# Polynomials : Exercise 2.2 (Mathematics NCERT Class 9th) Q.1В В В В В  Find the value of the polynomial $5x - 4{x^2} + 3$ at
В В В В В В В В В В В В В В  (i) x = 0 В  В  В В В В  (ii) x = вЂ“ 1 В  В  В В  (iii) x = 2
Sol.

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

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

Sol.

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

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

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

(i)В В В В  We have , p(x) = 3x + 1
В В В В В В В В В  At $x = - {1 \over 3},$ $p\left( { - {1 \over 3}} \right) = 3\left( { - {1 \over 3}} \right) + 1 = - 1 + 1 = 0$
В В В В В В В В
Therefore $- {1 \over 3}$ is a zero of polynomial = 3x + 1.
(ii)В В В  We have $p\left( x \right) = 5x - \pi$

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

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

В В В В В В В В В В В  At x = вЂ“1, $p\left( { - 1} \right) = \left( { - 1 + 1} \right)\left( { - 1 - 2} \right)$
В В В В В В В В В В В  $= \left( 0 \right)\left( { - 3} \right) = 0$
В В В В В В В В В В В  Therefore -1 is В a zero of p (x) .
В В В В В В В В В В В  Also , at x = 2 , $p\left( 2 \right) = \left( {2 + 1} \right)\left( {2 - 2} \right) = \left( 3 \right)\left( 0 \right) = 0$
В В В В В В В В В В В
Therefore 2 is zero of polynomial (x + 1)(x + 2).
(v)В В В В В В  We have, $p\left( x \right) = {x^2}$

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

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

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

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

Q.4В В В В В В В  Find the zero of the polynomial in each of the following cases :
В В В В В В В В В В В В В В
(i) $p\left( x \right) = x + 5$
В В В В В В В В В В В В В В
(ii) $p\left( x \right) = x - 5$
В В В В В В В В В В В В В В
(iii) $p\left( x \right) = 2x + 5$
В В В В В В В В В В В В В В
(iv) $p\left( x \right) = 3x - 2$
В В В В В В В В В В В В В В
(v) $p\left( x \right) = 3x$
В В В В В В В В В В В В В В
(vi) $p\left( x \right) = ax,a \ne 0$
В В В В В В В В В В В В В В
(vii) $p\left( x \right) = cx + d,c \ne 0,c,d$ are real numbers.
Sol.

(i)В В В В  We have to solve p(x) = 0
В В В В В В В В В  $\Rightarrow$ x + 5 = 0 $\Rightarrow$ x = - 5
В В В В В В В В В  Therefore - 5 is a zero of the polynomial x + 5 .
(ii)В В В В  We have to solve p(x) = 0
В В В В В В В В В В  $\Rightarrow$ x вЂ“ 5 = 0 $\Rightarrow$ x = 5
В В В В В В В В В В  Therefore 5 is a zero of the polynomial x вЂ“ 5.
(iii)В В В  We have to solve p(x) = 0
В В В В В В В В В В В  $\Rightarrow$ 2x + 5 = 0 $\Rightarrow$ $x = - {5 \over 2}$
В В В В В В В В В В В  Therefore $- {5 \over 2}$ is a zero of the polynomial 2x + 5.
(iv)В В В В  We have to solve p(x) = 0
В В В В В В В В В В В  $\Rightarrow$ 3x вЂ“ 2 = 0 $\Rightarrow$ $x = {2 \over 3}$
В В В В В В В В В В В  Therefore ${2 \over 3}$ is a zero of the polynomial 3x вЂ“ 2.
(v)В В В В В В  We have to solve p(x) = 0
В В В В В В В В В В В  $\Rightarrow$ 3x = 0 $\Rightarrow$ x = 0
В В В В В В В В В В В
Therefore 0 is a zero of the polynomial 3x.
(vi)В В В В В  We have to solve p(x) = ax, $a \ne 0$
В В В В В В В В В В В В  $\Rightarrow$ ax = 0 $\Rightarrow$ x = 0
В В В В В В В В В В В В  Therefore 0 is a zero of the polynomial ax.
(vii)В В В В  We have to solve p(x) = 0 , $c \ne 0$
В В В В В В В В В В В В  $\Rightarrow$ cx + d = 0 $\Rightarrow$ $x = - {d \over c}$
В В В В В В В В В В В В  Therefore $- {d \over c}$ is a zero of the polynomial cx + d.

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