ex 2.2 class 9

Question 1: 

Find the value of the polynomial at 5x-4x²+3 at

(i)x = 0

(ii) x = −1

(iii)x = 2

Solution 1:$$

(i) $p(x)=5x-4x^2+3$

    $p(0)=5(0)-4(0)2+3$

            $=3$

(ii)$p(x)=5x-4x^2+3$

   $p(-1)=5(-1)-4(-1{)}^2+3$

            $=-5-4(1)+3$

            $=-6$

(iii) $p(x)=5x-4x^2+3$

$p(2)=5(2)-4(2{)}^2+3$

        $=10-16+3$

        $=-3$

Question 2:

Find p(0), p(1) and p(2) for each of the following polynomials:

(i)p(y) = y² − y + 1

(ii)p(t) = 2 + t + 2t² − t³

(iii)p(x) = x³

(iv)p(x) = (x − 1) (x + 1)

Solution 2:$$

(i)$p(y)=y^2-y+1$

$p(0)=(0{)}^2-(0)+1$

        $=1$

$p(1)=(1{)}^2-(1)+1$

        $=1–1+1$

        $=1$

$p(2)=(2{)}^2-(2)+1$

        $=4-2+1$

        $=3$

(ii)$p(t)=2+t+2t^2-t^3$

$p(0)=2+0+2(0{)}^2-(0{)}^3$

        $=2$

$p(1)=2+(1)+2(1{)}^2-(1{)}^3$

        $=2+1+2-1$

        $=4$

$p(2)=2+2+2(2{)}^2-(2{)}^3$

        $=2+2+8-8$

        $=4$

(iii)$p(x)=x^2$

$p(0)=(0{)}^3=0$

$p(1)=(1{)}^3=1$

$p(2)=(2{)}^3=8$

(iv) $p(x)=(x-1)(x+1)$

$p(0)=(0-1)(0+1)$

        $=(-1)(1)$

        $=-1$

$p(1)=(1-1)(1+1)$

        $=0(2)$

        $=0$

$p(2)=(2-1)(2+1)$

        $=1(3)$

        $=3$

Question 3:

Verify whether the following are zeroes of the polynomial, indicated against them.

(i)$p(x)=3x+1,x={\displaystyle \frac{-1}{3}}$

(ii)$p(x)=5x-\pi ,x={\displaystyle \frac{4}{5}}$

(iii)$p(x)=x^2-1,x=1,-1$

(iv)$p(x)=(x+1)(x-2),x=-1,2$

(v)$p(x)=x^2,x=0$

(vi)$p(x)=lm+m,x={\displaystyle \frac{-m}{l}}$

(vii)$p(x)=3x^2-1,x={\displaystyle \frac{-1}{\surd 3}},{\displaystyle \frac{-2}{\surd 3}}$

(viii)$p(x)=2x+1,x={\displaystyle \frac{1}{2}}$

Solution 3:

(i) If $x={\displaystyle \frac{-1}{3}}$ is a zero of given polynomial p$({\displaystyle \frac{-1}{3}})=3x+1$, then, $p({\displaystyle \frac{-1}{3}})$ should be $0$.

Here, $p({\displaystyle \frac{-1}{3}})=3({\displaystyle \frac{-1}{3}})+1=-1+1=0$

Therefore, $p(x)=3x+1,x={\displaystyle \frac{-1}{3}}$, is a zero of the given polynomial.

(ii)If $x={\displaystyle \frac{4}{5}}$ is a zero of given polynomial

$p(x)=5x-\pi$ then $x=({\displaystyle \frac{4}{5}})$ should be $0$.

Here, $p({\displaystyle \frac{4}{5}})=5({\displaystyle \frac{4}{5}})-\pi =4-\pi$.

As $p({\displaystyle \frac{4}{5}})\ne 0$.

Therefore, $p(x)=5x-\pi$, is not a zero of the given polynomial.

(iii)If $x=1$ and $x=-1$ are zeroes of polynomial $p(x)=x^2-1$, then $p(1)$ and $p(-1)$ should be $0$.

Here, $p(1)=(1{)}^2-1=0$, and

$p(-1)=(-1{)}^2-1=0$

Hence, $x=1$ and $-1$ are zeroes of the given polynomial.

(iv)If $x=-1$ and $x=2$ are zeroes of polynomial $p(x)=(x+1)(x-2)$, then $p(-1)$ and $p(2)$ should be $0$.

Here, $p(-1)=(-1+1)(-1-2)=0(-3)=0$, and

$p(2)=(2+1)(2-2)=3(0)=0$

Therefore, $x=-1$ and $x=2$ are zeroes of the given polynomial.

(v)If $x=0$ is a zero of polynomial $p(x)=x^2$, then $p(0)$ should be zero.

Here, $p(0)=(0{)}^2=0$

Hence, $x=0$ is a zero of the given polynomial.

Question 4:

Find the zero of the polynomial in each of the following cases:

(i)p(x) = x + 5

(ii)p(x) = x – 5

(iii)p(x) = 2x + 5

(iv) p(x) = 3x – 2

(v)p(x) = 3x

(vi) p(x) = ax, a ≠ 0

(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.

Solution 4:

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as $0$.

(i)$p(x)=x+5$

Let $p(x)=0$

$x+5=0$

$x=-5$

Therefore, for $x=-5$, the value of the polynomial is $0$ and hence, $x=-5$ is a zero of the given polynomial

(ii)$p(x)=x-5$

Let $p(x)=0$

$x-5=0$

$x=5$

Therefore, for $x=5$, the value of the polynomial is $0$ and hence, $x=5$ is a zero of the given polynomial.

(iii)$p(x)=2x+5$

Let $p(x)=0$

$2x+5=0$

$2x=-5$

$x={\displaystyle \frac{-5}{2}}$

Therefore, $x={\displaystyle \frac{-5}{2}}$ for, the value of the polynomial is $0$ and hence,$x={\displaystyle \frac{-5}{2}}$ is a zero of the given polynomial.

(iv)$p(x)=3x–2$

$p(x)=0$

$3x-2=0$

Therefore, for $x={\displaystyle \frac{2}{3}}$, the value of the polynomial is $0$ and hence, $x={\displaystyle \frac{2}{3}}$ is a zero of the given polynomial.

(v)$p(x)=3x$

Let $p(x)=0$

$3x=0$

$x=0$

Therefore, for $x=0$, the value of the polynomial is $0$ and hence, $x=0$ is a zero of the given polynomial.

(vi)$p(x)=ax$

Let $p(x)=0$

$ax=0$

$x=0$

Therefore, for $x=0$, the value of the polynomial is $0$ and hence, $x=0$ is a zero of the given polynomial.

(vii)$p(x)=cx+d$

Let $p(x)=0$

$cx+d=0$

$x={\displaystyle \frac{-d}{c}}$

Therefore,for $x={\displaystyle \frac{-d}{c}}$ , the value of the polynomial is $0$ and hence, $x={\displaystyle \frac{-d}{c}}$ is a zero of the given polynomial.