Squares of Binomials

Learning goals

Expand $(a + b)^2$ as $a^2 + 2ab + b^2$
Subtract the middle term for a difference instead
Show why the middle term cannot be dropped
Square the whole term, coefficient included
Read the pattern backward to spot a perfect-square trinomial

Squaring a sum

To square any quantity is to multiply it by itself, so squaring the binomial $a + b$ means forming the product $(a + b)(a + b)$ . There is nothing new in expanding it; it is a product of two sums, exactly the kind you learned to multiply term by term. What is worth doing once, slowly, is watching where each piece lands, because the answer has a shape that repeats for every sum you will ever square.

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.

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.

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.

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, squaring a sum is never the same as adding the squares. The reason is that the middle term $2ab$ is real, and in that case it is not zero. Keep all three terms and you keep the truth.

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 $(2x + 3)^2$ and $(5 - 2x)^2$

For $(2x + 3)^2$ , 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 $x$ . For $(5 - 2x)^2$ the binomial is a difference, so use $(a - b)^2 = a^2 - 2ab + b^2$ with $a = 5$ and $b = 2x$ :

(5 - 2x)^2 = 5^2 - 2\cdot 5\cdot(2x) + (2x)^2 = 25 - 20x + 4x^2.

The outer terms $25$ and $4x^2$ are squares, and the middle term $-20x$ is twice the product $5\cdot 2x$ carrying a minus sign because the binomial is a difference.

Worked example 3 Square $(3x - 4)^2$ and $(2a + 5b)^2$

For $(3x - 4)^2$ , the difference pattern applies with $a = 3x$ and $b = 4$ :

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

The same pattern handles two different variables, since $a$ and $b$ never had to be numbers. For $(2a + 5b)^2$ , 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. Recognizing one lets you factor it instantly, with no searching through factor pairs. There is a short test with three parts. The first term must be a perfect square, with some square root $A$ . The last term must be a perfect square, with some square root $B$ . And the middle term must equal $2AB$ , up to its sign. When all three hold, the trinomial factors as $(A + B)^2$ if the middle term is positive, or $(A - B)^2$ if it is negative. If the middle term does not match $2AB$ , or if there is no middle term at all, the trinomial is not a perfect square.

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

Test the two outer terms first. The first term $9x^2$ is a perfect square, since $\sqrt{9x^2} = 3x$ , and the last term $4$ is a perfect square, since $\sqrt{4} = 2$ . Call these roots $A = 3x$ 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.

Worked example 5 Which trinomials are perfect squares?

Run each one through the same three conditions: a perfect-square first term, a perfect-square last term, and a middle term equal to twice the product of their roots.

For $16x^2 + 24x + 9$ , the roots of the outer terms are $\sqrt{16x^2} = 4x$ and $\sqrt{9} = 3$ , 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 perfect squares, with roots $2x$ and $3$ , 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 perfect square with root $x$ . But the last term $20$ is not a perfect square, since $\sqrt{20}$ is not a whole number. The pattern needs both outer terms to be perfect squares, so this trinomial does not fit either. Only the first of the three is a perfect square.

Check your understanding

Factor $25x^2 + 20x + 4$ .

(5x + 2)^2

(5x - 2)^2

(5x + 4)^2

(25x + 2)^2

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. Remember that $(3 + 4)^2 = 49$ , while $3^2 + 4^2 = 25$ .
Forgetting to double the middle. The middle term is $2ab$ , not $ab$ . There are two equal cross-products, so 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 $+b^2$ .
Confusing $(a - b)^2$ with $a^2 - b^2$ . A difference squared still has three terms, with a middle term and a positive last term. It is not the two-term expression $a^2 - b^2$ .
Calling a two-term expression a perfect-square trinomial. An expression like $4x^2 + 9$ has no middle term, so it is not one. Neither is a trinomial like $9x^2 + 15x + 25$ , whose middle term $15x$ is not twice the product $2\cdot 3x\cdot 5 = 30x$ of the outer roots.

Key takeaways

The two patterns are $(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$ . Both have the same first and last squares; the middle term is twice the product, added for a sum and subtracted for a difference.
The middle term $2ab$ is the crux. Squaring a sum is never $a^2 + b^2$ , as $(3 + 4)^2 = 49$ while $3^2 + 4^2 = 25$ .
Square the whole term, coefficient and all: $(2x + 3)^2 = 4x^2 + 12x + 9$ , because $(2x)^2 = 4x^2$ .
Read the patterns backward to factor: $a^2 + 2ab + b^2 = (a + b)^2$ and $a^2 - 2ab + b^2 = (a - b)^2$ .
A trinomial is a perfect square exactly when its first and last terms are perfect squares and its middle term equals $\pm 2$ times the product of their roots. For example $9x^2 - 12x + 4 = (3x - 2)^2$ , while $4x^2 + 9$ is not one because it has no middle term.

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)

Free Response Questions (FRQ)