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

Given:

Given expression is $ax^4 + 2x^3 - 3x^2 + bx - 4$.

$(x^2 - 4)$ is a factor of $ax^4 + 2x^3 - 3x^2 + bx - 4$.

To do:

We have to find the values of $a$ and $b$.

Solution:

We know that,

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

$x^2-4=x^2-2^2$

$=(x+2)(x-2)

This implies,

$x+2$ and $x-2$ are factors of $ax^4 + 2x^3 - 3x^2 + bx - 4$.

Therefore,

$f(-2)=0$

$\Rightarrow a(-2)^4 + 2(-2)^3 - 3(-2)^2 + b(-2) - 4=0$

$\Rightarrow 16a+2(-8)-3(4)-2b-4=0$

$\Rightarrow 16a-2b-16-12-4=0$

$\Rightarrow 16a-2b-32=0$

$\Rightarrow 2(8a-b-16)=0$

$\Rightarrow b=8a-16$........(i)

$f(2)=0$

$\Rightarrow a(2)^4 + 2(2)^3 - 3(2)^2 + b(2) - 4=0$

$\Rightarrow 16a+2(8)-3(4)+2b-4=0$

$\Rightarrow 16a+2b+16-12-4=0$

$\Rightarrow 16a+2b+16-16=0$

$\Rightarrow 2(8a+b)=0$

$\Rightarrow 8a+(8a-16)=0$........[From (i)]

$\Rightarrow 16a-16=0$

$\Rightarrow 16a=16$

$\Rightarrow a=1$

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

$\Rightarrow b=8-16=-8$

The values of $a$ and $b$ are $1$ and $-8$ respectively.      

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Updated on: 10-Oct-2022

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