2.2 Factoring Techniques

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Learning Objectives

By the end of this section, you will be able to:

identify when terms have factors in common and write the factored form
factor simple trinomials with and without nontrivial leading coefficients
complete the square
factor by grouping

Common Factors

We practiced some basic factoring in the previous section just by virtue of reverse engineering the distribution process, but let's codify some things here. Let's consider the factors of all the terms in this algebraic expression:

\[ 3x^2y + 12 xy - 9x y^2 \quad \rightarrow \quad 3\cdot x \cdot x \cdot y + 3 \cdot 4 \cdot x \cdot y - 3 \cdot 3 \cdot x \cdot y \cdot y \notag \]

Notice what they all have in common...

\[ \textcolor{red}{3\cdot x} \cdot x \cdot \textcolor{red}{y} + \textcolor{red}{3} \cdot 4 \cdot \textcolor{red}{ x \cdot y} - 3 \cdot \textcolor{red}{3 \cdot x \cdot y} \cdot y \notag \]

All three terms have at least one $3, x, $ and $y$ in them. In this case, we say that $3xy$ is the greatest common factor of these terms, and we can factor out $3xy$ by writing it in front of a set of parentheses. Inside the parentheses, we leave all the leftover factors from each term.

\[ 3x^2y + 12 xy - 9x y^2 = 3xy ( x + 4 - 3 y ) \notag \]

There are lots of reasons you want to be able to recognize when factoring like this is possible. You'll see stuff that looks like this all the time in Calc I...

Example $\PageIndex{1}$

Simplify the difference quotient $ \dfrac{ (x+h)^2 - x^2}{h} $.

Solution

In the numerator, we start by expanding the squared binomial and combine like terms (this should look familiar if you did the homework).

\[ \dfrac{ x^2 + 2xh + h^2 - x^2}{h} = \dfrac{ 2xh + h^2 }{h} \notag \]

Then we notice that both terms in the numerator have at least one $ h$, so we factor it out. Once the $h$ is out in front being multiplied by everything, it's legal to cancel!

\[ \dfrac{ h(2x + h) }{h} = \frac{h}{h}\cdot\frac{2x+h}{1} = 2x + h \notag \]

Exercise $\PageIndex{1}$

Factor out the greatest common factor from these algebraic expressions.

$ 9m^3 + 27m^2 - 3m$
$ 12ab^2 + 24 a^2b - 6 ab^3 $
$ 4p^2 - 2p^4 + 8p^6 $

Answer

$ 3m (3m^2 + 9m - 1) $
$ 6ab ( 2b + 4a - b^2 ) $
$ 2p^2 ( 2 - p^2 + 4p^4 )$

Factoring Trinomials With Leading Coefficient 1

All we're doing in this section is rewinding all of the FOILing we just did. In the equality

\[ (x+2)(x-3) = x^2 - x - 6 \notag \]

FOILing is going from left to right, while factoring is going from right to left. It's not terrible, but it takes practice to get a good feel for it.

To factor an expression of the form $ x^2 + bx + c$, where $b, c$ are real numbers, and the leading coefficient on $ x^2$ is (secretly) 1:

Set up two factors with blank spaces: $ (x \quad \quad )( x \quad \quad ) $
Think of pairs of numbers that multiply to give $c$, and check what their sums are. Check different options of positive and negative factors as well.
If you find a pair that multiply to give $c$ and add to give the number $b$, place them in the blanks, using $+$ for positive and $-$ for negative.
If you have time, check your answer by FOILing and make sure it yields the original expression.

Example $\PageIndex{2}$

Factor the expressions.

$ x^2 + 6x + 8 $
$ x^2 - x - 20 $
$ x^2 - 18x + 81 $
(Important!) $ x^4 - 4x^2 + 4 $

Solution

1. Identify that $c = 8$ and $b = 6$. I think of pairs of numbers that multiply to give 8: $ (1)(8), (-1)(-8), (2)(4), (-2)(-4) $. I want the pair that, when added, will give me a positive 6. The option that works is 2 with 4, so I write the factors $ (x+2)(x+4) $. You can FOIL and check.

