How do I tell a difference of squares from a perfect-square trinomial fast?

Count the terms. Two terms subtracted, both perfect squares, means difference of squares and the answer has the form (a+b)(a-b). Three terms with square outer terms means you should check whether the middle term equals twice the product of the roots; if it does, the answer is (a± b)^2.

Do I have to pull out the GCF first, or can I do it later?

Pull it out first. Expressions like 12x^2-27 don't look like any special form until the 3 comes out, and 50x^2-72 hides a difference of squares behind a factor of 2. Factoring the GCF first also keeps the numbers small, which reduces mistakes.

Is x^2-5 a difference of squares?

Not over the integers, which is what "factor" normally means in this course, because 5 is not a perfect square. In later courses you can write it as (x+sqrt(5))(x-sqrt(5)), but in Algebra 1 you should call it prime unless the problem specifically allows irrational coefficients.

Why does the sum of squares matter if it never factors?

Because recognizing it tells you when to stop, and it shows up as a leftover factor in complete factorizations like x^4-16=(x^2+4)(x+2)(x-2). It also connects to why some quadratic equations have no real solutions, which is a major idea later in the course.

ALG1-7.5

Factoring Special Forms

Learn to factor difference of squares and perfect-square trinomials, pull out the GCF first, factor completely, and see why a sum of squares won't factor.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Factoring Special Forms, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Factoring Special Forms free

What this lesson covers

Some quadratics don't need trial and error at all. If you can recognize their shape, the factorization is basically already written for you. Two shapes show up constantly: the difference of squares, like

x^2-49

, and the perfect-square trinomial, like

x^2+14x+49

. Both come straight from the special products you multiplied out earlier in this unit — factoring just runs those multiplications backward.

In this lesson you'll learn how to test whether an expression really fits one of these patterns, how to handle coefficients like

25x^2

9y^4

, why you should always pull out a common factor before you start pattern-matching, and why

x^2+49

stubbornly refuses to factor with real numbers. The payoff: problems that would take a full minute of guess-and-check become a five-second recognition.

Difference of Squares: $a^2-b^2=(a+b)(a-b)$

Multiply

(a+b)(a-b)

and the middle terms cancel:

a^2-ab+ab-b^2=a^2-b^2

. Read that backwards and you have a factoring rule.

An expression is a difference of squares when it has exactly two terms, both are perfect squares, and they are subtracted. To use the rule, ask: what squares to give the first term, and what squares to give the second term?

For

x^2-49

x

squares to

x^2

and

7

squares to

49

, so

x^2-49=(x+7)(x-7)

.

Coefficients and higher powers work the same way, as long as they are squares. Since

25x^2=(5x)^2

and

36y^2=(6y)^2

, you get

25x^2-36y^2=(5x+6y)(5x-6y)

. Even powers are always squares:

x^6=(x^3)^2

, so

x^6-1=(x^3+1)(x^3-1)

Expression	Is it a difference of squares?	Why
$x^2-64$	Yes	$(x)^2-(8)^2$
$9m^2-49n^2$	Yes	$(3m)^2-(7n)^2$
$x^2-20$	No (not over the integers)	$20$ is not a perfect square
$x^3-25$	No	$x^3$ is not a perfect square
$x^2+64$	No	The terms are added, not subtracted

The order matters for the sign inside the factors, but not for the answer overall:

(x+7)(x-7)

and

(x-7)(x+7)

are the same product. What you may not do is write

(x-7)(x-7)

— that multiplies out to

x^2-14x+49

, which has a middle term.

Perfect-Square Trinomials

The other two special products give the perfect-square patterns:

a^2+2ab+b^2=(a+b)^2 \qquad a^2-2ab+b^2=(a-b)^2

A trinomial is a perfect square when the first and last terms are perfect squares (and the last term is positive), and the middle term is exactly twice the product of their square roots. That middle-term check is the whole test — skipping it is the number-one error here.

Take

x^2+14x+49

. The square roots of the outer terms are

x

and

7

. Is the middle term

2\cdot x\cdot 7=14x

? Yes. So

x^2+14x+49=(x+7)^2

.

Now take

x^2+15x+49

. Same outer terms, but

15x\ne 14x

, so it is not a perfect square. (It doesn't factor over the integers at all.)

The sign of the middle term becomes the sign inside the binomial. Since

x^2-14x+49

has

-14x

, it factors as

(x-7)^2

. The last term stays positive either way, because a negative squared is positive.

With coefficients: for

9x^2-30x+25

, the roots are

3x

and

5

, and

2(3x)(5)=30x

, matching the middle term's size. So

9x^2-30x+25=(3x-5)^2

.

