RD Sharma 2022 Solutions for Class 10 Maths Chapter 2 - Polynomials

Answer:

(i) We have,

f(x) = x² − 2x − 8

f(x) = x² + 2x − 4x − 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

x² − 2x − 8 = 0

(x + 2) (x − 4) = 0

x + 2 = 0

x = −2

Or

x − 4 = 0

x = 4

Thus, the zeros of f(x) = x² − 2x − 8 are α = −2 and β = 4.

Now,

and

Therefore, the sum of the zeros =

Product of the zeros

= − 2 × 4

= −8

and

Therefore,

Product of the zeros =

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

(ii) Given

When have,

g(s) = 4s² − 4s + 1

g(s) = 4s² − 2s − 2s + 1

g(s) = 2s (2s − 1) − 1(2s − 1)

g(s) = (2s − 1) (2s − 1)

The zeros of g(s) are given by 

Or 

Thus, the zeros ofare and .

Now, sum of the zeros

and 

Therefore, the sum of the zeros =

Product of the zeros

and =

Therefore, the product of the zeros =

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

(iii) Given

We have,

$$\begin{gathered} h\left(t\right) = t^2 - 15 \\ h\left(t\right) = {\left(t\right)}^2 - {\left(\sqrt{15}\right)}^2 \\ h\left(t\right) = \left(t + \sqrt{15}\right) \left(t - \sqrt{15}\right) \end{gathered}$$

The zeros of are given by

$$\begin{gathered} h\left(t\right) = 0 \\ \left(t - \sqrt{15}\right) \left(t + \sqrt{15}\right) = 0 \\ \left(t - \sqrt{15}\right) = 0 \\ t = \sqrt{15} \\ \mathrm{or} \\ \left(t + \sqrt{15}\right) = 0 \\ t = -\sqrt{15} \end{gathered}$$

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

Now, 

Sum of the zeros

and =

Therefore, the sum of the zeros =

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, the sum of the zeros = 

Product of the zeros = α × β

and, = 

Product of zeros = 

Hence, the relation between the zeros and their coefficient is verified.

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

$$\begin{gathered} q\left(y\right)=\frac{1}{3}\left(21y^2-11y-2\right) \\ =\frac{1}{3}\left(21y^2-14y+3y-2\right) \\ =\frac{1}{3}\left[7y\left(3y-2\right)+1\left(3y-2\right)\right] \\ =\frac{1}{3}\left[\left(7y+1\right)\left(3y-2\right)\right] \end{gathered}$$

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 = $\alpha +\beta$
$$\begin{gathered} =\frac{-1}{7}+\frac{2}{3} \\ =\frac{11}{21} \end{gathered}$$

and

$$\begin{gathered} \frac{-\mathrm{Coefficient} \mathrm{of} y}{\mathrm{Coefficient} \mathrm{of} y^2} \\ =\frac{-\left(-{\displaystyle \frac{11}{3}}\right)}{7}=\frac{11}{21} \end{gathered}$$

Therefore, the sum of the zeros = 

Product of the zeros 
$$\begin{gathered} =\frac{-1}{7}\times \frac{2}{3} \\ =\frac{-2}{21} \end{gathered}$$

and

$$\begin{gathered} \frac{\mathrm{Constant} \mathrm{term}}{\mathrm{Coefficient} \mathrm{of} y^2} \\ =\frac{{\displaystyle \frac{-2}{3}}}{7} \\ =\frac{-2}{21} \end{gathered}$$

Therefore,

Product of the zeros = 
Hence, the relationship between the zeros and coefficient is verified.

(vi) We have, $\mathrm{\varphi }\left(x\right)=2x^2+\frac{7}{2}x+\frac{3}{4}$
$\begin{array}{rcl}\mathrm{\varphi }\left(x\right)& =& \frac{1}{4}\left(8x^2+14x+3\right)\\ & =& \frac{1}{4}\left(8x^2+12x+2x+3\right)\\ & =& \frac{1}{4}\left(4x\left(2x+3\right)+1\left(2x+3\right)\right)\\ & =& \frac{1}{4}\left(4x+1\right)\left(2x+3\right)\end{array}$
 

The zeros are given by $\mathrm{\varphi }$(x) = 0.

Thus, the zeros of $\mathrm{\varphi }\left(x\right)=\frac{1}{4}\left(4x+1\right)\left(2x+3\right) \mathrm{are} \alpha =\frac{-1}{4}\mathrm{and} \beta =\frac{-3}{2}$.