2. Identify that $c = -20$ and $b = -1$. Pairs of numbers that multiply to give $-20$ are: $ (1)(-20), (-1)(20), (-4)(5), (4)(-5), (-2)(10), (2)(-10) $. You don't have to think of all of them if you find one that works! Notice that $ (-5)+4 = -1 $ like I want, so I can stop there and write $ (x +4)(x-5) $.

3. Identify that $c = 81$ and $b = -18$. I want a pair of number that multiply to get $81$ and add to get $-18$. I quickly see that $ (-9)(-9) $ will do the trick. The answer is $ (x-9)(x-9) = (x-9)^2 $.

4. This one is tricky, because at first glance it doesn't look like the form I know. But look closer... if I give $x^2$ the nickname "$y$,"

\[ x^4 - 4x^2 + 4 = (x^2)^2 - 4(x^2) + 4 = y^2 - 4y + 4, \notag \]

suddenly I know what to do with this! I want two numbers that multiply to get $4$ and add to get $-4$. I take $-2$ and $-2$ and write $ (y-2)(y-2)$. But then remember that $y$ is just a nickname, so replace it with $x^2$. The final answer is $ (x^2 - 2)(x^2 - 2)$.

Keep an eye out for those special products we saw in the last section! You can always factor them this way to save time:

"Difference of Squares" : $A^2 - B^2 $	$ (A+B)(A-B) $
"Perfect Square" : $ A^2 + 2AB + B^2 $	$ (A+B)(A+B)$

Factoring by Grouping

Sometimes there isn't a greatest common factor among all the terms in an expression, but individually some groups of terms do have things in common. For example, there's no single common factor for all the terms here,

\[ x^3 + x^2 + 4x + 4 \notag \]

but if I look at the first two and the last two separately, I notice

\[ x^3 + x^2 \textcolor{red}{+ 4x + 4} = x^2(x+1) \textcolor{red}{ + 4(x+1)}. \notag \]

Then when I compare the black piece and the red piece, I see they both have $ (x+1)$ in common, and I can factor that guy out, to get $ (x+1) ( x^2 + 4) $. Cute.

Exercise $\PageIndex{2}$

Factor $ x^3 - 2x^2 - 9x + 18 $.

Answer: \[ \textcolor{red}{x^3 - 2x^2} - 9x + 18 = \textcolor{red}{x^2(x-2)} - 9(x-2) = (x-2)(x^2 - 9) = (x-2)(x+3)(x-3) \notag \]

Factoring Trinomials With Nontrivial Leading Coefficient

We know how to handle expressions that start with $ x^2$, but what about something like $2x^2 + 3x + 1$? Not to "uphill both ways" on you, but when I learned algebra, we factored these by "Guess & Check," setting up blanks $ (2x \quad \quad )(x \quad \quad) $ and trying different options to see what worked. I still do it that way, but students have shared with me this thing called The $ac$ Method.

To factor an expression of the form $ ax^2 + bx + c$ using the $ac$ method,

Compute the product $ac$.
Find a pair of numbers $m$ and $n$ that multiply to give $ac$ and add to give $b$.
Replace the $b$ with $(m+n)$ and split up the terms: $ ax^2 + bx + c = ax^2 + (m+n)x + c = ax^2 + mx + nx + c $.
Now that there are four terms, factor by grouping.

Example $\PageIndex{3}$

Factor $ 2x^2 - x - 3$.

Solution

Identify $a = 2, b= -1, c = -3 $ and compute the product $ ac = -6$. I think of some numbers that multiply to give $-6$: $-2$ and $3$, or $2$ and $-3$... Wait, $2 + (-3) = -1 $ which is $b$, so that's what I want. Now I split the middle term,

\[ 2x^2 + \textcolor{red}{ (-1)} x - 3 = 2x^2 + \textcolor{red}{(2-3)} x - 3 = 2x^2 + 2x - 3x - 3 \notag \]

and factor by grouping:

\[ 2x^2 + 2x - 3x - 3 = 2x(x+1) - 3(x + 1) = (x+1)(2x-3). \notag \]

Exercise $\PageIndex{3}$

Factor $ 3x^2 + 14x + 8 $.

