ORCCAOpen Resources for Community College Algebra

Example7.3.2

Factor \(x^2+13x+40\text{.}\) Since the leading coefficient is \(1\text{,}\) we are looking to write this polynomial as \((x+\mathord{?})(x+\mathord{?})\) where the question marks are two possibly different, possibly negative, numbers. We need these two numbers to multiply to \(40\) and add to \(13\text{.}\) How can you track these two numbers down? Since the numbers need to multiply to \(40\text{,}\) one method is to list all factor pairs of \(40\) in a table just to see what your options are. We'll write every pair of factors that multiply to \(40\text{.}\)

We wanted to find all factor pairs. To avoid missing any, we started using \(1\) as a factor, and then slowly increased that first factor. The table skips over using \(3\) as a factor, because \(3\) is not a factor of \(40\text{.}\) Similarly the table skips using \(6\) and \(7\) as a factor. And there would be no need to continue with \(8\) and beyond, because we already found “large” factors like \(8\) as the partners of “small” factors like \(5\text{.}\)

There is an entire second column where the signs are reversed, since these are also ways to multiply two numbers to get \(40\text{.}\) In the end, there are eight factor pairs.

We need a pair of numbers that also adds to \(13\text{.}\) All we need to do is check what each of our factor pairs add up to:

The winning pair of numbers is \(5\) and \(8\text{.}\) Again, what matters is that \(5\cdot8=40\text{,}\) and \(5+8=13\text{.}\) So we can conclude that \(x^2+13x+40=(x+5)(x+8)\text{.}\)

Example7.3.3

Factor \(y^2-11y+24\text{.}\) The negative coefficient is a small complication from Example 7.3.2, but the process is actually still the same.

Example7.3.4

Factor \(z^2+5z-6\text{.}\) The negative coefficient is again a small complication from Example 7.3.2, but the process is actually still the same.

Example7.3.6

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:

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

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:

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:

Our factorization passes the tests.

Example7.3.9

Factor \(2z^2-6z-80\text{.}\)

Example7.3.10

Factor \(-r^2+2r+24\text{.}\)

Example7.3.11

Factor \(p^2q^3+4p^2q^2-60p^2q\text{.}\)

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:

Our factorization passes the tests.