Section2.2Factoring Trinomials with Leading Coefficient One
In Section A.8, we learned how to multiply binomials like \((x+2)(x+3)\) and obtain the trinomial \(x^2+5x+6\text{.}\) In this section, we will learn how to undo that. So we’ll be starting with a trinomial like \(x^2+5x+6\) and obtaining its factored form \((x+2)(x+3)\text{.}\) The trinomials that we’ll factor in this section all have leading coefficient \(1\text{.}\)
So the idea is that if you need to factor \(x^2+5x+6\) and you somehow discover that \(2\) and \(3\) are special numbers (because \(2\cdot3=6\) and \(2+3=5\)), then you can conclude that \((x+2)(x+3)\) is the factored form of the given polynomial.
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.
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 \(x^2+13x+40=(x+5)(x+8)\text{.}\)
If the answer really is \((x+5)(x+8)\text{,}\) then notice how evaluating at \(-5\) would result in \(0\text{.}\) So the original expression should also result in \(0\) if we evaluate at \(\substitute{-5}\text{.}\) And similarly, if we evaluate it at \(\substitute{-8}\text{,}\)\(x^2+13x+40\) should be \(0\text{.}\)
We need a pair of numbers that multiply to \(-6\) and add to \(5\text{.}\) Note that we do care to keep track that they multiply to a negative product.
We need a pair of numbers that multiply to \(-40\) and add to \(-6\text{.}\) Note that we do care to keep track that they multiply to a negative product and sum to a negative total.
So far we have only factored examples of quadratic trinomials: trinomials whose highest power of the variable is \(2\text{.}\) However, this technique can also be used to factor trinomials where there is a larger highest power of the variable. It only requires that the highest power is even, that the next highest power is half of the highest power, and that the third term is a constant term.
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.
We need a pair of numbers that multiply to \(-50\) and add to \(-23\text{.}\) Note that we do care to keep track that they multiply to a negative product and sum to a negative total.
Sometimes factoring a polynomial will take two or more “stages.” Always begin factoring a polynomial by factoring out its greatest common factor, and then apply a second stage where you use a technique from this section. The process of factoring a polynomial is not complete until each of the factors cannot be factored further.
Now we are left with a factored expression that might factor more. Looking inside the parentheses, we ask ourselves, “what two numbers multiply to be \(-40\) and add to be \(-3\text{?}\)” Since \(5\) and \(-8\) do the job the full factorization is:
The three terms don’t exactly have a common factor, but as discussed in Section 1, when the leading term has a negative sign, it is often helpful to factor out that negative sign:
Looking inside the parentheses, we ask ourselves, “what two numbers multiply to be \(-24\) and add to be \(-2\text{?}\)” Since \(-6\) and \(4\) work here and the full factorization is shown:
Looking inside the parentheses, we ask ourselves, “what two numbers multiply to be \(-60\) and add to be \(4\text{?}\)” Since \(10\) and \(-6\) fit the bill, the full factorization can be shown below:
You might encounter a trinomial with two variables that can be factored using the methods we’ve discussed in this section. It can be tricky though: \(x^2+5xy+6y^2\) has two variables and it can factor using the methods from this section, but \(x^2+5x+6y^2\) also has two variables and it cannot be factored. So in examples of this nature, it is even more important to check that factorizations you find actually work.
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.
