Question

Find the values of $a$ and $b$ so that the polynomial $p\left(x\right)=x^2+x^3+8x^2-ax+b$ is exactly divisible by $x^2-1$.

Answer 1

We have,

$p\left(x\right)=x^4+x^3+8x^2-ax+b$

$x^2-1=0$

$x^2=1$

$x=\pm 1$

Since, polynomial $p\left(x\right)$ is exactly divisible by $x=\pm 1$.

$\Rightarrow p\left(1\right)=0$

$\Rightarrow p(-1)=0$

Put $x=1$, then

$1^4+1^3+8(1{)}^2-a(1)+b=0$

$1+1+8-a+b=0$

$10-a+b=0$

$a-b=10$ $.........\left(1\right)$

Put $x=-1$, then

$(-1{)}^4+(-1{)}^3+8(-1{)}^2-a(-1)+b=0$

$1-1+8+a+b=0$

$8+a+b=0$

$a+b=-8$ $.........\left(2\right)$

Add equations $\left(1\right)$ and $\left(2\right)$, we get

$a-b=10$

$a+b=-8$

$\overline{2a+0=2}$

$2a=2$

$a=1$

Subtract equation $\left(1\right)$ and $\left(2\right)$, we get

$a+b=-8$

$a-b=10$

$\overline{0+2b=-18}$

$2b=-18$

$b=-9$

$\therefore a=1,b=-9$

Hence, this is the answer.