Factoring The Difference Of Two Squares

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Factoring the difference of two squares

Here you will learn how to factor the difference of two square terms to problem solve. You have already learned other techniques for factoring, now, you can add to your factoring strategies, factoring the difference of perfect square terms.

Students first learn how to factor in $7$ th grade math where they learn how to factor the greatest common factor out of an expression. They extend that knowledge of factoring in an algebra $1$ class.

What is factoring the difference of two squares?

The Difference of two squares is an algebraic expression where the first expression and the second expression are perfect square terms with the second square term being subtracted from the first. The difference of two squares is always in the form of:

$a^2-b^2$

Algebraic expressions that are the difference of squares are factorable.

They factor to be:

$a^2-b^2=(a+b)(a-b)$

Let’s take a look algebraically and geometrically why the expression factors in this way.

If the expression were to be written in standard form of a quadratic, it would be written like this:

$a^2-b^2=a^2+0 a b-b^2$

You can use the trial and error method of factoring to factor the quadratic expression:

$a^2+0 a b-b^2$

The first term factors to $a \times a$ and the last term factors to $(- \, b) \times b$

$(a-b)(a+b)$

The inside terms multiply to $- \, ab$ and the outside terms multiply to be $+ \, ab$

Factoring the difference of two squares 1 US

When you add ab and $- \, ab,$ it equals $0ab$ or just $0$ (they cancel each other out). This means that when you factor, $a^2-b^2$ it factors to be $(a-b)(a+b).$ This works all the time with the difference of two perfect squares.

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Now, let’s look at why this works geometrically using squares.

Factoring the difference of two squares 2 US

Take a square with side length a. This has an area of $a^{2}.$

Factoring the difference of two squares 3 US

Let $b=\cfrac{1}{2} \, a.$ Draw a square of side length $b$ from a vertex (you may notice that a vertex of square $b$ is the centre of square $a$ ). Square $b$ has an area of $b^{2}.$

Factoring the difference of two squares 4 US

Remove the square with side length $b$ from the diagram. The remaining area is now $a^{2}-b^{2}$ as the area of $b^2$ has been subtracted from the area of $a^2.$

Factoring the difference of two squares 5 US

The width of the top of the $L$ -shape is $a-b$ as the length of $b$ has been subtracted from the side length of the square $a.$

Factoring the difference of two squares 6 US

This is the same as the furthest right vertical of the $L$ -shape, however as $b=\cfrac{1}{2} \, a, \, a-b=a-\cfrac{1}{2} \, a=\cfrac{1}{2} \, a=b.$ So, the furthest right vertical side has length $b.$

Factoring the difference of two squares 7 US

If you then split the $L$ -shape into a rectangle and a square, you can then move the square to make a longer rectangle. The area has not changed as nothing has been removed here:

Factoring the difference of two squares 8 US

The rectangle now has a width of $a-b,$ a height of $a+b$ and an area of $a^{2}-b^{2}$ which if substituted into the standard formula for the area of a rectangle is equivalent.

Factoring the difference of two squares 9 US

A=L\times{W}=(a+b)\times(a-b)=a^{2}-b^{2}

This geometrical representation with squares further proves that:

a^2-b^2=(a+b)(a-b)

What is the factoring the difference of two squares?

Common Core State Standards

How does this relate to high school math?

High School Algebra – Seeing Structure in Expressions: (HSA-SSE.A.2)
Use the structure of an expression to identify ways to rewrite it. For example, see $x^4 - y^4$ as $(x^2)^2 - (y^2)^2,$ thus recognizing it as a difference of squares that can be factored as $(x^2 - y^2)(x^2 + y^2).$

High School Algebra – Seeing Structure in Expressions: (HSA-SSE.B.3a)
Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression.★ a. Factor a quadratic expression to reveal the zeros of the function it defines.

How to factor the difference of two squares

In order to factor an algebraic expression using the difference of two squares:

Write down two sets of parentheses.
Square root the first term and write it on the left hand side of both parentheses.
Square root the last term and write it on the right hand side of both parentheses.
Put a $\textbf{“ +”}$ in the middle of one bracket and a $\textbf{“- “}$ in the middle of the other (the order doesn’t matter).

