#### Page No 2.33:

#### Answer:1

(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 2.33:

#### Answer:2

(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 2.33:

#### Answer:3

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 2.33:

#### Answer:4

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 2.33:

#### Answer:5

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:

#### Answer:6

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:

#### Answer:7

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:

#### Answer:8

Since and are the zeros of the quadratic polynomial

We have

Hence, the value of is .

#### Page No 2.34:

#### Answer:9

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:

#### Answer:10

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:

#### Answer:11

Since and are the zeros of the quadratic polynomial

= *p*

We have,

Hence, it is proved that is equal to .

#### Page No 2.34:

#### Answer:12

Given and are the zeros of the quadratic polynomial

We have,

Substituting and then we get,

Hence, the value of is .

#### Page No 2.34:

#### Answer:13

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:

#### Answer:14

Since and are the zeros of the quadratic polynomial

= 0

Hence, the Value of is .

#### Page No 2.34:

#### Answer:15

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 2.34:

#### Answer:16

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 2.34:

#### Answer:17

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 2.34:

#### Answer:18

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 2.34:

#### Answer:19

(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 2.34:

#### Answer:20

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 2.40:

#### Answer:1

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 2.40:

#### Answer:2

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 2.40:

#### Answer:3

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 2.40:

#### Answer:4

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 2.41:

#### Answer:5

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 2.41:

#### Answer:6

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 2.52:

#### Answer:1

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 2.52:

#### Answer:2

. 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 2.52:

#### Answer:3

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 2.52:

#### Answer:4

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 2.52:

#### Answer:5

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 2.52:

#### Answer:6

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 2.52:

#### Answer:7

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 2.52:

#### Answer:8

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 2.52:

#### Answer:9

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 2.52:

#### Answer:10

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 2.52:

#### Answer:11

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 2.52:

#### Answer:12

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 2.53:

#### Answer:1

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

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#### Answer:2

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.

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#### Answer:3

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

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#### Answer:4

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

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#### Answer:5

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

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#### Answer:6

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

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#### Answer:7

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

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#### Answer:8

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

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#### Answer:9

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

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#### Answer:10

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.

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#### Answer:11

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

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#### Answer:12

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

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#### Answer:13

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

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#### Answer:14

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

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#### Answer:15

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

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#### Answer:16

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

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#### Answer:17

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

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#### Answer:18

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.

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#### Answer:19

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

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#### Answer:20

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

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#### Answer:21

Using division algorithm, we have

Hence an example for polynomial,, and satisfying are

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#### Answer:22

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.

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#### Answer:23

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

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#### Answer:24

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

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#### Answer:25

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

Sum of the zeros =

Hence, the value of *a* is.

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#### Answer:26

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

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#### Answer:27

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

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#### Answer:28

Given is a factor of.

Let us now divide by.

We have,

Now, remainder

Hence, the value of *a* is

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#### Answer:29

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

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#### Answer:30

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

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#### Answer:31

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

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#### Answer:32

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

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#### Answer:33

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

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#### Answer:34

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.

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#### Answer:35

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

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#### Answer:36

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

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#### Answer:37

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.

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#### Answer:38

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. .

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#### Answer:39

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*

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#### Answer:40

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

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#### Answer:41

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

#### Page No 2.56:

#### Answer:1

Since and are the zeros of the quadratic polynomial

We have

The value of is

Hence, the correct choice is _{.}

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#### Answer:2

Since and are the zeros of the quadratic polynomial

We have

The value of is.

Hence, the correct choice is

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#### Answer:3

We are given then

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

1

Hence the correct choice is

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#### Answer:4

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

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#### Answer:5

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

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#### Answer:6

Since and are the zeros of quadratic polynomial

We have

The value of is.

Hence, the correct choice is

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#### Answer:7

Since and are the zeros of quadratic polynomial

We have

The value of *c* is.

Hence, the correct alternative is

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#### Answer:8

If has no real zeros and then

Hence, the correct choice is

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#### Answer:9

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

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#### Answer:10

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

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#### Answer:11

Since and are the zeros of quadratic polynomial

So we have

The value of *a* is

Hence, the correct alternative is.

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#### Answer:12

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

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#### Answer:13

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.

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#### Answer:14

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 _{.}

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#### Answer:15

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

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#### Answer:16

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

Now we have

Since

Therefore we have

Hence, the correct choice is

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#### Answer:17

We have to find the value of

Given be the zeros of the polynomial

We know that

So

Hence, the correct choice is

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#### Answer:18

We have to find the value of

Given be the zeros of the polynomial

Now

The value of

Hence, the correct choice is

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#### Answer:19

We have to find the value of

Given be the zeros of the polynomial

Now we calculate the expression

Hence, the correct choice is

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#### Answer:20

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

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#### Answer:21

Let and be the zeros of the polynomial

Therefore

The value of

Hence, the correct choice is

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#### Answer:22

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

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#### Answer:23

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

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#### Answer:24

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

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#### Answer:25

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.

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#### Answer:26

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

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#### Answer:27

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.

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#### Answer:28

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

Substitutingandin

We get

Hence, the correct choice is

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#### Answer:29

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

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#### Answer:30

We know that

Therefore,

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

Hence, the correct choice is.

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