Question

Find the value of $a$ and $b$ so the polynomial $(x^4+a{x}^3-7x^2-8x+b)$ exactly divisible by $(x+2)$ as well $(x+3)$.

Answer 1

Given : Polynomial $x^4+a{x}^3-7x^2-8x+b$ is exactly divisible

by $(x+2)$ as well $(x+3)$

$\Rightarrow x=-2,-3$ are the roots of polynomial

$\therefore (-2{)}^4+a(-2{)}^3-7(-2{)}^2-8(-2)+b=0$

$16-8a-28+16+b=0$

$-8a+b=-4$ ... (I)

$(-3{)}^4+a(-3{)}^3-7(-3{)}^2-8(-3)+b=0$

$81-27a-63+24+b=0$

$-27+b=-42$ .... (II)

(I) - (II)

Ref. Image

$a=2$

put $a=2$ in equation (I)

$-16+b=-4$

$b=-4+16$

$b=12$

$\therefore a=2,b=12$