Question

Question 9
Find P(0) , P(1) and P(-2) for the following polynomials:
(i) $P\left(x\right)=10x–4x^2–3$

(ii) $P\left(y\right)=(y+2)(y-2)$

Answer 1

(i)
$P\left(x\right)=10x–4x^2–3$ …… (i)
On putting x = 0 , 1 and -2, respectively in eq.(i) , we get;

$\Rightarrow P\left(0\right)=10\left(0\right)–4(0{)}^2–3$
$=0–0–3=-3$

$\Rightarrow P\left(1\right)=10\left(1\right)–4(1{)}^2–3$
$=10–4–3=10–7=3$

$\Rightarrow P(-2)=10(-2)–4(-2{)}^2–3$
$=-20–4\times 4–3$
$=-20–16–3=-39$

(ii)
Given , polynomial is
$P\left(y\right)=(y+2)(y–2).......\left(i\right)$
On putting y= 0, 1 and -2 respectively in eq.(i) we get;

$\Rightarrow P\left(0\right)=(0+2)(0–2)=-4$

$\Rightarrow P\left(1\right)=(1+2)(1-2)$
$=3\times (-1)=-3$

$\Rightarrow P(-2)=(-2+2)(-2-2)$
$=0(-4)=0$