body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

Dashboard

Algebra 1 View details

7. Factoring Quadratic Expressions Using Special Products

Lesson

Exercises

Tests

eCourses /

Algebra 1 /

Chapter 7

Factoring Quadratic Expressions Using Special Products

Understanding how to factor quadratic expressions can be a game-changer in solving various mathematical problems. This lesson delves into specialized techniques like using perfect square trinomials, conjugate binomials, and the difference of squares to make the process easier. These methods are particularly useful in algebra, calculus, and even in real-world applications like engineering and physics. By grasping these techniques, you can simplify complex equations, making them easier to solve or analyze. Whether you're a student looking to improve your grades or a professional seeking to enhance your problem-solving skills, these factoring methods offer valuable insights.

Lesson Settings & Tools

	10 Theory slides
	10 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Image Credits

Slide of 10

There are a few different methods for factoring binomials and trinomials. Some polynomials can be factored by reversing the rules for multiplying special case binomials. This lesson will focus on those special cases.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Challenge

Showing Equivalent Expressions By Factoring

Consider the following algebraic expressions.

(n + 2)^{2} - 4 and n^{2} + 4 n

These algebraic expressions are equivalent.

a Use the figures below to illustrate why the expressions are equivalent.

b Find some ways to algebraically verify the same result.

Discussion

Perfect Square Trinomials and Factoring Them

There can be different methods for factoring a quadratic expression, depending on its type. If the expression is the square of a binomial, it can be factored as a perfect square trinomial.

Concept

Perfect Square Trinomial

A perfect square trinomial is a trinomial that can be written as the square of a binomial. There are two general types of perfect square trinomials that can be useful when dealing with quadratic functions or quadratic equations.

Perfect Square Trinomial	Square of Binomial
$a^{2} + 2 a b + b^{2}$	$(a + b)^{2}$
$a^{2} - 2 a b + b^{2}$	$(a - b)^{2}$

A perfect square trinomial should have one of the forms shown in the table — that is, the first and the last terms are perfect squares and the middle term is two times the product of the square roots of the first and last terms.

Once a quadratic expression is recognized as a perfect square trinomial, it can be written as a square of a binomial by factoring the perfect square trinomial.

Method

Factoring a Perfect Square Trinomial

For a trinomial to be factorable as a perfect square trinomial, the first and last terms must be perfect squares and the middle term must be two times the square roots of the first and last terms. Consider the following expression.

16 x^{2} + 88 x + 121

To factor this trinomial, there are three steps.

Confirm That the First and Last Terms Are Perfect Squares

One good way to recognize if a trinomial is a perfect square trinomial is to look at its first and last terms. If they are both perfect squares, there is a good chance that it is a perfect square trinomial. In the given expression, the first and last terms can be written as the squares of

4 x

and

11,

respectively.

16 x^{2} + 88 x + 121 ⇓ (4 x)^{2} + 88 x + (11)^{2}

These perfect squares show that the expression could be a perfect square trinomial. However, this is not enough to decide yet.

Confirm That the Middle Term Is Twice the Product of the Square Roots of First and Last Terms

The next step is to check whether the middle term is two times the square roots of the first and last terms.

(4 x)^{2} + 88 x + (11)^{2} ⇓ (4 x)^{2} + 2 (4 x) (11) + (11)^{2}

It can be seen that the given expression satisfies this condition as well.

Write as a Square of a Binomial

Since the expression satisfies both conditions, it is a perfect square trinomial. Therefore, it can be written as a square of a binomial where

4 x

and

11

are the first and second terms of the binomial, respectively.

16 x^{2} + 88 x + 121 = (4 x + 11)^{2}

Example

Framed Paintings

Jordan has two copies of the same square painting, and she wants to frame these copies. The area of each painting is represented by a quadratic expression.

Before preparing the frames, Jordan needs to find the side length of each painting. What are the side lengths of the paintings in terms of

x ?

Hint

Begin by determining whether the expressions are perfect square trinomials.

Solution

The side length of a square can be found by taking the square root of its area.

