Example 7.3.6

Let's factor \(x^2+13x+40\) again (the polynomial from Example 7.3.2). As before, it is important to discover that \(5\) and \(8\) are important numbers, because they multiply to \(40\) and add to \(13\text{.}\) As before, listing out all of the factor pairs is one way to discover the \(5\) and the \(8\text{.}\)

Instead of jumping to the factored answer, we can show how \(x^2+13x+40\) factors in a more step-by-step fashion using \(5\) and \(8\text{.}\) Since they add up to \(13\text{,}\) we can write:

\begin{align*} x^2+\attention{13}x+40\amp=x^2+\attention{\overbrace{5x+8x}}+40\\ \end{align*}

We have intentionally split up the trinomial into an unsimplified polynomial with four terms. In Section 7.2, we handled such four-term polynomials by grouping:

\begin{align*} \amp=\left(x^2+5x\right)+(8x+40)\\ \end{align*}

Now we can factor out each group's greatest common factor:

\begin{align*} \amp=x(x+5)+8(x+5)\\ \amp=x\attention{\overbrace{(x+5)}}+8\attention{\overbrace{(x+5)}}\\ \amp=(x+5)(x+8) \end{align*}

And we have found that \(x^2+13x+40\) factors as \((x+5)(x+8)\) without memorizing the shortcut.

in-context