# Recognizing Products

Sometimes we can tell by looking at a polynomial that it's a product of a particular type. Not that we want to put polynomials into a box, but...it does preserve their freshness.

## Difference of Two Squares

Finding the difference of two squares is a two-way street. When we multiply a sum and difference of the same terms together, we have a difference of two squares. If you put your car in reverse and look in your rearview mirror, you see that a difference of two squares comes from the product of a sum and difference of the same terms. Now, no more algebra while driving. It isn't safe.

Any polynomial that looks like ▲^{2} – ■^{2} is the result of the polynomial multiplication (▲ + ■)(▲ – ■).

To factor the polynomial, we write it as the product of the sum and the difference.

### Sample Problem

Factor the polynomial *x*^{2} – 9.

This is a difference of two squares, so we know it can be factored as (▲ + ■)(▲ – ■) where ▲^{2} = *x*^{2} and ■^{2} = 9.

Therefore, we can take ▲ = *x* and ■ = 3 and factor the polynomial like so:

(*x* + 3)(*x* – 3)

See ya later, shapes. Go find a preschooler to play with.

### Sample Problem

Factor the polynomial 16*y*^{4} – *x*^{4}.

The variables have exponents greater than 2, but this polynomial is still a difference of two squares. It's trying really hard not to be recognized. Probably because it doesn't want the paparazzi all up in its face.

Since 16*y*^{4} – *x*^{4} = (4*y*^{2})^{2} – (*x*^{2})^{2}, we can factor the polynomial as:

(4*y*^{2} + *x*^{2})(4*y*^{2} – *x*^{2})

Beautiful. No cameras, please.

## Perfect-Square Trinomials

Sometimes you can figure out the factors of a polynomial by looking at it. If you have advanced telepathic powers, you may not even need to do *that* much.

We know what happens when we square a binomial, so we can recognize polynomials that are squares of binomials. Then we can work backwards to find the factors. If you want to challenge yourself, try working backwards, upside-down, *and* submerged underwater. We'll spot you.

Here's a quick reminder of what our perfect-square trinomials look like:

(▲ + ■)^{2} = ▲^{2} + 2▲■ + ■^{2}

(▲ – ■)^{2} = ▲^{2} – 2▲■ + ■^{2}

### Sample Problem

Factor the polynomial 4*y*^{2} + 12*y* + 9.

The second term is positive, so we'll see if we can get this polynomial by squaring a binomial of the form (▲ + ■)^{2}.

We need ▲^{2} = 4*y*^{2}, so ▲ = 2*y* is a good choice; nay, the choice of champions.

We also need ■^{2} = 9, so take ■ = 3.

It seems like the binomial we want is 2*y* + 3, but we need to check to make sure the middle term comes out right. The middle is important. If the ends were all that mattered, feature-length films would be a whole lot shorter.

When we square 2*y* + 3, we have a middle term of 2(2*y*)(3). Since 2(2*y*)(3) = 12*y*, the middle term of our original polynomial, our final answer is indeed:

4*y*^{2} + 12*y* + 9 = (2*y* + 3)^{2}

Whew. Beginning? Check. Middle? Check. End? Check. We just got ourselves either a correctly factored polynomial or the rough draft of a perfectly structured screenplay. See if you can get James Cameron on the horn.

### Sample Problem

Is there a binomial whose square equals 25*x*^{2} + 10*x* + 4? If so, what is it?

Let's see if this polynomial is the square of a binomial (▲ + ■)^{2}.

We would need ▲^{2} = 25*x*^{2}, so ▲ = 5*x*.

We also need ■^{2} = 4, so take ■ = 2.

It seems like the binomial we want is 5*x* + 2.

When we square 5*x* + 2, however, the middle term will be 2(5*x*)(2) = 20*x*. While we're sure 20*x* is a perfectly nice term and we wish it no ill will, it isn't what we were looking for since it differs from the middle term in our original polynomial. Sadly, or perhaps not *that* sadly, we can't have any binomial whose square is 25*x*^{2} + 10*x* + 4. We would go back to the drawing board if we had one. Ooh...we should put that on our birthday list.