One warning: if the last term is negative, as in

x^2+6x-9

, it can never be a perfect-square trinomial, no matter how tempting

9

and

6

look. Students often write

(x-3)^2

for that expression; expanding gives

x^2-6x+9

, which is not the original.

Common Factor First, Then Factor Completely

Always look for a greatest common factor before pattern-matching. Many expressions only reveal a special form after the GCF comes out.

Consider

12x^2-27

. Neither term is a perfect square, so it looks hopeless. But the GCF is

3

12x^2-27=3(4x^2-9)=3(2x+3)(2x-3)

.

Same idea with trinomials:

2x^2+20x+50=2(x^2+10x+25)=2(x+5)^2

. And with variable factors:

x^3-16x=x(x^2-16)=x(x+4)(x-4)

.

"Factor completely" means keep going until nothing left can be factored. Difference of squares can repeat. Look at

x^4-81

x^4-81=(x^2+9)(x^2-9)=(x^2+9)(x+3)(x-3)

The first difference of squares produced another one,

x^2-9

. Stopping at

(x^2+9)(x^2-9)

is incomplete. The other factor,

x^2+9

, is a sum of squares and stops there.

A reliable order of operations for any factoring problem in this unit:

Step	What to do
1	Pull out the GCF, including variables
2	Count terms: two terms suggests difference of squares
3	Three terms: test for a perfect square, then use general trinomial methods
4	Re-check every factor you wrote — can any factor again?
5	Multiply back to confirm

That last step catches almost every mistake. Expanding

3(2x+3)(2x-3)

gives

3(4x^2-9)=12x^2-27

, the original.

Why a Sum of Squares Doesn't Factor

x^2+9

cannot be factored using real numbers. Neither can

4x^2+25

x^2+1

. This is not a gap in your toolkit — it's a fact about the expression.

Here's the reasoning. If

x^2+9

factored into two linear factors with real coefficients, then setting it equal to zero would give real solutions. But

x^2+9=0

means

x^2=-9

, and no real number squares to a negative. So no such factors exist.

You can also test it directly. The only integer pairs multiplying to

9

are

1\cdot 9

and

3\cdot 3

(and their negatives), and none of them add to

0

, which is what the missing middle term requires. Attempts like

(x+3)(x+3)=x^2+6x+9

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

both miss.

The correct thing to write is that the expression is prime over the real numbers, or simply leave it as a final factor in your answer.

One caution: a sum of squares with a common factor is still partly factorable.

2x^2+50=2(x^2+25)

— the GCF comes out, then you stop. And a sum of cubes like

x^3+8

does factor, but that's a different pattern for a later course.

Where students go wrong most often: writing

x^2+16=(x+4)^2

. Expand it —

(x+4)^2=x^2+8x+16

, which has an extra

8x

. A binomial squared always produces a middle term, so a two-term sum of squares can never equal one.

Key terms

Difference of squares.: A binomial of the form $a^2-b^2$ , where both terms are perfect squares and they are subtracted; it factors as $(a+b)(a-b)$ .
Perfect-square trinomial.: A trinomial of the form $a^2\pm 2ab+b^2$ , which factors as $(a\pm b)^2$ . The middle term must equal twice the product of the square roots of the outer terms.
Perfect square (term).: An expression that is something squared, such as $49=7^2$ , $25x^2=(5x)^2$ , or $x^6=(x^3)^2$ . Every even power of a variable is a perfect square.
Greatest common factor (GCF).: The largest monomial that divides every term of a polynomial; it should be removed before applying any special-form pattern.
Factor completely.: To keep factoring until no factor can be broken down further, including re-factoring results such as $x^2-9$ inside a partially factored expression.
Sum of squares.: A binomial $a^2+b^2$ . It is prime over the real numbers, though a GCF may still be pulled out of it.
Prime (irreducible) polynomial.: A polynomial that cannot be written as a product of lower-degree polynomials with real (or integer) coefficients, such as $x^2+9$ .

Worked example

Factor completely:

32x^4-162

Step 1 — GCF. Both coefficients are even:

32=2\cdot 16

and

162=2\cdot 81

. There is no common variable. Factor out

2

32x^4-162=2(16x^4-81)

Step 2 — Identify the pattern. Inside the parentheses there are two terms, subtracted. Is each a perfect square?

