Question

A quadratic polynomial in $x$ leaves remainders as $4$ and $7$ ,respectively, when divided by $(x+1)$ and $(x-2).$ Also it is exactly divisible by $(x-1).$ Find the quadratic polynomial.

Answer 1

Let the quadratic polynomial be $p\left(x\right)=a{x}^2+bx+c$

Given $p(-1)=4,p\left(2\right)=7$ and $p\left(1\right)=0$

$p(-1)=a(-1{)}^2+b(-1)+c=4$

$a-b+c=4$ .....(1)

Now, $p\left(1\right)=0$ and $p\left(2\right)=7$

Therefore,

$a(1{)}^2+b(1)+c=0$ and

$a(2{)}^2+b(2)+c=7$

$⟹a+b+c=0$ ....(2)

$4a+2b+c=7$ ......(3)

Subtracting equation 1 from equation 2, we have

$2b=-4$

$b=-2$

Subtracting equation 2 from equation 3, we have

$3a+b=7$

$3a-2=7$

$a=3$

Substituting the values of a and b in 1, we get

$c=-1$

Hence, the required quadratic polynomial is $3x^2-2x-1$.