Question

The expression $a{x}^2+bx+c$ equals $-2$ when $x=0$; leaves remainder $3$, when divided by $(x-1)$ and a remainder $-3$, when divided by $(x+1)$. Find the value of $a,b,c$.

Answer 1

Given: $p\left(x\right)=a{x}^2+bx+c$, $p\left(0\right)=-2,$ $p\left(1\right)=3$ and $p(-1)=-3$

To find: $a,b,c=?$

Solution:

$p\left(0\right)=a(0{)}^2+b\times 0+c$

$-2=c$

$c=-2$

$p\left(1\right)=a(1{)}^2+b\times 1-2$

$3=a+b-2$

$a+b=5$ $------------\left(1\right)$

$p(-1)=a(-1{)}^2+b\times (-1)-2$

$-3=a-b-2$

$a-b=-1$ $------------\left(2\right)$

On adding eqn.(1) and (2) we get,

$2a=4$

$a=2$

Put value of $a=2$ in eqn.(1) we get

$2+b=5$

$b=3$

Value of $a=2,$ $b=3$ and $c=-2$

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