16x^4=(4x^2)^2

and

81=9^2

. Yes — this is a difference of squares with

a=4x^2

and

b=9

2(16x^4-81)=2(4x^2+9)(4x^2-9)

Step 3 — Check each factor again. The factor

4x^2+9

is a sum of squares, so it is prime and stops here. The factor

4x^2-9

is another difference of squares:

4x^2=(2x)^2

and

9=3^2

, so

4x^2-9=(2x+3)(2x-3)

32x^4-162=2(4x^2+9)(2x+3)(2x-3)

Step 4 — Verify by multiplying back.

(2x+3)(2x-3)=4x^2-9

. Then

(4x^2+9)(4x^2-9)=16x^4-81

. Finally

2(16x^4-81)=32x^4-162

. It matches the original.

The two places students stop too early: forgetting the

2

at the start (which makes the terms non-squares and the problem look impossible), and stopping at

2(4x^2+9)(4x^2-9)

without factoring the second binomial again.

Practice questions

Which expression is a perfect-square trinomial?

$4x^2+12x+9$
$4x^2+13x+9$
$4x^2+12x-9$
$4x^2+6x+9$

Answer: $4x^2+12x+9$

Test the middle term. The square roots of the outer terms are

2x

and

3

, and twice their product is

2(2x)(3)=12x

. Only the first choice has that middle term with a positive last term, so it equals

(2x+3)^2

. The choice with

13x

fails the middle-term test. The one with

-9

can never be a perfect square, since a squared binomial always ends with a positive constant. The one with

6x

uses the product

2x\cdot 3=6x

but forgets to double it.

Factor completely:

5y^3-45y

Answer: $5y(y+3)(y-3)$

Start with the GCF. Both terms share

5

and

y

, so pull out

5y

5y^3-45y=5y(y^2-9)

. Now

y^2-9

has two terms, both perfect squares, subtracted, so it factors as

(y+3)(y-3)

. The complete factorization is

5y(y+3)(y-3)

. Check:

(y+3)(y-3)=y^2-9

, and

5y(y^2-9)=5y^3-45y

. A common incomplete answer is

5y(y^2-9)

, which stops one step short.

A student writes

x^2+36=(x+6)^2

. Explain why this is incorrect, and state the correct factorization of

x^2+36

over the real numbers.

Answer: Expanding $(x+6)^2$ gives $x^2+12x+36$ , which has an extra $12x$ term, so it is not equal to $x^2+36$ . The expression $x^2+36$ is a sum of squares and is prime over the real numbers, so it cannot be factored.

Squaring a binomial always produces three terms, because

(x+6)^2=(x+6)(x+6)=x^2+6x+6x+36

. The two middle terms add rather than cancel, so the result can never be a two-term sum. Cancellation only happens with opposite signs,

(x+6)(x-6)

, and that gives

x^2-36

— a difference, not a sum. Since

x^2+36=0

would require

x^2=-36

, which no real number satisfies, no real linear factors exist.

FAQ

How do I tell a difference of squares from a perfect-square trinomial fast?: Count the terms. Two terms subtracted, both perfect squares, means difference of squares and the answer has the form $(a+b)(a-b)$ . Three terms with square outer terms means you should check whether the middle term equals twice the product of the roots; if it does, the answer is $(a\pm b)^2$ .
Do I have to pull out the GCF first, or can I do it later?: Pull it out first. Expressions like $12x^2-27$ don't look like any special form until the $3$ comes out, and $50x^2-72$ hides a difference of squares behind a factor of $2$ . Factoring the GCF first also keeps the numbers small, which reduces mistakes.
Is $x^2-5$ a difference of squares?: Not over the integers, which is what "factor" normally means in this course, because $5$ is not a perfect square. In later courses you can write it as $(x+\sqrt{5})(x-\sqrt{5})$ , but in Algebra 1 you should call it prime unless the problem specifically allows irrational coefficients.
Why does the sum of squares matter if it never factors?: Because recognizing it tells you when to stop, and it shows up as a leftover factor in complete factorizations like $x^4-16=(x^2+4)(x+2)(x-2)$ . It also connects to why some quadratic equations have no real solutions, which is a major idea later in the course.

Learn this with a teacher, not a page

The Crimsora tutor teaches Factoring Special Forms live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

Try a live session free

What you'll do in this lesson

What this lesson covers

Difference of Squares: a2−b2=(a+b)(a−b)a^2-b^2=(a+b)(a-b)a2−b2=(a+b)(a−b)

Perfect-Square Trinomials

Common Factor First, Then Factor Completely

Why a Sum of Squares Doesn't Factor

Key terms

Worked example

Practice questions

FAQ

Learn this with a teacher, not a page

Difference of Squares: $a^2-b^2=(a+b)(a-b)$