Side Length = A

If the given expressions are perfect square trinomials, each can be factored and written as a square of a binomial. Therefore, it first needs to be determined whether the given expressions are perfect square trinomials. To do so, begin by checking whether the first and last terms of the expressions are perfect squares.

Expression	First Term	Last Term
$49 x^{2} - 56 x + 16$	$49 x^{2} = (7 x)^{2}$	$16 = (4)^{2}$
$16 x^{2} + 24 x + 9$	$16 x^{2} = (4 x)^{2}$	$9 = (3)^{2}$

Since the first and last terms are perfect squares, the trinomials are probably perfect square trinomials. To be sure, check whether the middle terms are two times the square roots of the first and last terms.

Expression	First Term	Last Term	Middle Term
$49 x^{2} - 56 x + 16$	$49 x^{2} = (7 x)^{2}$	$16 = (4)^{2}$	$56 x = 2 \cdot 7 x \cdot 4$
$16 x^{2} + 24 x + 9$	$16 x^{2} = (4 x)^{2}$	$9 = (3)^{2}$	$24 x = 2 \cdot 4 x \cdot 3$

The expressions satisfy the conditions to be a perfect square trinomials. From here, each trinomial can be written as a square of a binomial.

49 x^{2} - 56 x + 16 16 x^{2} + 24 x + 9 = (7 x - 4)^{2} = (4 x + 3)^{2}

By taking the square roots of these expressions, the side lengths of the paintings can be found.

\underline{Side Length of Larger Painting} (7 x - 4)^{2} = 7 x - 4 \underline{Side Length of Smaller Painting} (4 x + 3)^{2} = 4 x + 3

Example

Factoring a Quadratic Expression and Analyzing Errors

Ali and Jordan are factoring the following quadratic expression.

121 m^{2} ℓ^{2} - 44 m ℓ c + 4 c^{2}

However, they obtained different results.

By factoring the given expression, determine who is correct and analyze the error.

Answer

See solution.

Hint

Determining if the trinomial is a perfect square trinomial, then factor it.

Solution

Looking at the given answers, both Ali and Jordan consider that the given expression is a perfect square trinomial because both find the square of a binomial.

Therefore, it would be a good idea to start by checking if the given expression is indeed a perfect square trinomial. To do this, check if the first and last terms of the expression are perfect squares.

121 m^{2} ℓ^{2} - 44 m ℓ c + 4 c^{2} ⇓ (11 m ℓ)^{2} - 44 m ℓ c + (2 c)^{2}

As it can be seen, the first and last terms are perfect squares. This means that the given expression could be a perfect square trinomial. To be sure, check whether the middle term is two times the square roots of the first and last terms.

(11 m ℓ)^{2} - 44 m ℓ c + (2 c)^{2} ⇓ (11 m ℓ)^{2} - 2 (11 m ℓ) (2 c) + (2 c)^{2}

Since the expression satisfies the conditions, it is a perfect square trinomial and can be written as a square of a binomial.

121 m^{2} ℓ^{2} - 44 m ℓ c + 4 c^{2} ⇓ (11 m ℓ)^{2} - 2 (11 m ℓ) (2 c) + (2 c)^{2} ⇓ (11 m ℓ - 2 c)^{2}

With this information, it can be concluded that Jordan is correct. On the other hand, Ali only took the square root of the coefficients of the first and last terms, so he did not get the correct answer.

Discussion

Factoring an Expression as a Difference of Two Squares

Note that the product of two conjugate binomials results in a difference of two squares. Therefore, the difference of two squares can be factored out using reverse thinking.

Method

Factoring a Difference of Two Squares

The product of a conjugate pair of binomials results in a difference of two squares. Using this relationship, the difference of two squares can be factored as the product of the sum and difference of two quantities.

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

As an example, the following expression will be factored.

9 x^{2} - 121

There are two steps to factor the expression as a difference of two squares.

Examine the Terms of the Expression

To factor an expression as a difference of two squares, the terms of the expression should be perfect squares.

9 x^{2} - 121 ⇓ (3 x)^{2} - (11)^{2}

As it can be seen, the terms of the expression are perfect squares.