Now,

Sum of the zeros = $\alpha +\beta$
                           $$\begin{gathered} =\frac{-1}{4}+\frac{-3}{2} \\ =\frac{-1+\left(-6\right)}{4} \\ =\frac{-7}{4} \end{gathered}$$

and

$\begin{array}{rcl}\frac{-\mathrm{Coefficient} \mathrm{of} x}{\mathrm{Coefficient} \mathrm{of} x^2}& =& \frac{-\left({\displaystyle \frac{7}{2}}\right)}{2}\\ & =& -\frac{7}{4}\end{array}$

Therefore, the sum of the zeros = $\begin{array}{rcl}& & \frac{-\mathrm{Coefficient} \mathrm{of} x}{\mathrm{Coefficient} \mathrm{of} x^2}\end{array}$ .

Product of the zeros 
                                 $$\begin{gathered} =\frac{-1}{4}\times \left(\frac{-3}{2}\right) \\ =\frac{3}{8} \end{gathered}$$

and

$\begin{array}{rcl}\frac{\mathrm{Constant} \mathrm{term}}{\mathrm{Coefficient} \mathrm{of} x^2}& =& \frac{{\displaystyle \frac{3}{4}}}{2}\\ & =& \frac{3}{8}\end{array}$

Therefore, the product of the zeros = $\begin{array}{rcl}& & \frac{\mathrm{Constant} \mathrm{term}}{\mathrm{Coefficient} \mathrm{of} x^2}\end{array}$.

Hence, the relationship between the zeros and coefficient is verified.

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 $f\left(x\right)=k\left(x^2+\frac{8}{3}x+\frac{4}{3}\right)$
$$\begin{gathered} f\left(x\right)=k\left(x^2+\frac{8}{3}x+\frac{4}{3}\right) \\ =\frac{k}{3}\left(3x^2+8x+4\right) \\ =\frac{k}{3}\left(3x^2+6x+2x+4\right) \\ =\frac{k}{3}\left(3x\left(x+2\right)+2\left(x+2\right)\right) \\ =\frac{k}{3}\left(3x+2\right)\left(x+2\right) \end{gathered}$$
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 $f\left(x\right)=k\left(x^2-\frac{21}{8}x+\frac{5}{16}\right)$
$$\begin{gathered} f\left(x\right)=k\left(x^2-\frac{21}{8}x+\frac{5}{16}\right) \\ =\frac{k}{16}\left(16x^2-42x+5\right) \\ =\frac{k}{16}\left(16x^2-40x-2x+5\right) \\ =\frac{k}{16}\left(16x^2-2x-40x+5\right) \\ =\frac{k}{3}\left(2x\left(8x-1\right)-5\left(8x-1\right)\right) \\ =\frac{k}{3}\left(8x-1\right)\left(2x-5\right) \end{gathered}$$
Now, the zeroes are given by f(x) = 0.
Thus, $x=\frac{1}{8} \mathrm{and} x=\frac{5}{2}$

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.

Answer:

Since α and β are the zeros of the quadratic polynomial

We have,

By substituting and we get ,

Hence, the value of is.

Answer:

Since and are the zeros of the quadratic polynomial

= 0

Hence, the Value of is .

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

Answer:

(i) 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 relationship between the zeros and coefficient are verified.

(ii) Given

We have,

The zeros of g(x) are given by 

Or

$$\begin{gathered} \sqrt{3}x+7=0 \\ \Rightarrow x=-\frac{\sqrt{7}}{3} \end{gathered}$$

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 relationship between the zeros and coefficient are verified.

(iii) 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 relationship between the zeros and coefficient are verified.

(iv) 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 relationship between the zeros and coefficient are verified.

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

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

The zeros of h(s) are given by 

h(s) = 0

$$\begin{gathered} 2s^2-\left(1+2\sqrt{2}\right)s+\sqrt{2}=0 \\ \left(2s-1\right)\left(s-\sqrt{2}\right)=0 \\ \left(2s-1\right)=0 \mathrm{or} \left(s-\sqrt{2}\right)=0 \\ s=\frac{1}{2} \mathrm{or} s=\sqrt{2} \end{gathered}$$

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

$$\begin{gathered} \frac{-\mathrm{Coefficient} \mathrm{of} s}{\mathrm{Coefficient} \mathrm{of} s^2} \\ =-\left(\frac{-\left(1+2\sqrt{2}\right)}{2}\right) \\ =\frac{1+2\sqrt{2}}{2} \\ =\frac{1}{2}+\sqrt{2} \end{gathered}$$