Factor the difference of two squares examples

Example 1: x² variable has a coefficient of 1

Factor the expression $x^{2}-9.$

Write down two sets of parentheses.

(\qquad)(\qquad)

2Square root the first term and write it on the left hand side of both parentheses.

$x^2$ is the first term and $\sqrt{x^2}=x$

Each set of parentheses starts with $x.$

(x\quad)(x\quad)

3Square root the last term and write it on the right hand side of both parentheses.

$9$ is the second term and $\sqrt{9}=3$

Each set of parentheses ends with $3.$

(x\quad{3})(x\quad{3})

4Put a $\textbf{“ +”}$ in the middle of one bracket and a $\textbf{“- “}$ in the middle of the other (the order doesn’t matter).

(x+3)(x-3)

Remember, you can always check to see if your answer is correct by multiplying the factored expression back out.

Factoring the difference of two squares 10 US

x^{2}+3x-3x-9=x^{2}-9

$x^{2}-9$ factors to be $(x+3)(x-3)$ .

Note: The order of the $“+”$ and $“-”$ does not matter.

Example 2: y² variable has a coefficient of 1

Factor the expression $64-y^{2}$ .

Write down two sets of parentheses.

(\qquad)(\qquad)

Square root the first term and write it on the left hand side of both parentheses.

The first term is $64$ and $\sqrt{64}=8$ .

Each set of parentheses starts with $8.$

$({8}\quad)({8}\quad)$

Square root the last term and write it on the right hand side of both parentheses.

The second term is $y^2$ and $\sqrt{y^2}=y$ .

Each set of parentheses ends with $y.$

$({8}\quad{y})({8}\quad{y})$

Put a $\textbf{“ +”}$ in the middle of one bracket and a $\textbf{“- “}$ in the middle of the other (the order doesn’t matter).

(8-y)(8+y)

Check:

Factoring the difference of two squares 11 US

$64+8y-8y-y^{2}=64-y^{2}$

$64-y^{2}$ factors to be $(8-y)(8+y)$ .

Note: The order of the $“+”$ and $“-”$ does not matter.

Example 3: x² variable has a coefficient greater than 1

Factor the expression $25x^{2}-16$ .

Write down two sets of parentheses.

(\qquad)(\qquad)

Square root the first term and write it on the left hand side of both parentheses.

The first term is $25x^2$ and $\sqrt{25x^{2}}=5x$

Each set of parentheses starts with $5x.$

$(5x\quad)(5x\quad)$

Square root the last term and write it on the right hand side of both parentheses.

The second term is $16$ and $\sqrt{16}=4$

Each set of parentheses ends with $4.$

$(5x\quad{4})(5x\quad{4})$

Put a $\textbf{“ +”}$ in the middle of one bracket and a $\textbf{“- “}$ in the middle of the other (the order doesn’t matter).

(5 x+4)(5 x-4)

Check:

Factoring the difference of two squares 12 US

$25x^{2}+20x-20x-16=25x^{2}-16$

$25x^{2}-16$ factors to be $(5x+4)(5x-4)$ .

Example 4: coefficient of both terms is greater than 1

Factor the expression fully.

4x^{2}-81y^{2}

Write down two sets of parentheses.

(\qquad)(\qquad)

Square root the first term and write it on the left hand side of both parentheses.

The first term is $4x^2$ and $\sqrt{4x^{2}}=2x$

Each set of parentheses starts with $2x.$

$(2x\quad)(2x\quad)$

Square root the last term and write it on the right hand side of both parentheses.

The second term is $81 y^2$ and $\sqrt{81 y^2}=9 y$

Each set of parentheses ends with $9y.$

$(2x\quad{9y})(2x\quad{9y})$

Put a $\textbf{“ +”}$ in the middle of one bracket and a $\textbf{“- “}$ in the middle of the other (the order doesn’t matter).

(2x+9y)(2x-9y)

Check:

Factoring the difference of two squares 13 US

$4x^{2}+18xy-18xy-81y^{2}=4x^{2}-81y^{2}$

$4x^{2}-81y^{2}$ factors to be $(2x+9y)(2x-9y)$ .