Write as a Product of a Conjugate Pair of Binomials

Since the terms of the expression are perfect squares, it can be factored as the product of a conjugate pair of binomials. The terms of the binomials will be

3 x

and

11 .

(3 x)^{2} - (11)^{2} ⇓ (3 x - 11) (3 x + 11)

Extra

Justification of Factorization

The factored form of the expression can be expanded to justify the factorization.

(3 x - 11) (3 x + 11)

▼

Simplify

Distr

Distribute $(3 x - 11)$

3 x (3 x - 11) + 11 (3 x - 11)

Distr

Distribute $3 x$

3 x \cdot 3 x - 3 x \cdot 11 + 11 (3 x - 11)

Distr

Distribute $11$

3 x \cdot 3 x - 3 x \cdot 11 + 11 \cdot 3 x - 11 \cdot 11

Multiply

9 x^{2} - 33 x + 33 x - 121

AddTerms

Add terms

9 x^{2} - 121

Example

Writing Expressions for the Dimensions of a Rectangular Prism

The volume of a rectangular prism is given by the expression $81 x^{4} - 16 .$

Recall that the volume of a rectangular prism is the product of its dimensions. Find the possible dimensions of the prism if the dimensions are binomials with integer coefficients.

Answer

$(9 x^{2} + 4),$ $(3 x + 2),$ and $(3 x - 2)$

Hint

Factor the expression that represents the volume of the prism.

Solution

To find the dimensions of the prism, the given expression should be factored. It seems like it could be factored as a difference of two squares. To be sure, check whether the terms are perfect squares.

81 x^{4} - 16

▼

Rewrite

81 x^{4} - 16

(9 x^{2})^{2} - 4^{2}

SplitIntoFactors

Split into factors

9 \cdot 9 \cdot x^{2} \cdot x^{2} - 4 \cdot 4

ProdToPowTwoFac

$a \cdot a = a^{2}$

9^{2} \cdot (x^{2})^{2} - 4^{2}

ProdPow

$a^{m} \cdot b^{m} = (a \cdot b)^{m}$

(9 x^{2})^{2} - 4^{2}

The given expression is also equal to the square of

9 x^{2}

minus the square of

4 .

From here, the expression can be written as the product of a conjugate pair of binomials.

V = (9 x^{2})^{2} - (4)^{2} ⇓ V = (9 x^{2} + 4) (9 x^{2} - 4)

The volume of the prism can be thought of as the product of its base area and height. Notice that the binomial

9 x^{2} + 4

cannot be factored further. However, the other binomial is also a product of a conjugate pair of binomials and can therefore be further factored as a difference of two squares.

9 x^{2} - 4

▼

Rewrite

9 x^{2} - 4

(3 x)^{2} - 2^{2}

SplitIntoFactors

Split into factors

3 \cdot 3 \cdot x \cdot x - 2 \cdot 2

ProdToPowTwoFac

$a \cdot a = a^{2}$

3^{2} \cdot (x)^{2} - 2^{2}

ProdPow

$a^{m} \cdot b^{m} = (a \cdot b)^{m}$

(3 x)^{2} - 2^{2}

Therefore,

(9 x^{2} - 4)

can be written as the product of the binomials

3 x + 2

and

3 x - 2 .

81 x^{4} - 16 ↙ ↘ (9 x^{2} + 4) (9 x^{2} - 4) ↙ ↘ (9 x^{2} + 4) (3 x + 2) (3 x - 2)

The dimensions of the rectangular prism are

9 x^{2} + 4,

3 x + 2,

and

3 x - 2 .

Example

Finding the Dimensions of a Rectangular Room

Tadeo's older brother is in college. He currently lives in a one-room square apartment, from which a small square is cut out as shown in the diagram.

He wants to move to a rectangular room with the same area as his current room.

a Write an expression for the area of his current room.

b Find the possible dimensions of the rectangular room where Tadeo's brother wants to live. Assume that the dimensions of the room must be represented by binomials with integers coefficients and that its length is greater than its width.

Hint

a The area of a square is the square of its side length.

b Factor the expression that represents the area of the room.