Therefore, sum of the zeros = 

Product of the zeros 
$=\frac{1}{2}\times \sqrt{2}=\frac{1}{\sqrt{2}}$

and

$$\begin{gathered} \frac{\mathrm{Constant} \mathrm{term}}{\mathrm{Coefficient} \mathrm{of} s^2} \\ =\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \end{gathered}$$

Therefore,

Product of the zeros = 

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

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

$$\begin{gathered} f\left(v\right)=v^2+5\sqrt{3}v-\sqrt{3}v-15 \\ =v^2-\sqrt{3}v+5\sqrt{3}v-15 \\ =v\left(v-\sqrt{3}\right)+5\sqrt{3}\left(v-\sqrt{3}\right) \\ =\left(v-\sqrt{3}\right)\left(v+5\sqrt{3}\right) \end{gathered}$$

The zeros of f(v) are given by 

f(v) = 0

$$\begin{gathered} v^2+4\sqrt{3}v-15=0 \\ \left(v+5\sqrt{3}\right)\left(v-\sqrt{3}\right)=0 \\ \left(v-\sqrt{3}\right)=0 \mathrm{or} \left(v+5\sqrt{3}\right)=0 \\ v=\sqrt{3} \mathrm{or} v=-5\sqrt{3} \end{gathered}$$

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

$$\begin{gathered} \frac{-\mathrm{Coefficient} \mathrm{of} v}{\mathrm{Coefficient} \mathrm{of} v^2} \\ =\frac{-4\sqrt{3}}{1}=-4\sqrt{3} \end{gathered}$$

Therefore, sum of the zeros = 

Product of the zeros 
$=\sqrt{3}\times \left(-5\sqrt{3}\right)=-15$

and

$$\begin{gathered} \frac{\mathrm{Constant} \mathrm{term}}{\mathrm{Coefficient} \mathrm{of} v^2} \\ =\frac{-15}{1}=-15 \end{gathered}$$

Therefore,

Product of the zeros = 

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

(vii) $p\left(y\right)=y^2+\frac{3\sqrt{5}}{2}y-5$
$$\begin{gathered} p\left(y\right)=\frac{1}{2}\left(2y^2+4\sqrt{5}y-\sqrt{5}y-10\right) \\ =\frac{1}{2}\left[2y\left(y+2\sqrt{5}\right)-\sqrt{5}\left(y+2\sqrt{5}\right)\right] \\ =\frac{1}{2}\left[\left(2y-\sqrt{5}\right)\left(y+2\sqrt{5}\right)\right] \end{gathered}$$

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

$$\begin{gathered} \frac{-\mathrm{Coefficient} \mathrm{of} y}{\mathrm{Coefficient} \mathrm{of} y^2} \\ =\frac{{\displaystyle \frac{-3\sqrt{5}}{2}}}{1}=\frac{-3\sqrt{5}}{2} \end{gathered}$$

Therefore, sum of the zeros = 

Product of the zeros 
$$\begin{gathered} =\frac{\sqrt{5}}{2}\times -2\sqrt{5} \\ =-5 \end{gathered}$$

and

$$\begin{gathered} \frac{\mathrm{Constant} \mathrm{term}}{\mathrm{Coefficient} \mathrm{of} y^2} \\ =\frac{-5}{1}=-5 \end{gathered}$$

Therefore,

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

(viii) $\mathrm{\varphi }\left(x\right)=4x^2+5\sqrt{2}x-3$
$\begin{array}{rcl}\mathrm{\varphi }\left(x\right)& =& 4x^2+6\sqrt{2}x-\sqrt{2}x-3\\ & =& 2\sqrt{2}x\left(\sqrt{2}x+3\right)-1\left(\sqrt{2}x+3\right)\\ & =& \left(2\sqrt{2}x-1\right)\left(\sqrt{2}x+3\right)\end{array}$
The zeros are given by $\mathrm{\varphi }$(x) = 0.
Thus, the zeros of $\mathrm{\varphi }\left(x\right)=\left(2\sqrt{2}x-1\right)\left(\sqrt{2}x+3\right) \mathrm{are} \alpha =\frac{1}{2\sqrt{2}}\mathrm{and} \beta =-\frac{3}{\sqrt{2}}$.

