# Polynomials : Exercise 2.2 (Mathematics NCERT Class 9th)

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Q.1      Find the value of the polynomial $5x - 4{x^2} + 3$ at
(i) x = 0          (ii) x = – 1        (iii) x = 2
Sol.

Let $p\left( x \right) = 5x - 4{x^2} + 3$
(i)    At x = 0 : $p\left( 0 \right) = 5\left( 0 \right) - 4{\left( 0 \right)^2} + 3$
$= 0 - 0 + 3 = 3$
(ii)   At x = –1 : $p\left( { - 1} \right) = 5\left( { - 1} \right) - 4{\left( { - 1} \right)^2} + 3$
$= - 5 - 4 + 3 = - 6$
(iii)  At x = 2 : $p\left( 2 \right) = 5\left( 2 \right) - 4{\left( 2 \right)^2} + 3 = 10 - 16 + 3$
$= 13 - 16 = - 3$

Q.2       Find p (0) , p (1) and p(2) for each of the following polynomials :
(i) $p\left( y \right) = {y^2} - y + 1$
(ii) $p\left( t \right) = 2 + t + 2{t^2} - {t^3}$
(iii) $p\left( x \right) = {x^3}$
(iv) p(x) = (x–1) (x + 1)

Sol.

(i)    We have , $p\left( y \right) = {y^2} - y + 1$
$p\left( 0 \right) = {\left( 0 \right)^2} - 0 + 1 = 0 - 0 + 1 = 1$
$p\left( 1 \right) = {\left( 1 \right)^2} - 1 + 1 = 1 - 1 + 1 = 1$
and $p\left( 2 \right) = {\left( 2 \right)^2} - 2 + 1 = 4 - 2 + 1 = 3$
(ii)   We have $p\left( t \right) = 2 + t + 2{t^2} - {t^3}$
Therefore $p\left( 0 \right) = 2 + 0 + 2{\left( 0 \right)^2} - {\left( 0 \right)^3}$
$= 2 + 0 + 0 - 0 = 2$
$p\left( 1 \right) = 2 + 1 + 2{\left( 1 \right)^2} - {\left( 1 \right)^3}$
$= 2 + 1 + 2 - 1 = 5 - 1 = 4$
and $p\left( 2 \right) = 2 + 2 + 2{\left( 2 \right)^2} - {\left( 2 \right)^3}$
$= 2 + 2 + 8 - 8 = 4$
(iii)   We have, $p\left( x \right) = {x^3}$
Therefore $p\left( 0 \right) = {\left( 0 \right)^3} = 0$
$p\left( 1 \right) = {\left( 1 \right)^3} = 1$
and $p\left( 2 \right) = {\left( 2 \right)^3} = 8$
(iv)    We have $p\left( x \right) = \left( {x - 1} \right)\,\left( {x + 1} \right)$
Therefore  $p\left( 0 \right) = \left( {0 - 1} \right)\,\left( {0 + 1} \right) = \left( { - 1} \right)\left( 1 \right) = - 1$
$p\left( 1 \right) = \left( {1 - 1} \right)\,\left( {1 + 1} \right) = \left( 0 \right)\left( 2 \right) = 0$
and $p\left( 2 \right) = \left( {2 - 1} \right)\,\left( {2 + 1} \right) = \left( 1 \right)\left( 3 \right) = 3$

Q.3        Verify whether the following are zeroes of the polynomial, indicated against them.
(i) $p\left( x \right) = 3x + 1,x = - {1 \over 3}$

(ii) $p\left( x \right) = 5x - \pi ,x = {4 \over 5}$
(iii) $p\left( x \right) = {x^2} - 1,x = 1, - 1$
(iv) $p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right),x = - 1,2$
(v) $p\left( x \right) = {x^2},x = 0$
(vi) $p\left( x \right) = l x + m,x = - {m \over \ell }$
(vii) $p\left( x \right) = 3{x^2} - 1,x = - {1 \over {\sqrt 3 }},{2 \over {\sqrt 3 }}$
(viii) $p\left( x \right) = 2x + 1,x = {1 \over 2}$
Sol.

