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{align*} (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{align*}
\(h^8\) \(7\)
\(h^8\) \(h^{16}\) \(7h^8\)
\(15\) \(15h^8\) \(105\)

Our factorization passes the tests.

in-context