Find the values of $p$ and $q$ so that $x^4 + px^3 + 2x^2 - 3x + q$ is divisible by $(x^2 - 1)$.

Given:

Given expression is $x^4 + px^3 + 2x^2 - 3x + q$.

$x^4 + px^3 + 2x^2 - 3x + q$ is divisible by $(x^2 - 1)$.

To do:

We have to find the values of $p$ and $q$.

Solution:

We know that,

If $(x-m)$ is a root of $f(x)$ then $f(m)=0$.

$x^2-1=x^2-1^2$

$=(x+1)(x-1)$

This implies, $x+1$ and $x-1$ are factors of $x^4 + px^3 + 2x^2 - 3x + q$.

Therefore,

$f(-1)=0$

$\Rightarrow (-1)^4+p(-1)^3+2(-1)^2 - 3(-1) + q=0$

$\Rightarrow 1-p+2+3+q=0$

$\Rightarrow p=6+q$...............(i)

$f(1)=0$

$\Rightarrow (1)^4+p(1)^3+2(1)^2 - 3(1) + q=0$

$\Rightarrow 1+p+2-3+q=0$

$\Rightarrow 6+q+q=0$                   [From (i)]

$\Rightarrow 6+2q=0$

$\Rightarrow 2q=-6$

$\Rightarrow q=-3$

$\Rightarrow p=6+(-3)=6-3=3$

The values of $p$ and $q$ are $3$ and $-3$ respectively.