Page No 2.64:

#### Question 12:

In Fig. 2.17, the graph of a polynomial *p*(*x*) is given. Find the zeros of the polynomial.

#### Answer:

Just see the point of intersection of the curve and *x*-axis and find out the *x*-coordinate of these points. These *x*-coordinates will be the zeros of the polynomial

Since the intersection points are

Hence, the zeros of the polynomial is

#### Page No 2.64:

#### Question 13:

The graph of a polynomial *y* = *f*(*x*), shown in Fig. 2.18. Find the number of real zeros of *f*(*x*).

#### Answer:

A real number is a zero of polynomial, if

In the above figure the curve intersects *x*-axis at one point and touches at one point

When a curve touches *x*-axis at one point, it means it has two common zeros at that point

Hence the number of real zeroes is

#### Page No 2.65:

#### Question 14:

The graph of the polynomial *f*(*x*) = *ax*^{2} + *bx* + *c* is as shown below (Fig. 2.19). Write the signs of '*a*' and *b*^{2} − 4*ac*.

#### Answer:

Clearly, represent a parabola opening upwards. Therefore,

Since the parabola cuts *x*-axis at two points, this means that the polynomial will have two real solutions

Hence

Hence and

#### Page No 2.65:

#### Question 15:

The graph of the polynomial *f*(*x*) = *ax*^{2} + *bx* + *c* is as shown in Fig. 2.20. Write the value of *b*^{2} − 4*ac* and the number of real zeros of *f*(*x*).

#### Answer:

The graph of the polynomial or the curve touches *x*−axis at point. The *x*-coordinate of this point gives two equal zeros of the polynomial and.

Hence the number of real zeros of is 2 and

#### Page No 2.65:

#### Question 16:

In Q. No. 14, write the sign of *c*.

#### Answer:

The parabola cuts *y*-axis at point P which lies on y-axis. Putting in , we get *y = c*. So the coordinates of P are. Clearly, P lies on OY. Therefore

Hence, the sign of *c* is

#### Page No 2.65:

#### Question 17:

In Q. No. 15, write the sign of *c*.

#### Answer:

The parabola cuts *y*-axis at P which lies on OY.

Putting in, we get y=c. So the coordinates of P are. Clearly, P lies on. Therefore

#### Page No 2.65:

#### Question 18:

The graph of a polynomial *f*(*x*) is as shown in Fig. 2.21. Write the number of real zeros of *f*(*x*).

#### Answer:

The graph of a polynomial touches *x*−axis at two points

We know that if a curve touches the *x*-axis at two points then it has two common zeros of .

Hence the number of zeros of, in this case is 2.

#### Page No 2.65:

#### Question 19:

If *x* = 1 is a zero of the polynomial *f*(*x*) = *x*^{3} − 2*x*^{2} + 4*x* + *k*, write the value of *k*.

#### Answer:

We have to find the value of *K* if is a zero of the polynomial *f*(*x*) = *x*^{3} − 2*x*^{2} + 4*x* + *k*.

Hence, the value of *k* is

#### Page No 2.65:

#### Question 20:

State division algorithm for polynomials.

#### Answer:

If and are any two polynomials with then we can always find polynomials and such that, where or degree degree

#### Page No 2.65:

#### Question 21:

Give an example of polynomials *f*(*x*), *g*(*x*), *q*(*x*) and *r*(*x*) satisfying *f*(*x*) = *g*(*x*), *q*(*x*) + *r*(*x*), where degree *r*(*x*) = 0.

#### Answer:

Using division algorithm, we have

Hence an example for polynomial,, and satisfying are

#### Page No 2.65:

#### Question 22:

Write a quadratic polynomial, sum of whose zeros is $2\sqrt{3}$ and their product is 2.

#### Answer:

Let S and P denotes respectively the sum and product of the zeros of a polynomial are and.

The required polynomial g(x) is given by

Hence, the quadratic polynomial is where *k* is any non-zeros real number.

#### Page No 2.65:

#### Question 23:

If fourth degree polynomial is divided by a quadratic polynomial, write the degree of the remainder.

#### Answer:

Here represent dividend and represent divisor.

=quadratic polynomial

Therefore degree of

Degree of

The quotient q(x) is of degree

The remainder is of degree or less.

Hence, the degree of the remainder is equal to or less than

#### Page No 2.65:

#### Question 24:

If *f*(*x*) = *x*^{3} + *x*^{2} − *ax* + *b* is divisible by *x*^{2} −* x* write the value of *a* and *b*.

#### Answer:

We are given is exactly divisible by then the remainder should be zero

Therefore Quotient and

Remainder

Now, Remainder

Equating coefficient of x, we get

.

Equating constant term

Hence, the value of *a* and *b* are

#### Page No 2.65:

#### Question 25:

If* a* − *b*, *a* and *b* are zeros of the polynomial *f*(*x*) = 2*x*^{3}^{ }− 6*x*^{2} + 5*x* − 7, write the value of *a*.

#### Answer:

Let *a − b*, *a* and *a + b* be the zeros of the polynomial then

Sum of the zeros =

Hence, the value of *a* is.

#### Page No 2.66:

#### Question 26:

Write the coefficient of the polynomial *p*(*z*) = *z*^{5} − 2*z*^{2} + 4.