Solution

a The area of the room is the difference between the areas of the squares.

Therefore, the area of the larger square and the area of the smaller square need to be found first. Recall that area of a square is the square of its side length.

Square	Side Length, $s$	Area, $s^{2}$
Larger Square	$4 n + 1$	$(4 n + 1)^{2}$
Smaller Square	$5$	$5^{2}$

By finding the difference between the areas, the area of the room can be found.

(4 n + 1)^{2} - 5^{2}

Note that this is an example expression. An equivalent expression can also be written by expanding the squares.

(4 n + 1)^{2} - 5^{2}

ExpandPosPerfectSquare

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

(4 n)^{2} + 2 (4 n) (1) + 1^{2} - 5^{2}

▼

Simplify

CalcPow

Calculate power

16 n^{2} + 2 (4 n) (1) + 1 - 25

Multiply

16 n^{2} + 8 n + 1 - 25

SubTerm

Subtract term

16 n^{2} + 8 n - 24

b Recall that the area of the room was found as a difference of two squares in Part A.

\underline{Area of the Room} (4 n + 1)^{2} - 5^{2}

To find the dimensions of a rectangular room with the same area as his current room, the expression needs to be factored. Note that when the difference of two squares is factored, the result is the product of a conjugate pair of binomials.

(4 n + 1)^{2} - 5^{2}

FacDiffSquares

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

(4 n + 1 + 5) (4 n + 1 - 5)

AddSubTerms

Add and subtract terms

(4 n - 4) (4 n + 6)

Therefore,

4 n - 4

and

4 n + 6

should be the dimensions of the rectangular room. Since

4 n + 6

is greater than

4 n - 4,

4 n + 6

is the length of the room.

Length : Width : 4 n + 6 4 n - 4

Example

Deciding Which Method to Use for Factoring Expressions

Two square windows of a house and their areas are shown in the given image.

Write an expression that represents the difference between the areas of the windows. Show two different ways to find the solution.

a How can the problem be solved without factoring?

b How can the factored forms of the areas be used to find the difference between the areas of the windows?

Answer

a See solution.

b See solution.

Hint

a Find the difference between the given expressions.

b Begin by factoring the given expressions. Then find the difference between the factored forms of the expressions.

Solution

a There are two different ways to find the difference in the areas of the windows. One way to find the difference is to subtract one expression from the other.

- 25 x^{2} - 30 x + 9 25 x^{2} - 14 x + 49 24 x^{2} - 16 x - 40

The resulting expression represents the difference of the areas.

b Another way to find the difference in the areas starts with factoring the given expressions. Both expressions seem like perfect square trinomials. To be sure, check whether the first and last terms of the expressions are perfect squares.

Expression	First Term	Last Term
$25 x^{2} - 30 x + 9$	$25 x^{2} = (5 x)^{2}$	$9 = (3)^{2}$
$x^{2} - 14 x + 49$	$x^{2} = (x)^{2}$	$49 = (7)^{2}$

Since the first and last terms are perfect squares, there is a good chance that these expressions are perfect square trinomials. Now, check whether the middle terms are two times the square roots of the first and last terms.

Expression	First Term	Last Term	Middle Term
$25 x^{2} - 30 x + 9$	$25 x^{2} = (5 x)^{2}$	$9 = (3)^{2}$	$30 x = 2 \cdot 5 x \cdot 3$
$x^{2} - 14 x + 49$	$x^{2} = (x)^{2}$	$49 = (7)^{2}$	$14 x = 2 \cdot x \cdot 7$

The expressions satisfy the conditions to be a perfect square trinomial. From here, they can be written as a square of a binomial.

25 x^{2} - 30 x + 9 x^{2} - 14 x + 49 = (5 x - 3)^{2} = (x - 7)^{2}

Since the expressions are in factored form, the difference in the areas can be found.

(5 x - 3)^{2} - (x - 7)^{2}

As can be seen, the resulting expression is a difference of two squares. This can be factored further and written as a product of a conjugate pair of binomials.

