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

AcademicMathematicsNCERTClass 9

To do: 

We have to find \( p(0), p(1) \) and \( p(2) \) for each of the given polynomials.

Solution:

To find the value of the polynomial $f(x)$ at $x=a$, we have to substitute $x=a$ in $f(x)$.

Therefore,

(i) \( p(y)=y^{2}-y+1 \)

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

$= 0-0+1$

$= 1$

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

$= 1-1+1$

$= 1$

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

$= 4-2+1$

$= 3$

Hence, $p(0), p(1), p(2)$ for the given polynomial are $1,1$ and $3$ respectively.

(ii) \( p(t)=2+t+2 t^{2}-t^{3} \)

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

$= 2+2(0)-0$

$= 2+0$

$=2$

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

$= 3+2(1)-1$

$= 4$

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

$= 4+2(4)-8$

$=-4+8$

$=4$

Hence, $p(0), p(1), p(2)$ for the given polynomial are $2,2$ and $4$ respectively.

(iii) \( p(x)=x^{3} \)

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

$=0$

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

$= 1$

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

$= 8$

Hence, $p(0), p(1), p(2)$ for the given polynomial are $0,1$ and $8$ respectively.

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

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

$=(-1)(1)$

$=-1$

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

$=(0)(2)$

$=0$

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

$=(1)(3)$

$=3$

Hence, $p(0), p(1), p(2)$ for the given polynomial are $-1,0$ and $3$ respectively.

raja
Updated on 10-Oct-2022 13:39:07

Advertisements