Factoring quartic equations

Question

factorize $$x^4+4abx^2-(a^2-b^2)^2$$ I am getting problem in factoring this. Anyone help me. I tried synthetic division but that doesn't work. Can anyone suggest me any other ways

Answer 1

HINT: $$x=a+b$$ and $$x=-(a+b)$$ are two Solutions and now you can divide your equation $$x^4-4abx^2-(a^2-b^2)^2$$ by $$x^2+(a+b)^2$$ note that your equation can factorized into $$- \left( {a}^{2}-2\,ab+{b}^{2}+{x}^{2} \right) \left( b+a+x \right) \left( b+a-x \right) $$

Answer 2

$\color{red}{x^4-4abx^2}-(a^2-b^2)^2$

The red part is almost a square. What we need is adding $4a^2b^2$

But if we add then we modify the polynomial and this is not legal, so we must subtract the same quantity to be fair. Therefore we get

$\color{red}{x^4-4abx^2+4a^2b^2}\color{blue}{-4a^2b^2-(a^2-b^2)^2}=\color{red}{x^4-4abx^2+4a^2b^2}\color{blue}{-a^4-2 a^2 b^2-b^4}=$

$=\color{red}{x^4-4abx^2+4a^2b^2}-\color{blue}{(a^4+2 a^2 b^2+b^4)}=$

The red is perfect square and the blue is a perfect square either so we can write

$=\color{red}{(x^2-2ab)^2}-\color{blue}{(a^2+b^2)^2}=$

Now remember the basic $P^2-Q^2=(P+Q)(P-Q)$

here $P\to x^2-2ab$ and $Q\to a^2+b^2$ so we have

$(\color{red}{(x^2-2ab)}+\color{blue}{(a^2+b^2)})(\color{red}{(x^2-2ab)}-\color{blue}{(a^2+b^2)})=\\=(x^2-2ab+a^2+b^2)(x^2-2ab-a^2-b^2)$

We are almost done

The second parenthesis can be written in this way

$x^2-(2ab+a^2+b^2)=x^2-(a+b)^2=(x+a+b)(x-a-b)$

Now the factoring is complete

$$x^4-4abx^2-(a^2-b^2)^2=(x^2-2ab+a^2+b^2)(x+a+b)(x-a-b)$$

I tried to explain best than I could

Hope it is useful