(i)     We have , p(x) = 3x + 1
At $x = - {1 \over 3},$ $p\left( { - {1 \over 3}} \right) = 3\left( { - {1 \over 3}} \right) + 1 = - 1 + 1 = 0$

Therefore $- {1 \over 3}$ is a zero of polynomial = 3x + 1.
(ii)    We have $p\left( x \right) = 5x - \pi$

At  $x = {4\over 5}$
$P\left( {{4 \over 5}}\right) = 5\left( {{4\over 5}}\right)-\pi=4-\pi$
Therefore ${4\over 5}$ is not a zero of polynomial $5x - \pi$
(iii)   We have, $p\left( x \right) = {x^2} - 1$

At x = 1, $p\left( 1 \right) = {\left( 1 \right)^2} - 1 = 1 - 1 = 0$
Therefore 1 is a zero of p (x).
Also , at x = -1 $p\left( { - 1} \right) = {\left( { - 1} \right)^2} - 1$
$= 1 - 1 = 0$
Therefore -1 is a zero of polynomial $x^2- 1$
(iv)     We have , $p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right)$

At x = –1, $p\left( { - 1} \right) = \left( { - 1 + 1} \right)\left( { - 1 - 2} \right)$
$= \left( 0 \right)\left( { - 3} \right) = 0$
Therefore -1 is  a zero of p (x) .
Also , at x = 2 , $p\left( 2 \right) = \left( {2 + 1} \right)\left( {2 - 2} \right) = \left( 3 \right)\left( 0 \right) = 0$

Therefore 2 is zero of polynomial (x + 1)(x + 2).
(v)       We have, $p\left( x \right) = {x^2}$

At x = 0 , $p\left( 0 \right) = {\left( 0 \right)^2} = 0$
Therefore 0 is a zero of polynomial $x^2$.
(vi)      We have, $p\left( x \right) = \ell x + m$

At $x = - {m \over \ell }$ $p\left( { - {m \over \ell }} \right) = \ell \left( {{{ - m} \over \ell }} \right) + m$

$= - m + m = 0$
Therefore ${{ - m} \over \ell }$ is a zero of polynomial lx + m.
(vii)     We have $p\left( x \right) = 3{x^2} - 1$

At $x = - {1 \over {\sqrt 3 }}$ $p\left( { - {1 \over {\sqrt 3 }}} \right) = 3{\left( {{1 \over {\sqrt 3 }}} \right)^2} - 1$
$= 3 \times -{1 \over 3} - 1 = 1 - 1 = 0$
Therefore ${1\over {\sqrt 3 }}$ is a zero of polynomial $3x^2- 1$
At $x = {2 \over {\sqrt 3 }}$ , $p\left( {{2 \over {\sqrt 3 }}} \right) = 3{\left( {{2 \over {\sqrt 3 }}} \right)^2} - 1 = 3 \times {4 \over 3} - 1$
$= 4 - 1 = 3$
Therefore ${2\over {\sqrt 3 }}$ is not a zero of polynomial $3x^2- 1$
(viii)    We have $p\left( x \right) = 2x + 1$

At $x = {1 \over 2}$ $p\left( {{1 \over 2}} \right) = 2\left( {{1 \over 2}} \right) + 1 = 1 + 1 = 2$
Therefore ${1 \over 2}$ is not a zero of polynomial 2x + 1.

Q.4        Find the zero of the polynomial in each of the following cases :

(i) $p\left( x \right) = x + 5$

(ii) $p\left( x \right) = x - 5$

(iii) $p\left( x \right) = 2x + 5$

(iv) $p\left( x \right) = 3x - 2$

(v) $p\left( x \right) = 3x$

(vi) $p\left( x \right) = ax,a \ne 0$

(vii) $p\left( x \right) = cx + d,c \ne 0,c,d$ are real numbers.
Sol.

(i)     We have to solve p(x) = 0
$\Rightarrow$ x + 5 = 0 $\Rightarrow$ x = - 5
Therefore - 5 is a zero of the polynomial x + 5 .
(ii)     We have to solve p(x) = 0
$\Rightarrow$ x – 5 = 0 $\Rightarrow$ x = 5
Therefore 5 is a zero of the polynomial x – 5.
(iii)    We have to solve p(x) = 0
$\Rightarrow$ 2x + 5 = 0 $\Rightarrow$ $x = - {5 \over 2}$
Therefore $- {5 \over 2}$ is a zero of the polynomial 2x + 5.
(iv)     We have to solve p(x) = 0
$\Rightarrow$ 3x – 2 = 0 $\Rightarrow$ $x = {2 \over 3}$
Therefore ${2 \over 3}$ is a zero of the polynomial 3x – 2.
(v)       We have to solve p(x) = 0
$\Rightarrow$ 3x = 0 $\Rightarrow$ x = 0

Therefore 0 is a zero of the polynomial 3x.
(vi)      We have to solve p(x) = ax, $a \ne 0$
$\Rightarrow$ ax = 0 $\Rightarrow$ x = 0
Therefore 0 is a zero of the polynomial ax.
(vii)     We have to solve p(x) = 0 , $c \ne 0$
$\Rightarrow$ cx + d = 0 $\Rightarrow$ $x = - {d \over c}$
Therefore $- {d \over c}$ is a zero of the polynomial cx + d.