Now,

Sum of the zeros = $\alpha +\beta$
                           $$\begin{gathered} =\frac{1}{2\sqrt{2}}-\frac{3}{\sqrt{2}} \\ =\frac{1-6}{2\sqrt{2}} \\ =\frac{-5}{2\sqrt{2}} \end{gathered}$$

and

$\begin{array}{rcl}\frac{-\mathrm{Coefficient} \mathrm{of} x}{\mathrm{Coefficient} \mathrm{of} x^2}& =& -\frac{5\sqrt{2}}{4}\\ & =& -\frac{5}{2\sqrt{2}}\end{array}$

Therefore, sum of the zeros = $\begin{array}{rcl}& & \frac{-\mathrm{Coefficient} \mathrm{of} x}{\mathrm{Coefficient} \mathrm{of} x^2}\end{array}$.

Product of the zeros 
                                $$\begin{gathered} =\frac{1}{2\sqrt{2}}\times \frac{\left(-3\right)}{\sqrt{2}} \\ =\frac{-3}{4} \end{gathered}$$

and

$\frac{\mathrm{Constant} \mathrm{term}}{\mathrm{Coefficient} \mathrm{of} x^2}=\frac{-3}{4}$

Therefore, the product of the zeros = $\frac{\mathrm{Constant} \mathrm{term}}{\mathrm{Coefficient} \mathrm{of} x^2}$.

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

Answer:

(i) We have, sum of zeroes = $-2\sqrt{3}$ and product of zeroes = −9.
So, the required quadratic polynomial will be $f\left(x\right)=k\left(x^2+2\sqrt{3}x-9\right)$.
$$\begin{gathered} f\left(x\right)=k\left(x^2+2\sqrt{3}x-9\right) \\ =k\left(x^2+3\sqrt{3}x-\sqrt{3}x-9\right) \\ =k\left(x+3\sqrt{3}\right)\left(x-\sqrt{3}\right) \end{gathered}$$
Now, the zeroes are given by f(x) = 0.
Thus, $x=-3\sqrt{3} \mathrm{and} x=\sqrt{3}$.

(ii) We have, sum of zeroes = $\frac{-3}{2\sqrt{5}}$ and product of zeroes = $\frac{-1}{2}$
So, the required quadratic polynomial will be $f\left(x\right)=k\left(x^2+\frac{3}{2\sqrt{5}}x-\frac{1}{2}\right)$.
$$\begin{gathered} f\left(x\right)=\frac{k}{2\sqrt{5}}\left(2\sqrt{5}{x}^2+3x-\sqrt{5}\right) \\ =\frac{k}{2\sqrt{5}}\left(2\sqrt{5}{x}^2+5x-2x-\sqrt{5}\right) \\ =\frac{k}{2\sqrt{5}}\left[\sqrt{5}x\left(2x+\sqrt{5}\right)-1\left(2x+\sqrt{5}x\right)\right] \\ =\frac{k}{2\sqrt{5}}\left(2x+\sqrt{5}\right)\left(\sqrt{5}x-1\right) \end{gathered}$$

Now, the zeroes are given by f(x) = 0.
Thus, $x=\frac{-\sqrt{5}}{2} \mathrm{and} x=\frac{1}{\sqrt{5}}$.

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.

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 .

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.

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.

Answer:

Let $\alpha \mathrm{and} \beta$ be the roots of the polynomial px² + qx + r.
$$\begin{gathered} \therefore \alpha +\beta =\frac{-\mathrm{Coefficient} \mathrm{of} x}{\mathrm{Coefficient} \mathrm{of} x^2}=\frac{-q}{p} .....\left(1\right) \\ \alpha \beta =\frac{\mathrm{Constant} \mathrm{term}}{\mathrm{Coefficient} \mathrm{of} x^2}=\frac{r}{p} .....\left(2\right) \end{gathered}$$

Since roots of the required polynomial are negative of the given polynomials, the roots of the required polynomial will be $-\alpha \mathrm{and} -\mathrm{\beta }$.

$\begin{array}{rcl}\mathrm{Required} \mathrm{polynomial}& =& x^2-\left(\mathrm{Sum} \mathrm{of} \mathrm{zeros}\right)x+\left(\mathrm{Product} \mathrm{of} \mathrm{zeros}\right)\\ & =& x^2-\left(-\alpha -\beta \right)x+\alpha \beta \\ & =& x^2+\left(\alpha +\beta \right)x+\alpha \beta \\ & =& x^2+\left(\frac{-q}{p}\right)x+\frac{r}{p}\\ & =& \frac{p{x}^2-qx+r}{p}\end{array}$

Thus, the required polynomial is $\frac{p{x}^2-qx+r}{p}$.
 

