Once you start solving algebraic equations that involve polynomials, the ability to recognize special, easily factored forms of polynomials becomes very useful. One of the most useful "easy-factor" polynomials to spot is the perfect square, or the trinomial that results from squaring a binomial. Once you've identified a perfect square, factoring it into its individual components is often a vital part of the problem-solving process.

## Identifying Perfect Square Trinomials

Before you can factor a perfect square trinomial, you have to learn to recognize it. A perfect square can take on either of two forms:

*a*^{2}+ 2_ab_ +*b*^{2}, which is the product of (*a*+*b*)(*a*+*b*) or (*a*+*b*)^{2}*a*^{2}– 2_ab_ +*b*^{2}, which is the product of (*a*–*b*)(*a*–*b*) or (*a*–*b*)^{2}

Some examples of perfect squares that you might see in the "real world" of math problems include:

*x*^{2}+ 8_x_ + 16 (This is the product of (*x*+ 4)^{2})*y*^{2}– 2_y_ + 1 (This is the product of (*y*– 1)^{2})- 4_x_
^{2}+ 12_x_ + 9 (This one is a little sneakier; it's the product of (2_x_ + 3)^{2})

What's the key to recognizing these perfect squares?

## Check the First and Third Terms

Check the first and third terms of the trinomial. Are they both squares? If yes, figure out what they are squares of. For example, in the second "real world" example given above, *y*^{2} – 2_y_ + 1, the term *y*^{2} is obviously the square of *y.* The term 1 is, perhaps less obviously, the square of 1, because 1^{2} = 1.

## Multiply the Roots

Multiply the roots of the first and third terms together. To continue the example, that's *y* and 1, which gives you *y* × 1 = 1_y_ or simply *y*.

Next, multiply your product by 2. Continuing the example, you have 2_y._

## Compare to the Middle Term

Finally, compare the result of the last step to the middle term of the polynomial. Do they match? In the polynomial *y*^{2} – 2_y_ + 1, they do. (The sign is irrelevant; it'd also be a match if the middle term were +2_y_.)

Because the answer in Step 1 was "yes" and your result from Step 2 matches the middle term of the polynomial, you know you're looking at a perfect square trinomial.

## Factoring a Perfect Square Trinomial

Once you know you're looking at a perfect square trinomial, the process of factoring it is quite straightforward.

## Identify the Roots

Identify the roots, or the numbers being squared, in the first and third terms of the trinomial. Consider another of your example trinomials that you already know is a perfect square, *x*^{2} + 8_x_ + 16. Obviously the number being squared in the first term is *x*. The number being squared in the third term is 4, because 4^{2} = 16.

## Write out Your Terms

Think back to the formulas for perfect square trinomials. You know your factors will take either the form (*a* + *b*)(*a* + *b*) or the form (*a* – *b*)(*a* – *b*), where *a* and *b* are the numbers being squared in the first and third terms. So you can write your factors out thusly, omitting the signs in the middle of each term for now:

(*a* ? *b*)(*a* ? *b*) = *a*^{2} ? 2_ab_ + *b*^{2}

To continue the example by substituting the roots of your current trinomial, you have:

(*x* ? 4)(*x* ? 4) = *x*^{2} + 8_x_ + 16

## Examine the Middle Term

Check the middle term of the trinomial. Does it have a positive sign or a negative sign (or, to put it another way, is it being added or subtracted)? If it has a positive sign (or is being added), then both factors of the trinomial have a plus sign in the middle. If it has a negative sign (or is being subtracted), both factors have a negative sign in the middle.

The middle term of the current example trinomial is 8_x_ – it's positive – so you've now factored the perfect square trinomial:

(*x* + 4)(*x* + 4) = *x*^{2} + 8_x_ + 16

## Check Your Work

Check your work by multiplying the two factors together. Applying the FOIL or first, outer, inner, last method gives you:

*x*^{2} + 4_x_ + 4_x_ + 16

Simplifying this gives the result *x*^{2} + 8_x_ + 16, which matches your trinomial. So the factors are correct.