Why is $|x^2 - a^2| = |x - a||x + a|$?

Question

By the definition of absolute value, we have \begin{equation} |x^2 - a^2| = \begin{cases} (x - a)(x + a) \ if \ x^2 \geq a^2\\ -(x - a)(x + a) \ if \ x^2 \leq a^2. \end{cases} \end{equation} How do we conclude that the right hand side equals $|x - a||x + a|$?

Answer 1

It's a combination of difference of two squares and the fact that $$ |ab| = |a||b|$$

Answer 2

Assume that $a\ge0$.

Note that $x^2\ge a^2$ if and only if $x\le -a$ or $x\ge a$.

If $x\le-a$, then $x+a\le0$ and $x-a\le0$. So $|x+a||x-a|=[-(x+a)][-(x-a)]=(x+a)(x-a)$.

If $x\ge a$, then $x+a\ge0$ and $x-a\ge0$. So $|x+a||x-a|=(x+a)(x-a)$.

We also have $x^2\le a^2$ if and only if $-a\le x\le a$. In such case, $x+a\ge0$ and $x-a\le 0$. So $|x+a||x-a|=(x+a)[-(x-a)]=-(x+a)(x-a)$.

Therefore, $|x^2-a^2|=|x+a||x-a|$.

If $a<0$, then $-a>0$. So we have $|x^2-a^2|=|x^2-(-a)^2|=|x+(-a)||x-(-a)|=|x-a||x+a|$.