Section 3.4 Factoring
We often need to factor polynomial or other algebraic expressions. This means to rewrite sums and differences as a product. A monomial is a single term of a polynomial expression. When factoring polynomials, you might first look for the greatest common monomial of each term.
Example 3.8. Factoring out a Monomial Term.
Factor
\begin{equation*}
3x^3 y^3 - 9 x^2 y^5 + 12 x y^7.
\end{equation*}
Solution.
Each term has at least one factor of \(3\text{,}\) one factor of \(x\text{,}\) and three factors of \(y\text{.}\) So we can factor the greatest common factor \(3 x y^3\)
\begin{equation}
3x^3 y^3 - 9 x^2 y^5 + 12 x y^7
= 3xy^3\left(x^2-3xy^2+4y^4\right)\tag{3.1}
\end{equation}
Mentally redistribute the right side of
(3.1) to confirm the result is correct.
A common problem is factoring a quadratic polynomial
\(ax^2+bx+c\text{.}\) My recommendation is using a little trial and error while trying to reverse FOIL’ing. (See
Example 1.15.)
Example 3.9. Factoring a Quadratic.
Factor
\begin{equation*}
x^2-2 x-15.
\end{equation*}
Solution.
The product of the first terms must give us \(x^2 = x\cdot x\text{,}\) so we start with
\begin{equation*}
x^2-2 x-15 = (x - ??)(x-??).
\end{equation*}
The product of the last two must give us \(-15\text{.}\) Put aside the sign for a moment and consider all the integer factorization of \(15\text{:}\)
\begin{equation*}
15 = 1\times 15, \quad 3\times 5.
\end{equation*}
Try each one, adjusting the sign of each term. For example, none of
\begin{equation*}
(x - 1)(x-15), \quad (x+1)(x-15), \quad (x-1)(x+15), \quad (x+1)(x+15)
\end{equation*}
could work. The outer and inner terms all require adding or subtracting \(1x\) and \(15x\) which will never result in our middle term of \(-2x\text{.}\) So let’s try
\begin{equation*}
(x\pm 3)(x \pm 5)
\end{equation*}
which looks more promising since \(3x\) and \(5x\) will combine with the right sign to get \(-2x\text{.}\) Think about each possibility of sign to conclude
\begin{equation*}
x^2-2 x-15 = (x + 3)(x - 5).
\end{equation*}
When the leading coefficient is not \(a=1\text{,}\) the possibilities increase. While there are nice methods for helping you factor, I still recommend trial and error.
Example 3.10. Factoring a Quadratic.
Factor
\begin{equation*}
2 x^2-13 x+15.
\end{equation*}
Solution.
The first term factors as \(2x\cdot x\text{,}\) so we start with
\begin{equation*}
(2x - ??)(x - ??).
\end{equation*}
You should still consider the possibilities \(15 = 1\times 15\) and \(15 = 3\times 5\text{,}\) but observe now that the factor of \(2\) in the first term will double whatever we put in the second factor. I quickly ruled out using \(15 = 1\times 15\text{.}\) The remaining cases are of the form
\begin{equation*}
(2x \pm 3)(x \pm 5), \quad (2x \pm 5)(x \pm 3).
\end{equation*}
Try each one by FOIL’ing to find
\begin{equation*}
2 x^2-13 x+15 = (2x - 3)(x-5).
\end{equation*}
Look for special forms, like a difference of squares
\begin{equation}
a^2 - b^2 = (a-b)(a+b).\tag{3.2}
\end{equation}
Example 3.11. Factoring a Difference of Squares.
Factor
\begin{equation*}
x^6 - 4y^2.
\end{equation*}
Solution.
Each term is a square
\begin{equation*}
x^6 - 4y^2 = (x^3)^2 - (2 y)^2.
\end{equation*}
So that
\begin{equation}
x^6 - 4y^2 = (x^3-2y)(x^3 + 2y).\tag{3.3}
\end{equation}
Again, mentally distribute the right side of
(3.3) to confirm the result is correct.
Example 3.12. Combining Methods.
Factor
\begin{equation*}
3 x^4+3 x^3-36 x^2.
\end{equation*}
Solution.
There is a common monomial factor of \(3x^2\text{:}\)
\begin{equation*}
3 x^4+3 x^3-36 x^2 = 3x^2 (x^2+x-12).
\end{equation*}
Next, factor the quadratic
\begin{equation*}
3 x^4+3 x^3-36 x^2 = 3x^2 (x-3)(x+4).
\end{equation*}
Example 3.13. Combining Methods.
Factor
\begin{equation*}
x^4-10 x^2+9.
\end{equation*}
Solution.
This is like a quadratic if we think of the first term as \(x^4 = (x^2)^2\text{.}\) Each factor is a difference of squares, resulting in
\begin{equation*}
x^4-10 x^2+9 = (x-3)(x+3)(x-1)(x+1).
\end{equation*}
By creatively “splitting” the middle term of a quadratic you can sometimes create interesting factoring opportunities. One such method is called factoring by grouping.
Example 3.14. Factor by Grouping.
Factor
\begin{equation*}
9x^2 + 18x+ 8,
\end{equation*}
by observing that \(18 x = 6x + 12x\text{.}\)
Solution.
First, using the advice
\begin{equation*}
9x^2 + 18x+ 8 = 9x^2 + 6x + 12x + 8.
\end{equation*}
Now the first two terms have a common factor of \(3x\) and the second two terms have a common factor of \(4\text{.}\)
\begin{equation*}
9x^2 + 18x+ 8 = 3x(3x + 2) + 4(3x + 2).
\end{equation*}
Cool! Each term now has a factor \(3x+2\text{,}\) so that
\begin{equation*}
9x^2 + 18x+ 8 = (3x+2)(3x+4).
\end{equation*}
