Squares of Binomials

Learning goals

Expand $(a + b)^2$ as $a^2 + 2ab + b^2$
Use $(a - b)^2 = a^2 - 2ab + b^2$ for a difference, keeping the last term positive
Show why the middle term cannot be dropped
Square the whole term, coefficient included
Factor a perfect-square trinomial by reading the pattern backward, pulling out a common factor first

Squaring a sum

To square any quantity is to multiply it by itself, so squaring the binomial $a + b$ means forming the product $(a + b)(a + b)$ . Before working through the algebra, look at a picture of it: cut a square into pieces and watch where each term in the answer comes from.

The picture makes the middle term impossible to miss. A square with side $a + b$ has area $(a + b)^2$ . Cut it at the point where the length $a$ meets the length $b$ , in both directions. Those two cuts split the square into four regions: a square of area $a^2$ , a square of area $b^2$ , and two identical rectangles each of area $ab$ . The two rectangles are exactly why the middle term is $2ab$ and not just $ab$ . There are genuinely two of them, one above the small square and one beside it. This picture needs $a$ and $b$ to be positive lengths; the algebra below proves the identity for any real numbers at all, including negative ones.

A square of side a plus b divides into a square of area a squared, a square of area b squared, and two equal rectangles each of area a b. The two rectangles are why the middle term is twice a b.

The picture shows why the middle term is there. Algebra confirms it for every real number, not just positive lengths:

Why $(a + b)^2 = a^2 + 2ab + b^2$ #

Squaring a quantity means multiplying it by itself, so $(a + b)^2$ is shorthand for the product $(a + b)(a + b)$ . Expand it the way you expand any product of two sums, multiplying every term of the first factor by every term of the second:

(a + b)(a + b) = a\cdot a + a\cdot b + b\cdot a + b\cdot b.

The four products are $a^2$ , then $ab$ , then $ba$ , then $b^2$ . Because the order of multiplication does not matter, $ab$ and $ba$ are the same quantity, so the two middle products are equal and combine into a single $2ab$ :

a^2 + ab + ba + b^2 = a^2 + 2ab + b^2.

So squaring a sum gives three terms, not two. The three are the square of the first term, the square of the last term, and a middle term twice the product of the sum’s two terms. That doubled middle term is there because a sum has two matching cross-products, one from each factor, and they land on top of each other.

Squaring a difference

A difference squared works in the very same way, and only one thing changes. Write $(a - b)^2$ as the product $(a - b)(a - b)$ and expand it, carrying every sign along.

Why $(a - b)^2 = a^2 - 2ab + b^2$ #

Write the square as a product and multiply every term of the first factor by every term of the second, keeping track of the signs:

(a - b)(a - b) = a\cdot a - a\cdot b - b\cdot a + b\cdot b.

The first product is $a^2$ , and the last is $(-b)(-b) = b^2$ , which is positive because a negative times a negative is positive. The two middle products are $-ab$ and $-ba$ , both negative, and they combine into $-2ab$ :

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

Only the middle term changed. Squaring a difference gives the same first square $a^2$ and the same last square $b^2$ as squaring a sum, but the middle term is now subtracted. The final term stays positive because the last term of $a - b$ is $-b$ , and squaring $-b$ gives $+b^2$ .

The two patterns are worth holding side by side, because they differ in exactly one place. In both, the first and last terms are the squares $a^2$ and $b^2$ . The middle term is twice the product of the two terms, added when the binomial is a sum and subtracted when it is a difference.

Worked example 1 Expand $(x + 5)^2$ and $(x - 7)^2$

For $(x + 5)^2$ , use $(a + b)^2 = a^2 + 2ab + b^2$ with $a = x$ and $b = 5$ . Square the first term, square the last term, and put twice their product in the middle:

(x + 5)^2 = x^2 + 2\cdot x\cdot 5 + 5^2 = x^2 + 10x + 25.

For $(x - 7)^2$ , use the difference pattern $(a - b)^2 = a^2 - 2ab + b^2$ with $a = x$ and $b = 7$ . The first and last terms are still squares, and only the middle term turns negative:

(x - 7)^2 = x^2 - 2\cdot x\cdot 7 + 7^2 = x^2 - 14x + 49.

Notice that the last term is $+49$ in the second one too. Squaring the $-7$ makes it positive, so a difference squared never ends in a negative constant.