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.

Answer:

Given and are the zeros of the quadratic polynomial

We have, 

Substituting and then we get,

Hence, the value of is .

Answer:

Since and are the zeros of the quadratic polynomial

= p

We have,

Hence, it is proved that is equal to .

Answer:

(i)  Given α and are the zeros of the quadratic polynomial .

=

We have,

By substituting and we get ,

Hence, the value of is $\frac{b}{ac}$.

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

=

We have,

By substituting and we get ,

Hence, the value of is $\frac{-2}{a}$.

(iii) Since α and are the zeros of the quadratic polynomial .

=

We have,

By substituting and we get ,

Hence, the value of is b.

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.

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.

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.

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 .

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 .

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 _.

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.

Answer:

Given 4 is a zero of a cubic polynomial $x^3-3x^2-10x+24$
$\Rightarrow \left(x-4\right)$is the factor of polynomial $x^3-3x^2-10x+24$

Therefore, we have


To find the other two zeroes of the given polynomial, we need to find the zeroes of the quotient $x^2+x-6$.
$$\begin{gathered} \mathrm{i}.\mathrm{e}. x^2+x-6=0 \\ \Rightarrow x^2+3x-2x-6=0 \\ \Rightarrow x\left(x+3\right)-2\left(x+3\right)=0 \\ \Rightarrow \left(x+3\right)\left(x-2\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x+3=0 \mathrm{or} x-2=0 \\ \Rightarrow x=-3 \mathrm{or} x=2 \end{gathered}$$
Hence, the other two zeroes of the given polynomial are 2 and $-3$.

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

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 .

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.

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 .

Answer:

Let f(x) = 3x⁴ – 9x³ + x² + 15x + k

It is given that f(x) is completely divisible by 3x² – 5.

Therefore, one factor of f(x) is (3x² – 5).

We get another factor of f(x) by dividing it with (3x² – 5).

On division, we get the quotient x² – 3x + 2 and the remainder k + 10.

Since, f(x) = 3x⁴ – 9x³ + x² + 15x + k completely divisible by 3x² – 5

Therefore, remainder must be zero.

$$\begin{gathered} \Rightarrow k+10=0 \\ \Rightarrow k=-10 \end{gathered}$$

Hence, the value of k is –10.

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.

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 .

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 .

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 .

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 .

​

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

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

$$\begin{gathered} f\left(x\right)=\left(x-\sqrt{5}\right)\left(x^2-2\sqrt{5}x+3\right)+0 \\ =\left(x-\sqrt{5}\right)\left[x-\left(\sqrt{5}+\sqrt{2}\right)\right]\left[x-\left(\sqrt{5}-\sqrt{2}\right)\right] \end{gathered}$$

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

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 .

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.

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

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.

Answer:

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

Answer:

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

Answer:

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

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

Answer:

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

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

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

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.

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

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 (−3.5, 0) and (−1, 0).

Hence, the zeros of the polynomial is −3.5 and −1.

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

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.

Answer:

We have to find the value of K if is a zero of the polynomial f(x) = x³ − 2x² + 4x + k.

Hence, the value of k is

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.

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

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

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

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

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.

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

Answer:

Given is a factor of.

Let us now divide by.

We have, 

Now, remainder

Hence, the value of a is

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

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

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

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.

Answer:

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

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.

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. .

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

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

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

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

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

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

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

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

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

Answer:

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

Answer:

Using division algorithm, we have

Hence an example for polynomial,, and satisfying are

Answer:

Quadratic polynomial with given zeroes is 
x² – (sum of zeroes)x + (product of zeroes)

$$\begin{gathered} =x^2-\left(-4+3\right)x+\left(-4\times 3\right) \\ =x^2-\left(-1\right)x+\left(-12\right) \\ =x^2+x-12 \end{gathered}$$

Hence, Quadratic polynomials, whose zeroes are –4 and 3 are given by x² + x – 12.

Answer:

Quadratic polynomial with given zeroes is 
x² – (sum of zeroes)x + (product of zeroes)

$$\begin{gathered} =x^2-\left(-3+2\right)x+\left(-3\times 2\right) \\ =x^2-\left(-1\right)x+\left(-6\right) \\ =x^2+x-6 \end{gathered}$$

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

Answer:

Let f(x) =  x² + x – 6

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

Hence, the zeroes of the quadratic polynomial x² + x – 6 are 2 and –3.