# Polynomials : Exercise - 2.3 (Mathematics NCERT Class 10th) Like the video?  Subscribe Now  and get such videos daily!

Q.1       Divide the polynomial p(x) by the polynomial g (x) and find the quotient and remainder in each of the following :
(i) $p\left( x \right) = {x^3} - 3{x^2} + 5x - 3,\,g\left( x \right) = {x^2} - 2$
(ii) $p\left( x \right) = {x^4} - 3{x^2} + 4x + 5,\,g\left( x \right) = {x^2} + 1 - x$
(iii) $p\left( x \right) = {x^4} - 5x + 6,\,g\left( x \right) = 2 - {x^2}$
Sol.      (i) We have, Therefore, the quotient is x – 3 and the remainder is 7 x – 9

(ii) Here, the dividend is already in the standard form and the divisor is also in the standard form.
We have, Therefore, the quotient is ${x^2} + x - 3$ and the remainder is 8.

(iii) To carry out the division, we first write divisor in the standard form.
So, divisor = $- {x^2} + 2$
We have, Therefore, the quotient is $- {x^2} + 2$
and the remainder is – 5x + 10.

Q.2         Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial :
(i) ${t^2} - 3,\,2{t^4} + 3{t^3} - 2{t^2} - 9t - 12$
(ii) ${x^2} + 3x + 1,\,3{x^4} + 5{x^3} - 7{x^2} + 2x + 2$
(iii) ${x^3} - 3x + 1,\,{x^5} - 4{x^3} + {x^2} + 3x + 1$
Sol.      (i) Let us divide $2{t^4} + 3{t^3} - 2{t^2} - 9t - 12\,\,by\,{t^2} - 3$.
We have, Since the remainder is zero, therefore, ${t^2} - 3$ is a factor of $2{t^4} + 3{t^3} - 2{t^2} - 9t - 12$.

(ii) Let us divide $3{x^4} + 5{x^3} - 7{x^2} + 2x + 2\,by\,{x^2} + 3x + 1$.
We get, Since the remainder is zero, therefore ${x^2} + 3x + 1$ is a factor of
$3{x^4} + 5{x^3} - 7{x^2} + 2x + 2$

(iii) Let us divide ${x^5} - 4{x^3} + {x^2} + 3x + 1$ by ${x^3} - 3x + 1$
We get, Here remainder is 2($\ne$ 0). Therefore,  ${x^3} - 3x + 1$ is not a factor of
${x^5} - 4{x^3} + {x^2} + 3x + 1$.

Q.3      Obtain all the zeroes of $3{x^4} + 6{x^3} - 2{x^2} - 10x - 5$ if two of its zeroes are $\sqrt {{5 \over 3}} \,and\, - \sqrt {{5 \over 3}}$.
Sol.      Since two zeroes are $\sqrt {{5 \over 3}} \,and\, - \sqrt {{5 \over 3}}$, so $\left( {x - \sqrt {{5 \over 3}} \,} \right)and\,\left( {x + \sqrt {{5 \over 3}} } \right)$ are the factors of the given polynomial.
Now, $\left( {x - \sqrt {{5 \over 3}} \,} \right)\,\left( {x + \sqrt {{5 \over 3}} } \right) = {x^2} - {5 \over 3}$
$\Rightarrow$ $\left( {3{x^2} - 5} \right)$ is a factor of the given polynomial.
Applying the division algorithm to the given polynomial and ${3{x^2} - 5}$, we have Therefore, $3{x^4} + 6{x^3} - 2{x^2} - 10x - 5$$\left( {3{x^2} - 5} \right)\left( {{x^2} + 2x + 1} \right)$
Now, ${x^2} + 2x + 1$ $= {x^2} + x + x + 1$
$= x\left( {x + 1} \right) + 1\left( {x + 1} \right)$
$= \left( {x + 1} \right)\left( {x + 1} \right)$
So, its other zeroes are – 1 and – 1.
Thus,  all the zeroes of the given fourth degree polynomial are
$\sqrt {{5 \over 3}} , - \sqrt {{5 \over 3}} , - 1\,and\, - 1$.

Q.4       On dividing ${x^3} - 3{x^2} + x + 2$ by a polynomial g(x) , the quotient and remainder were x – 2 and – 2x + 4 respectively. Find g(x).
Sol.         Since on dividing ${x^3} - 3{x^2} + x + 2$ by a polynomial (x – 2) × g(x), the quotient and remainder were (x – 2) and (– 2x + 4) respectively, therefore,
Therefore, Quotient × Divisor + Remainder = Dividend
$\Rightarrow$ $\left( {x - 2} \right) \times g\left( x \right) + \left( { - 2x + 4} \right)$  $= {x^3} - 3{x^2} + x - 2$
$\Rightarrow$ (x – 2) × g(x) $= {x^3} - 3{x^2} + x - 2 + 2x - 4$
$\Rightarrow$ $g\left( x \right) = {{{x^3} - 3{x^2} + 3x - 2} \over {x - 2}}$                  ... (1)
Let us divide ${x^3} - 3{x^2} + 3x - 2\,by\,x - 2$. We get Therefore, equation (1) gives g (x) $= {x^2} - x + 1$

Q.5     Give examples of polynomials p(x), g(x) , q (x) and r (x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r (x)
(iii) deg r (x) = 0
Sol.     There can be several examples for each of (i), (ii) and (iii).
However, one example for each case may be taken as under :
(i) $p\left( x \right) = 2{x^2} - 2x + 14,g\left( x \right) = 2$,
$q\left( x \right) = {x^2} - x + 7,r\left( x \right) = 0$
(ii)  $p\left( x \right) = {x^3} + {x^2} + x + 1,g\left( x \right) = {x^2} - 1$,
$q\left( x \right) = x + 1\,,\,r\left( x \right) = 2x + 2$
(iii) $p\left( x \right) = {x^3} + 2{x^2} - x + 2$,
$g\left( x \right) = {x^2} - 1,q\left( x \right) = x + 2,\,r\left( x \right) = 4$