Unit 11 · Lesson 11.3

Special patterns

Two kinds of factoring show up so often that it pays to recognize them on sight, without running the full two-number search each time. The first is a difference of squares, like x²−9. The second is a perfect-square trinomial, like x²+6x+9. Learning to spot them saves you the whole two-number search.

Start with the difference of squares, and build it rather than just state it. Multiply (a+b)(a−b) with the area box from Unit 10:

$$\begin{array}{c|c|c} & a & -b \\ \hline a & a^2 & -ab \\ \hline b & ab & -b^2 \end{array}\qquad\Rightarrow\qquad a^2 - ab + ab - b^2 = a^2-b^2$$

Look at the two middle cells: +ab and −ab. They add to zero, so the middle term goes to zero and disappears, leaving just a²−b². Run that backward and you have your factoring rule.

How do you recognize one? Two perfect squares with a minus between them. Then a is the square root of the first, b is the square root of the last, and it splits into (a+b)(a−b):

$$x^2-9=(x)^2-(3)^2=(x+3)(x-3)$$

Now the warning. A sum of squares, x²+9, with a plus, does not factor over the integers. Difference of squares needs the minus. You can see why with the unit's own multiply-back move: tempted to write (x+3)(x−3) for x²+9? Multiply it back and you get x²−9, a minus, so that product factors x²−9, never x²+9. The two-number search says the same thing: you'd need two numbers multiplying to +9 and adding to 0, and no integer pair does that. So x²+9 is prime.

Now the perfect-square trinomial. These are what you get when you square a binomial: (x+3)² works out to x²+6x+9. How do you recognize one? The first and last terms are both perfect squares, and the middle term equals 2 times the two square roots multiplied.

Check x²+6x+9: the square root of x² is x, the square root of 9 is 3, and 2·x·3 = 6x, which matches the middle, so it's (x+3)². The sign in the middle becomes the sign inside the parentheses: x²−10x+25 is (x−5)².

These two patterns are really just special cases of Lesson 11.2. For a perfect square the two numbers are equal, and for a difference of squares they're opposites. So the multiply-back check is the same one you already know, and it's just as mandatory here.

Putting it together: GCF first, then the pattern

Here's where "always try the GCF first" pays off most. Pulling a common factor can turn a messy expression into a clean pattern that wasn't visible before, and forgetting to keep that GCF in your final answer is the classic "not fully factored" slip. Two worked-through examples, each done end to end.

GCF, then a perfect square: 11.3.w6 $$2x^2+12x+18 \;=\; 2(x^2+6x+9) \;=\; 2(x+3)^2 \qquad \text{Check: } 2(x^2+6x+9)=2x^2+12x+18$$ Pull the 2, since every coefficient is even. The inside, x²+6x+9, is now a perfect square (√(x²)=x, √9=3, 2·x·3=6x). The 2 stays out front: 2(x+3)² is fully factored, and (x+3)² alone would be missing the 2.

GCF, then a difference of squares: 11.3.w7 $$3x^2-12 \;=\; 3(x^2-4) \;=\; 3(x+2)(x-2) \qquad \text{Check: } 3(x+2)(x-2)=3(x^2-4)=3x^2-12$$ Pulling out 3 exposes x²−4, a difference of squares. Neither step was visible before the GCF came out, which is exactly why the GCF goes first.

So how do you know you're done? A factorization is finished only when no piece can be factored again. After each split, look at every factor and ask: can this one be factored? Is there a leftover difference of squares, a trinomial that still factors, a GCF you missed? Stop only when each piece is a single term, a prime trinomial, or a binomial that can't be broken down further.