Why the middle term cannot be dropped

The single most common error with these patterns is to square each piece and stop, writing $(a + b)^2 = a^2 + b^2$ . That throws away the two cross-products entirely, and it is simply false. A quick number check settles it once and for all. Take $a = 3$ and $b = 4$ :

(3 + 4)^2 = 7^2 = 49, \qquad 3^2 + 4^2 = 9 + 16 = 25.

The two results are not equal, and the gap between them is $49 - 25 = 24$ , which is exactly the missing middle term $2ab = 2\cdot 3\cdot 4 = 24$ . Whenever $a$ and $b$ are both nonzero, $2ab$ is nonzero too, so squaring a sum is never the same as adding the squares. Keep all three terms, or the total comes out wrong.

Check your understanding

In the area model, a square of side $a + b$ splits into a square of area $a^2$ , a square of area $b^2$ , and two matching rectangles. Why does dropping those two rectangles make $a^2 + b^2$ the wrong total?

A Because the two rectangles are real area, each worth

ab

, so leaving them out loses

2ab

of the total B Because

a^2

and

b^2

are never actually part of the square's area C Because the two rectangles only exist when

a

and

b

happen to be equal D Because the total area of a square is always just its side length

Check your understanding

Expand $(x + 6)^2$ .

x^2 + 36

x^2 + 12x + 36

x^2 + 6x + 36

x^2 + 12x + 6

Squaring binomials that carry coefficients

Nothing about the patterns requires $a$ and $b$ to be single letters. They can be terms with coefficients, and the rule is unchanged: square the whole term, coefficient and variable together. The trap is to square the variable but leave the coefficient behind. In $(2x)^2$ the entire $2x$ is squared, so $(2x)^2 = 2^2 x^2 = 4x^2$ , not $2x^2$ . Once each term is squared correctly and their product is doubled for the middle, everything proceeds as before.

Worked example 2 Square $(2x + 3)^2$

Take $a = 2x$ and $b = 3$ . Square the first term, square the last term, and double their product:

(2x + 3)^2 = (2x)^2 + 2\cdot(2x)\cdot 3 + 3^2 = 4x^2 + 12x + 9.

The first term is $(2x)^2 = 4x^2$ , with the $2$ squared as well as the $x$ . Nothing else about the pattern changes: $2x$ is treated as one whole term, the way a bare $x$ was treated before.

Worked example 3 Square $(2a + 5b)^2$

The pattern also handles a binomial built from two different variables instead of one. Take the first term to be $2a$ and the last to be $5b$ :

(2a + 5b)^2 = (2a)^2 + 2\cdot(2a)\cdot(5b) + (5b)^2 = 4a^2 + 20ab + 25b^2.

Each outer term is squared in full, $(2a)^2 = 4a^2$ and $(5b)^2 = 25b^2$ , and the middle term $20ab$ is twice their product.

Check your understanding

Expand $(3x - 5)^2$ .

9x^2 + 30x + 25

3x^2 - 30x + 25

9x^2 - 30x + 25

9x^2 - 15x + 25

Reading the pattern backward: perfect-square trinomials

Every identity in algebra can be read in both directions. Read forward, $a^2 + 2ab + b^2$ is what you get by squaring $a + b$ . Read backward, it tells you that any trinomial of that exact shape factors as a single square:

a^2 + 2ab + b^2 = (a + b)^2 \qquad \text{and} \qquad a^2 - 2ab + b^2 = (a - b)^2.

A trinomial that factors this way is called a perfect-square trinomial, and recognizing one lets you factor it instantly, with no searching through factor pairs. As in every example in this lesson, look for terms with whole-number coefficients, not any real number. Here is what to look for, worked through on one trinomial before the rule is stated in general.

Worked example 4 Factor $9x^2 - 12x + 4$

Test the two outer terms first. The first term $9x^2$ is the square of $3x$ , since $(3x)^2 = 9x^2$ , and the last term $4$ is the square of $2$ , since $2^2 = 4$ . Call these two terms $A = 3x$ and $B = 2$ .

Now the middle term decides everything. For a perfect square it must be twice the product of $A$ and $B$ :

2AB = 2\cdot(3x)\cdot 2 = 12x.

