Question

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

Answer 1

(i)

$p\left(y\right)=y^2-y+1$
Substitute $t=0,1,2$ in $p\left(y\right)$ to get the required values.
$p\left(0\right)=(0{)}^2-(0)+1=0-0+1=1$
$p\left(1\right)=(1{)}^2-(1)+1=1-1+1=1$
$p\left(2\right)=(2{)}^2-(2)+1=4-2+1=3$

Similar way,

(ii)

$p\left(t\right)=2+t+2t^2-t^3$

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

$p\left(1\right)=2+\left(1\right)+2(1{)}^2-(1{)}^3=2+1+2-1=4$

$p\left(2\right)=2+\left(2\right)+2(2{)}^2-(2{)}^3=2+2+8-8=4$

(iii)

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

$p\left(0\right)=(0{)}^3=0$

$p\left(1\right)=(1{)}^3=1$

$p\left(2\right)=(2{)}^3=8$

(iv)

$p\left(x\right)=(x-1)(x+1)$

$p\left(0\right)=(0-1)(0+1)=(-1)\left(1\right)=-1$

$p\left(1\right)=(1-1)(1+1)=\left(0\right)\left(2\right)=0$

$p\left(2\right)=(2-1)(2+1)=\left(1\right)\left(3\right)=3$