Example 5: exponents greater than 2

Factor the expression completely.

x^{4}-100

Write down two sets of parentheses.

(\qquad)(\qquad)

Square root the first term and write it on the left hand side of both parentheses.

The first term is $x^4$ and $\sqrt{x^4}=x^2$

Each set of parentheses starts with $x^2.$

$\left(x^2\quad\right)\left(x^2\quad\right)$

Square root the last term and write it on the right hand side of both parentheses.

The second term is $100$ and $\sqrt{100}=10$

Each set of parentheses ends with $10.$

$\left(x^{2}\quad{10}\right)\left(x^{2}\quad{10}\right)$

Put a $\textbf{“ +”}$ in the middle of one bracket and a $\textbf{“- “}$ in the middle of the other (the order doesn’t matter).

\left(x^2+10\right)\left(x^2-10\right)

Check:

Factoring the difference of two squares 14 US

$x^{4}+10x^{2}-10x^{2}-100=x^{4}-100$

$x^{4}-100$ factors to be $\left(x^{2}-10\right)\left(x^{2}+10\right)$ .

Example 6: exponents greater than 2

Factor the expression completely.

2m^6-72

Write down two sets of parentheses.

As both of the terms in the expression are a multiple of $2,$ factor this first.

$2(m^{6}-36)$

When writing the two pairs of parentheses, place the factor of $2$ infront.

$2(\qquad)(\qquad)$

The remaining expression to factor is $m^{6}-36.$

Square root the first term and write it on the left hand side of both parentheses.

The first term is $m^6$ and $\sqrt{m^6}=m^3$

Each set of parentheses starts with $m^3.$

$\left(m^3\right)\left(m^3\right)$

Square root the last term and write it on the right hand side of both parentheses.

The second term is $36$ and $\sqrt{36}=6$

Each set of parentheses ends with $6.$

$\left(m^3\quad{6}\right)\left(m^3\quad{6}\right)$

Put a $\textbf{“ +”}$ in the middle of one bracket and a $\textbf{“- “}$ in the middle of the other (the order doesn’t matter).

\left(m^3-6\right)\left(m^3+6\right)

Check:

Factoring the difference of two squares 15 US

$2\left(m^{6}-6m^{3}+6m^{3}-36\right)=2\left(m^{6}-36\right)=2m^{6}-72$

$2m^6-72$ factors fully to be $2\left(m^3-6\right)\left(m^3+6\right)$ .

Teaching tips for factoring the difference of squares

Use visual tools such as algebra tiles or digital algebra tiles so students can connect the area model with arrays to factoring.

Although worksheets have their place, having students work in collaborative groups to problem solve is essential to the math classroom.

Easy mistakes to make

Forgetting that in the parentheses the signs are different
There must be a + in one bracket, and a – in the other, the order doesn’t matter for example $(\quad+\quad)(\quad-\quad)$ or $(\quad-\quad)(\quad+\quad)$ .

Not finding the square root when factoring the difference of squares.
For example, factoring $16x^{2}-4$ to be $(8x-2)(8x+2).$ The correct factored expression should be $(4x-2)(4x+2)$ as $\sqrt{16x^{2}}=4x.$

Practice factoring the difference of two squares questions

(x-5)(x-5)

(x+5)(x-5)

(x+1)(x-25)

(x-1)(x+25)

The expression $x^2-25$ is the difference of two squares. To factor it, take the square root of both the first and the second expressions and place the square roots in two sets of parentheses. The middle sign of one set of parentheses is $“+”$ and the second one is $“-”.$

The first term in each pair of parentheses is $\sqrt{x^2}=x$

The second term in each pair of parentheses is $\sqrt{25}=5$

$(x+5)(x-5)$ is the factored expression.

