###### Example7.3.7

Factor $$h^{16}+22h^8+105\text{.}$$ This polynomial is one of the examples above where using factor pairs will help. We find that $$7\cdot15=105\text{,}$$ and $$7+15=22\text{,}$$ so the numbers $$7$$ and $$15$$ can be used:

\begin{align*} h^{16}+22h^8+105\amp=h^{16}+\overbrace{7h^8+15h^8}+105\\ \amp=\left(h^{16}+7h^8\right)+\left(15h^8+105\right)\\ \amp=h^8\left(h^8+7\right)+15\left(h^8+7\right)\\ \amp=\left(h^8+7\right)\left(h^8+15\right) \end{align*}

Actually, once we settled on using $$7$$ and $$15\text{,}$$ we could have concluded that $$h^{16}+22h^8+105$$ factors as $$\left(h^8+7\right)\left(h^8+15\right)\text{,}$$ if we know which power of $$h$$ to use. We'll always use half the highest power in these factorizations.

In any case, to confirm that this is correct, we should check by multiplying out the factored form:

\begin{aligned}[t] (h^8+7)(h^8+15)\amp=(h^8+7)\cdot h^8+(h^8+7)\cdot15\\ \amp=h^{16}+7h^8+15h^8+105\\ \amp\stackrel{\checkmark}{=}h^{16}+22h^8+15\end{aligned}

 $$h^8$$ $$7$$ $$h^8$$ $$h^{16}$$ $$7h^8$$ $$15$$ $$15h^8$$ $$105$$

Our factorization passes the tests.

in-context