

Binomials, Trinomials, and Squares, Oh My

Keep your eyes on the trail up Polynomial Mountain. Just ahead we can see some binomials that have been squared.

Our polynomial tour guide tells us that one special case of factoring involves squared binomials. Our guide leans in close and whispers another little trick in our ears (because we just don't have enough of those):

"The product of a squared binomial is a perfect square trinomial."

She then passes us a note with the following on it:

(a + b)2 = (a + b)(a + b) = a2 + 2ab + b2

(a – b)2 = (a – b)(a – b) = a2 – 2ab + b2

She then nods, factors herself into a squared binomial, and rolls down the mountain. Well, that was weird.

Look carefully at that note, mathronaut. Notice that the first and last terms of the trinomial are both perfect squares, and that the trinomial's middle term is . Also note that the sign before the middle term matches the sign in the binomial.

It's perfect. It has squares. It's a trinomial. Hence the name.

We could figure all of this out just by multiplying the squared binomial out. The important part is that this relationship works in both directions; the reverse is also true:

A perfect square trinomial factors to a binomial squared. Check it out:

a2 + 2ab + b2 = (a + b)2

a2 – 2ab + b2 = (a – b)2

How about we put this to work?

Sample Problem

Factor x2 – 2x + 1.

Okay, mathronaut. Is it a perfect square trinomial?

Well, the first and last term are perfect squares.

Does the middle term equal ?

Yep. The square root of x2 is just x, the square root of 1 is 1, and 2 times all that is 2(x)(1) = 2x. This factors out to a squared binomial. How convenient.

Just take the square root of the first term and the square root of the last term, throw a "–" sign between them, and square the whole shebang.

x2 – 2x + 1 = (x – 1)2

Let's keep this train moving.

Sample Problem

Factor 25x2 + 20x + 4.

We can see that the first and last terms are perfect squares, of 5x and 2. Looking good so far.

How about the middle term? Does 20x = 2(5x)(2)? Hyup, it sure does.

This is a perfect square trinomial, which factors to a squared binomial.

What's a mathronaut to do? Panic? No, we'll take the square roots of the first and last terms, throw in a "+" sign between them, and then square the whole thing.

25x2 + 20x + 4 = (5x + 2)2

Okay, just one more to go before we move on.

Sample Problem

Factor 9x4 – 6x2 + 1.

Well, 9x4 is the square of 3x2, and 1 is the square of…1. They seem pretty perfect to us, so that checks out.

Does 6x2 = 2(3x2)(1)? Why yes indeed. The middle term matches our expectations. This is a perfect square trinomial. We know how it factors from here.

9x4 – 6x2 + 1 = (3x2 – 1)2

Great, we've done it, and all without running into any lions, tigers, or bears. Now we can move on.

Domestic Square Dispute, Or, The Difference Between Two Squares

Next, look to the left of the trail up Polynomial Mountain. We can see Captain Kirk climbing the mountain—why is he climbing the mountain?

Next to him, we see another special case of polynomial factoring: the difference of squares.

Our polynomial friends can barely contain themselves. They have another little trick to show us:

(a + b)(a – b) = a2 – b2

(a – b)(a + b) = a2 – b2

In words, a more math-y person might say, "The product of a sum and a difference is a difference of two squares." A less math-y person might say, "Golly, that's a lot of letters."

Just like the squared binomials, this works both ways.

a2 – b2 = (a + b)(a – b)

Sample Problem

Factor 9x2 – 4.

We definitely have a difference here; that's subtraction to most folks. 3x squares to 9x2, and 2 squares to 4, and you square to you2. Now there are two of you, so you can enjoy Shmoop twice as much.

We can now take the square root terms and multiply them together as a sum and a difference.

9x2 – 4 = (3x – 2)(3x + 2)

Done. Now let's trek on a little further and apply this concept to solving equations.

Sample Problem

Solve 25x2 – 36 = 0 for x.

Our square roots terms are 5x and 6, so we plug those into our difference machine.

(5x + 6)(5x – 6) = 0

From here, it's business as usual for finding the roots for the equation.

5x + 6 = 0 and 5x – 6 = 0

Our business is concluded, so we'll be moving on before Kirk starts asking where all these zeros came from.