Question

Find the values of $a$ and $b$, if $x^2-4$ is a factor of $a{x}^2+2x^3-3x^2+bx-4$.

Answer 1

Let us first factorize $x^2-4$ as follows:

$x^2-4=x^2-2^2=(x-2)(x+2)(\because a^2-b^2=(a-b\left)\right(a+b\left)\right)$

It is given that $x^2-4$ is a factor of the polynomial $f\left(x\right)=a{x}^4+2x^3-3x^2+bx-4$ that is $(x-2)(x+2)$ are the factors of $f\left(x\right)=a{x}^4+2x^3-3x^2+bx-4$ and therefore, $x=-2$ and $x=2$ are the zeroes of $f\left(x\right)$.

Now, we substitute $x=-2$ and $x=2$ in $f\left(x\right)=a{x}^4+2x^3-3x^2+bx-4$ as shown below:

$f(-2)=a(-2{)}^4+2(-2{)}^3-3(-2{)}^2+b(-2)-4\Rightarrow 0=16a-16-12-2b-4\Rightarrow 0=16a-2b-32\Rightarrow 2(8a-b-16)=0\Rightarrow 8a-b-16=0\Rightarrow 8a-b=16....(1)$

$f\left(2\right)=a(2{)}^4+2(2{)}^3-3(2{)}^2+b(2)-4\Rightarrow 0=16a+16-12+2b-4\Rightarrow 0=16a+2b\Rightarrow 2(8a+b)=0\Rightarrow 8a+b=0....(2)$

Adding equations 1 and 2:

$(8a+8a)+(b-b)=16+0\Rightarrow 16a=16\Rightarrow a=1$

Substituting the value of $a$ in equation 1, we get:

$8a-b=16\Rightarrow (8\times 1)-b=16\Rightarrow 8-b=16\Rightarrow -b=16-8\Rightarrow -b=8\Rightarrow b=-8$

Hence, $a=1$ and $b=-8$.