Answer

Identify $a = 3, b = 14, c = 8$ and compute the product $ac = 24$. The $m$ and $n$ should be 12 and 2. Now I split my $bx$ term,

\[3x^2 + 14x + 8 = 3x^2 + (12+2)x + 8 = 3x^2 + 12x + 2x + 8 \notag \]

and factor by grouping:

\[ (3x^2 + 12x) + (2x + 8) = 3x(x+4) + 2(x+4) = (x+4)(3x+2). \notag \]

Caution

Don't make extra work for yourself if it's not necessary! If you see $ax^2 + bx + c$ with $a \neq 1$, first check for common factors from all terms and pull those out! For example, you don't need the $ac$ method for $ 5x^2 + 20x - 60 $. All of the terms have a factor of $5$ in common! If you take that out first to get $5(x^2 + 4x - 12) $, you can use the un-FOIL-ing method on the expression inside the parentheses.

Completing the Square

Completing the square is a technique that you will need if you ever want to take the equation of a circle and figure out its center and radius, or take the equation of a parabola and figure out its vertex, or solve a certain type of quadratic equation, or use substitution tricks in Calculus II, or... It's called "completing the square" because we're going to force a perfect square like $ (a+b)^2$ to appear. Let's introduce it here and get it simmering on the back burner.

I'm starting with the polynomial expression $x^2 - 4x + 7$, and I want to end up with an equivalent expression in the form $ (x-h)^2 + k $, where $h$ and $k$ are some constants (plain ol' numbers). Why? I'll tell you when you're older. I just really wish I had a perfect square like $ (x-h)^2 $ in my life. First, I write my expression down with a big space, putting the 7 in time out and ignoring him for now:

\[ x^2 - 4x \quad \quad \quad + 7 \notag \]

I do a little scratch work, taking the middle term's coefficient $-4$, cutting him in half, and squaring the result. That is, $-4 \overset{\div 2}{\longrightarrow} -2 \overset{\text{squared}}{\longrightarrow} 4 $. I add and subtract $4$ to my expression, which technically changes nothing!

\[ x^2 - 4x \textcolor{red}{+4} + 7 \textcolor{red}{-4} \notag \]

Now I look at the first three terms and factor them as usual, and simplify the constants hanging out at the end.

\[ (x^2 -4x + 4) + 7-4 = (x-2)(x-2) + 3 = (x-2)^2 + 3 \notag \]

Hey presto! It worked!

Completing the Square

Starting with the form $ x^2 + bx + c$, and aiming for the form $ (x-h)^2 + k$, where $b, c, h,$ and $k$ are constants...

1. Write out $ x^2 + bx \quad \quad \quad + c $, leaving space.

2. Compute $ \frac{b}{2}$ and $\left(\frac{b}{2}\right)^2 $.

3. Add $\left(\frac{b}{2}\right)^2 $ in the blank space and subtract $\left(\frac{b}{2}\right)^2 $ from the end:

\[ x^2 + bx + \left(\frac{b}{2}\right)^2 + c - \left(\frac{b}{2}\right)^2 \notag \]

4. Factor the first three terms as a perfect square $ \left(x+ \frac{b}{2}\right)^2 $, and simplify the constants at the end.

People don't generally love completing the square the first eight thousand or so times they see it, but it will come in handy if you remember it.

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Head over to the exercises section and practice your factoring skills!

"Difference of Squares" : \(A^2 - B^2 \)	\( (A+B)(A-B) \)
"Perfect Square" : \( A^2 + 2AB + B^2 \)	\( (A+B)(A+B)\)

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Completing the Square