Question

Find the value of $a$ and $b$ so that the polynomial $x^3+10x^2+ax+b$ is exactly divisible by $(x-1)$ as well as by $x+2$.

Answer 1

Polynomial $=x^3+10x^2+ax+b$ is divisible by both $(x-1)\&(x+2)$

which means, $1\&-2$ are the zeroes of the polynomial

$\therefore$ If $p\left(x\right)=x^3+10x^2+ax+b$

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

$p\left(1\right)=1+10+a+b=0\Rightarrow a+b=-11\to \left(1\right)$

$p(-2)={(-2)}^3+10{(-2)}^2-2\left(a\right)+b=0\Rightarrow -8+40-2a+b=0=2a-b=32\to \left(1\right)$

Now, we have two equations solving both, $3a=21$

$\Rightarrow a=7$

$\therefore b=-18.$

Hence, the answer is $-18.$