The actual middle term is $-12x$ , which is $-2AB$ , so the sign is negative and the trinomial fits the difference pattern:

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

Expanding $(3x - 2)^2$ back to $9x^2 - 12x + 4$ confirms the factorization.

That example passed a short test with three parts, and every perfect-square trinomial passes the same one. The first term must be the square of some whole-number term $A$ . The last term must be the square of some whole-number term $B$ . And the middle term must equal $2AB$ , up to its sign: positive gives $(A + B)^2$ , negative gives $(A - B)^2$ . If the middle term does not match $2AB$ , or if there is no middle term at all, the trinomial is not a perfect square this way. (Sticking to whole-number coefficients is what keeps $A$ and $B$ easy to spot by inspection.)

Worked example 5 Which trinomials are perfect squares?

Run each one through the same three conditions: the first term is the square of some term, the last term is the square of some term, and the middle term equals twice the product of those two terms.

For $16x^2 + 24x + 9$ , the outer terms are squares, since $(4x)^2 = 16x^2$ and $3^2 = 9$ , and twice their product is $2\cdot 4x\cdot 3 = 24x$ , which matches the middle term. So it is a perfect square:

16x^2 + 24x + 9 = (4x + 3)^2.

For $4x^2 + 9$ , the outer terms are still squares, $(2x)^2 = 4x^2$ and $3^2 = 9$ , but there is no middle term at all. A perfect square would need a middle term of $2\cdot 2x\cdot 3 = 12x$ , so without it this is not a perfect-square trinomial.

For $x^2 + 10x + 20$ , the first term is a square, $(x)^2 = x^2$ . But no whole number squares to give $20$ , so the last term is not a perfect square. The pattern needs both outer terms to be perfect squares, so this trinomial does not fit either. Only the first of the three passes.

Before running the test, make sure the trinomial is in lowest terms: if every coefficient shares a common whole-number factor, pull that factor out first, the same way you would before any other factoring, and then run the test on what remains.

Worked example 6 Factor $3x^2 + 18x + 27$

Before testing anything, check for a common numerical factor. Here $3$ , $18$ , and $27$ are all multiples of $3$ , so pull it out first:

3x^2 + 18x + 27 = 3\left(x^2 + 6x + 9\right).

Now run the three-part test on what is left inside the parentheses. The outer terms are squares, $(x)^2 = x^2$ and $3^2 = 9$ , and twice their product is $2\cdot x\cdot 3 = 6x$ , which matches the middle term with a positive sign:

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

The factor of $3$ pulled out at the start does not disappear. It stays in front of the answer:

3x^2 + 18x + 27 = 3(x + 3)^2.

Testing the original coefficients $3$ , $18$ , $27$ directly against the pattern would fail, even though the trinomial is a perfect square once the common factor is removed.

Check your understanding

Factor $2x^2 + 20x + 50$ completely.

2(x + 5)^2

(2x + 10)^2

(x + 5)^2

2(x + 5)

Common mistakes

Dropping the middle term. The square of a sum is $a^2 + 2ab + b^2$ , not $a^2 + b^2$ . The cross-product $2ab$ is a real part of the answer, and skipping it is the number-one error.
Forgetting to double the middle. The middle term is $2ab$ , not $ab$ : there are two equal cross-products, and writing $a^2 + ab + b^2$ counts only one of them.
Not squaring the coefficient. In $(2x)^2$ the whole term is squared, so $(2x)^2 = 4x^2$ , not $2x^2$ . Square the coefficient and the variable together.
Putting the wrong sign on the middle. For a difference, $(a - b)^2 = a^2 - 2ab + b^2$ : only the middle term is negative. The last term $b^2$ stays positive, because squaring $-b$ gives $+b^2$ , so $(a - b)^2$ is not the two-term expression $a^2 - b^2$ .
Testing a trinomial before pulling out a common factor. If every coefficient shares a factor, remove it first. Testing $3x^2 + 18x + 27$ directly fails; testing $x^2 + 6x + 9$ after dividing out the $3$ succeeds.
Calling a two-term expression a perfect-square trinomial. An expression like $4x^2 + 9$ has no middle term, so it is not one.

Squares of Binomials

Learning goals

Squaring a sum

Squaring a difference

Why the middle term cannot be dropped

Squaring binomials that carry coefficients

Reading the pattern backward: perfect-square trinomials

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice