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

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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.

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