# How do you factor (10x+24)^2−x^4?

How do you factor $\left(10x+24{\right)}^{2}-{x}^{4}$
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Evelin Castillo
$\begin{array}{}\text{(1)}& {a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)\end{array}$
may be directly applied to
$\begin{array}{}\text{(2)}& \left(10x+24{\right)}^{2}-{x}^{4}\end{array}$
and we obtain
$\begin{array}{}\text{(3)}& \left(10x+24{\right)}^{2}-{x}^{4}=\left(10x+24+{x}^{2}\right)\left(10x+24-{x}^{2}\right);\end{array}$
but $10x+24+{x}^{2}$ and $10x+24-{x}^{2}$ can be factored as well! We have
$\begin{array}{}\text{(4)}& 10x+24+{x}^{2}=\left(x+4\right)\left(x+6\right),\end{array}$
Apparently $\left(10x+24{\right)}^{2}-{x}^{4}$ may be completely reduced over $\mathbb{Z}\left[x\right]$ to linear factors:
$\begin{array}{}\text{(6)}& \left(10x+24{\right)}^{2}-{x}^{4}=-\left(x+4\right)\left(x+6\right)\left(x+2\right)\left(x-12\right),\end{array}$
and its roots are −6,−4,−2 and 12. Cute, the way $10x+24$ is always a perfect square for any of these integers; but, then again, that's factoring in $\mathbb{Z}\left[x\right]$ for you.