(y+9)(y-9)

(y-9)(y-9)

(y+1)(y-81)

(y-1)(y+81)

The expression $y^2-81$ is the difference of two squares. To factor it, take the square root of both the first and the second expressions and place the square roots in two sets of parentheses. The middle sign of one set of parentheses is $“+”$ and the second one is $“-”.$

The first term in each pair of parentheses is $\sqrt{y^2}=y$

The second term in each pair of parentheses is $\sqrt{81}=9$

$(y+9)(y-9)$ is the factored expression.

(7-y)(7-y)

(49+y)(1-y)

(7+y)(7-y)

(1+y)(49-y)

The expression $49-y^2$ is the difference of two squares. To factor it, take the square root of both the first and the second expressions and place the square roots in two sets of parentheses. The middle sign of one set of parentheses is $“+”$ and the second one is $“-”.$

The first term in each pair of parentheses is $\sqrt{49}=7$

The second term in each pair of parentheses is $\sqrt{y^{2}}=y$

$(7+y)(7-y)$ is the factored expression.

(4+x)(1-x)

\left(x^2+2\right)\left(x^2-2\right)

\left(1+x^2\right)\left(4-x^2\right)

\left(2+x^2\right)\left(2-x^2\right)

The expression $4-x^4$ is the difference of two squares. To factor it, take the square root of both the first and the second expressions and place the square roots in two sets of parentheses. The middle sign of one set of parentheses is $“+”$ and the second one is $“-”.$

The first term in each pair of parentheses is $\sqrt{4}=2$

The second term in each pair of parentheses is $\sqrt{x^{4}}=x^{2}$

$(2+x^{2})(2-x^{2})$ is the factored expression.

(3x+10)(x-10)

(3x-10)(3x-10)

(9x+100)(x-1)

(9x+1)(x-100)

The expression $9x^{2}-100$ is the difference of two squares. To factor it, take the square root of both the first and the second expressions and place the square roots in two sets of parentheses. The middle sign of one set of parentheses is $“+”$ and the second one is $“-”.$

The first term in each pair of parentheses is $\sqrt{9x^{2}}=3x$

The second term in each pair of parentheses is $\sqrt{100}=10$

$(3x+10)(3x-10)$ is the factored expression.

(11-3y)(11+3y)

\left(121+y^2\right)\left(1-9y^2\right)

\left(11+9 y^2\right)\left(11-9 y^2\right)

3\left(11+3 y^2\right)\left(11-3 y^2\right)

The expression $363-27y^{4}$ is the difference of two squares. To factor it, first take out a common factor of $3$ from the two terms in the expression, then take the square root of both the first and the second expressions and place the square roots in two sets of parentheses. The middle sign of one set of parentheses is $“+”$ and the second one is $“-”.$

Take out the common factor of $3\text{:}$

3\left(121-9y^{4}\right)

The first term in each pair of parentheses is $\sqrt{121}=11$

The second term in each pair of parentheses is $\sqrt{9y^{4}}=3y^{2}$

$3(11+3y^{2})(11-3y^{2})$ is the factored expression.

(81x+y)(x-4y)

(x+4y)(81x-y)

(9x+4y)(9x-4y)

(9x-4y)(9x-4y)

The expression $81x^{2}-16y^{2}$ is the difference of two squares. To factor it, take the square root of both the first and the second expressions and place the square roots in two sets of parentheses. The middle sign of one set of parentheses is $“+”$ and the second one is $“-”.$

The first term in each pair of parentheses is $\sqrt{81x^{2}}=9x$

The second term in each pair of parentheses is $\sqrt{16y^{2}}=4y$

$(9x+4y)(9x-4y)$ is the factored expression.

Factoring the difference of two squares FAQs

What factoring (factorising) strategies can you use to factor polynomials?

When factoring polynomials, you can apply the same strategies used for factoring trinomials, like quadratics. Many times when factoring polynomials, you will have to apply more than one factoring strategy.

Can you factor the sum of squares?

Binomials that are the sum of squares cannot be factored. They are considered to be prime expressions. For example, if asked to factor $x^2+y^2,$ you would leave that binomial as the final answer.

Can fractions be in a perfect square binomial expression?

Yes, some fractions are considered perfect squares, for example, $\cfrac{1}{4} \, x^2-1$ is the difference of two perfect squares, which can be factored.

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