#### Answer:

We have to find the co-efficient of the polynomial

Co-efficient of

Co-efficient of

Co-efficient of

Co-efficient of

Co-efficient of

Constant term

Hence, the co-efficient of and constant term is

#### Page No 2.66:

#### Question 27:

Write the zeros of the polynomial *x*^{2} − *x* − 6.

#### Answer:

We have to find the zeros of the polynomial

We know that if is a factor of then is a zero of polynomial

Therefore we have

Also

Hence, the zeros of polynomial is

#### Page No 2.66:

#### Question 28:

If (*x* + *a*) is a factor of 2*x*^{2} + 2*ax* + 5*x* + 10, find *a*.

#### Answer:

Given is a factor of.

Let us now divide by.

We have,

Now, remainder

Hence, the value of *a* is

#### Page No 2.66:

#### Question 29:

For what value of *k*, −4 is a zero of the polynomial *x*^{2} − *x* − (2*k* + 2)?

#### Answer:

We know that if is zero polynomial then is a factor of

Since is zero of

Therefore is a factor of

Now, we divide by to find the value of *k*

Now, Remainder

Hence, the value of *k* is

#### Page No 2.66:

#### Question 30:

If 1 is a zero of the polynomial *p*(*x*) = *ax*^{2} − 3(*a* − 1) *x* − 1, then find the value of *a*.

#### Answer:

We know that if is a zero of polynomial then is a factor of

Since is zero of

Therefore, is a factor of

Now, we divide by *x* − 1.

Now, Remainder

Hence, the value of a is

#### Page No 2.66:

#### Question 31:

If α, β are the zeros of a polynomial such that α + β = −6 and αβ = −4, then write the polynomial.

#### Answer:

Let S and P denotes respectively the sum and product of the zeros of a polynomial

We are given S = and P =. Then

The required polynomial is given by

Hence, the polynomial is

#### Page No 2.66:

#### Question 32:

If α, β are the zeros of the polynomial 2*y*^{2} + 7*y* + 5, write the value of α + β + αβ.

#### Answer:

Let and are the zeros of the polynomial .Then

The sum of the zeros The product of the zeros =

Then the value of is

Hence, the value of is

#### Page No 2.66:

#### Question 33:

For what value of *k*, is 3 a zero of the polynomial 2*x*^{2} + *x* + *k*?

#### Answer:

We know that if is zero polynomial, and then is a factor of

Since is zero of

Therefore is a factor of

Now, we divide by to find the value of *k*

Now, remainder

Hence, the value of *k* is

#### Page No 2.66:

#### Question 34:

For what value of *k*, is −3 a zero of the polynomial *x*^{2} + 11*x* + *k*?

#### Answer:

We know that if is zeros polynomial, then is a factor of

Since is zero of . Therefore is a factor of

Now, we divide by to find the value of *k.*

Now, Remainder

Hence, the value of *k* is.

#### Page No 2.66:

#### Question 35:

For what value of *k*, is −2 a zero of the polynomial 3*x*^{2} + 4*x* + 2*k*?

#### Answer:

We know that if is zero polynomial then is a factor of

Since is a factor of .Therefore is a factor of

Now, we divide by to find the value of *k*

Now, Remainder

Hence, the value of *k* is

#### Page No 2.66:

#### Question 36:

If a quadratic polynomial* f*(*x*) is factorizable into linear distinct factors, then what is the total number of real and distinct zeros of *f*(*x*)?

#### Answer:

If a quadratic polynomial is factorized into linear polynomials then the total number of real and distinct zeros of will be.

#### Page No 2.66:

#### Question 37:

If a quadratic polynomial *f*(*x*) is a square of a linear polynomial, then its two zeros are coincident. (True/False).

#### Answer:

The polynomial has two identical factors. The curve cuts *X* axis at two coincident points that is exactly at one point.

Hence, quadratic polynomial is a square of linear polynomial then its two zeros are coincident.

#### Page No 2.66:

#### Question 38:

If a quadratic polynomial *f*(*x*) is not factorizable into linear factors, then it has no real zero. (True/False)

#### Answer:

When polynomial is not factorizable then the curve does not touch *x*-axis. Parabola open upwards above the *x*-axis or open downwards below *x*-axis where or

Hence, if quadratic polynomial is not factorizable into linear factors then it has no real zeros. .

#### Page No 2.66:

#### Question 39:

If *f*(*x*) is a polynomial such that *f*(*a*) *f*(*b*) < 0, then what is the number of zeros lying between *a* and *b*?

#### Answer:

If is a polynomial such that then this means the value of the polynomial are of different sign for *a* to *b*

Hence, at least one zero will be lying between *a* and *b*

#### Page No 2.66:

#### Question 40:

If graph of quadratic polynomial *ax*^{2} + *bx* + *c* cuts positive direction of *y*-axis, then what is the sign of *c*?

#### Answer:

If graph of quadratic polynomial cuts positive direction of *y*−axis, then

Put *x* = 0 for the point of intersection of the polynomial and *y*−axis

We have

Since the point is above the *x*-axis

Hence, the sign of *c* is positive, that is

#### Page No 2.66:

#### Question 41:

If the graph of quadratic polynomial *ax*^{2} + *bx* + *c* cuts negative direction of *y*-axis, then what is the sign of *c*?

#### Answer:

Since graph of quadratic polynomial cuts negative direction of *y*−axis

So put *x*=0 to find the intersection point on y-axis

So the point is

Now it is given that the quadratic polynomial cuts negative direction of *y*

So

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