(5 x - 3)^{2} - (x - 7)^{2}

FacDiffSquares

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

(5 x - 3 + x - 7) (5 x - 3 - (x - 7))

Distr

Distribute $- 1$

(5 x - 3 + x - 7) (5 x - 3 - x + 7)

AddSubTerms

Add and subtract terms

(6 x - 10) (4 x + 4)

The final form of the expression can be obtained by multiplying the binomials.

(6 x - 10) (4 x + 4)

▼

Multiply parentheses

Distr

Distribute $(6 x - 10)$

4 x (6 x - 10) + 4 (6 x - 10)

Distr

Distribute $4 x$

24 x^{2} - 40 x + 4 (6 x - 10)

Distr

Distribute $4$

24 x^{2} - 40 x + 24 x - 40

AddTerms

Add terms

24 x^{2} - 16 x - 40

The result is the same as subtracting the original polynomials.

Closure

Combining the Factorization of Special Products

Throughout the lesson, two methods have been covered for factoring special products. These methods together can be used to solve the challenge presented at the beginning of the lesson. Recall that there were two algebraic expressions.

n^{2} + 4 n and (n + 2)^{2} - 4

These expressions are known to be equivalent.

a Use the figures below to illustrate why the expressions are equivalent.

Squares with 5,12,21 and 32 dots in each

b Find some ways to algebraically verify the same result.

Answer

a See solution.

b See solution.

Hint

a Let

n

be the number of the figure. Then, count the dots in each figure in terms of

n

in two different ways such that one represents

n^{2} + 4 n

and the other represents

(n + 2)^{2} - 4 .

b Begin by factoring the given expressions. Then find the difference between the factored forms of the expressions.

Solution

a Let

n

be the number of the figure such that

n = 1

for the leftmost figure.

The number of dots in each figure in terms of $n$ can be obtained in two different ways, one represented by $n^{2} + 4 n$ and the other represented by $(n + 2)^{2} - 4 .$ To illustrate $n^{2} + 4 n,$ assume that $n^{2}$ is the inside full square of dots and $4 n$ is the four outside borders with $n$ dots each.

Notice that the number of dots in the outside border and side length of the inside square, in terms of $n,$ match the figure number. Now, to illustrate $(n + 2)^{2} - 4,$ imagine the larger square with the four additional dots filled in at the corners. Then, $(n + 2)^{2}$ would be the number of dots in the larger square, since the missing $4$ dots were added.

As it can be seen, the number of dots in $n^{th}$ figure can be represented by $n^{2} + 4 n$ and by $(n + 2)^{2} - 4 .$ Therefore, these expressions are equivalent.

b Another way of showing that the expressions are equivalent is by rewriting

(n + 2)^{2} - 4

n^{2} + 4 n .

To do so,

(n + 2)^{2} - 4

can be factored as a difference of two squares.

(n + 2)^{2} - 4

WritePow

Write as a power

(n + 2)^{2} - 2^{2}

FacDiffSquares

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

(n + 2 + 2) (n + 2 - 2)

AddSubTerms

Add and subtract terms

(n + 4) (n)

Distr

Distribute $n$

n \cdot n + 4 \cdot n

Multiply

n^{2} + 4 n

As shown, the expressions are equivalent.

Level 1

Level 2

Level 3

Factoring Quadratic Expressions Using Special Products

Factor each polynomial by using the perfect square trinomial pattern and the difference of squares.

x^{a + 2} + 10 x^{a + 1} + 25 x^{a}

a^{b} x^{2 a} - a^{b} x^{2 b}

We need to rewrite the given sum as a product to factor it. Let's start by rewriting the first term by using the Product of Powers Property.

After factoring our x^a from the expression, we ended up with a trinomial inside the parentheses. If we factorize the middle term 10x and write the last term 25 as a power, the trinomial can be further factorized by using the perfect square trinomial pattern.

Once again, we will start by looking for a common factor. Since a^b is a common factor, we will first factor it out.

We will then write the expression inside the parentheses as a difference of two squares to factor it completely.

Factoring Quadratic Expressions Using Special Products

Recommended exercises

To get personalized recommended exercises, please login first.

Return to lesson.

Loading content