Example7.3.12

Factor \(x^2+5xy+6y^2\text{.}\) This is a trinomial, and the coefficient of \(x\) is \(1\text{,}\) so maybe we can factor it. We want to write \((x+\mathord{?})(x+\mathord{?})\) where the question marks will be something that makes it all multiply out to \(x^2+5xy+6y^2\text{.}\)

Since the last term in the polynomial has a factor of \(y^2\text{,}\) it is natural to wonder if there is a factor of \(y\) in each of the two question marks. If there were, these two factors of \(y\) would multiply to \(y^2\text{.}\) So it is natural to wonder if we are looking for \((x+\mathord{?}y)(x+\mathord{?}y)\) where now the question marks are just numbers.

At this point we can think like we have throughout this section. Are there some numbers that multiply to \(6\) and add to \(5\text{?}\) Yes, specifically \(2\) and \(3\text{.}\) So we suspect that \((x+2y)(x+3y)\) might be the factorization.

To confirm that this is correct, we should check by multiplying out the factored form:

\begin{align*} (x+2y)(x+3y)\amp=(x+2y)\cdot x+(x+2y)\cdot3y\\ \amp=x^2+2xy+3xy+6y^2\\ \amp\stackrel{\checkmark}{=}x^2+5xy+6y^2 \end{align*}
\(x\) \(2y\)
\(x\) \(x^2\) \(2xy\)
\(3y\) \(3xy\) \(6y^2\)

Our factorization passes the tests.

in-context