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Course: Digital SAT Math > Unit 12

Lesson 1: Factoring quadratic and polynomial expressions: advanced

Factoring quadratic and polynomial expressions | Lesson

A guide to factoring quadratic and polynomial expressions on the digital SAT

What are factoring quadratic and polynomial expressions questions?

Factoring quadratic and polynomial expressions questions ask you to rewrite polynomials in their equivalent, factored form.

For example,

x^{2} + 3 x + 2

can be rewritten as

(x + 1) (x + 2)

, and

x^{2} - 9

can be rewritten as

(x - 3) (x + 3)

In this lesson, we'll learn to:

Factor polynomial expressions
Use knowledge of factoring to evaluate expressions

You can learn anything. Let's do this!

How do I factor quadratic expressions?

Factoring quadratic expressions

Note: We cover how to factor quadratic expressions in the form

x^{2} + b x + c

in detail in the Solving quadratic equations lesson.

Quadratic expressions that have an

x^{2}

-coefficient that is not

1

are more difficult to factor. We should try to factor out any common factors if possible.

For example, in the expression

3 x^{2} + 12 x - 15

, we can factor out a

3

first and then factor the quadratic expression

x^{2} + 4 x - 5

To factor a quadratic expression in the form

x^{2} + b x + c

Find two numbers with a product equal to $c$ ‍ and a sum equal to $b$ ‍.
The two factors of the expression are each the sum of $x$ ‍ and one of the numbers from Step 1.

Example: Factor

x^{2} + 9 x - 10

When we can't factor out an integer, we can use a technique called factoring by grouping.

Factoring quadratics by grouping

If you'd like to review this technique, we recommend watching this video before proceeding!

Khan Academy video wrapper

Factoring quadratics by groupingSee video transcript

How do we factor by grouping?

The first step of factoring

a x^{2} + b x + c

is familiar: we're looking for two integers with a product equal to

a c

and a sum equal to

b

For example, let's factor

2 x^{2} + 9 x - 5

We are looking for two numbers that meet the following criteria:

Their product is equal to $(2) (- 5) = - 10$ ‍.
Their sum is equal to $9$ ‍.

The numbers

10

and

- 1

work:

$(10) (- 1) = - 10$ ‍
$10 + (- 1) = 9$ ‍

These two numbers alone do not give us the factors. Instead, they tell us how to split up the

x

-term of the expression. We can rewrite

9 x

10 x - x

2 x^{2} + 9 x - 5 = 2 x^{2} + 10 x - x - 5

Next, we group the terms into two pairs:

2 x^{2} + 10 - x - 5

Notice that we can factor each pair of terms:

\begin{aligned} 2 x^{2} + 10 x - x - 5 \\ = 2 x (x + 5) - 1 (x + 5) \end{aligned}

When we do this, we see that both of the initial pairs contain the factor

x + 5

. This means we can factor out the

x + 5

from the overall expression:

\begin{aligned} 2 x (x + 5) - 1 (x + 5) \\ = x + 5 \cdot (\frac{2 x (x + 5)}{x + 5} - \frac{1 (x + 5)}{x + 5}) \\ = (x + 5) (2 x - 1) \end{aligned}

Therefore,

2 x^{2} + 9 x - 5 = (x + 5) (2 x - 1)

Note: Factoring by grouping can be difficult, and there are often other strategies that can get you to the answer on test day. For example, you might be able to simply test multiple choice options by using FOIL, or plug in simple values for

x

to find a match!

To factor a quadratic expression in the form

a x^{2} + b x + c

Factor out any integers if possible. If this results in the product of an integer and a quadratic expression in the form $x^{2} + b x + c$ ‍, follow the steps for factoring $x^{2} + b x + c$ ‍ shown above.
Find two numbers with a product equal to $a c$ ‍ and a sum equal to $b$ ‍.
Use the two numbers from Step 2 to split $b x$ ‍ into two $x$ ‍-terms.
Group the resulting expression into two pairs of terms: one pair should have an $x^{2}$ ‍-term and an $x$ ‍-term, and the other pair should have an $x$ ‍-term and a constant term.
Factor out an expression containing $x$ ‍ from the pair with an $x^{2}$ ‍-term and an $x$ ‍-term. Factor out a constant from the pair with an $x$ ‍-term and a constant term. These two pairs should now share a binomial factor.
The shared binomial factor is one factor of the quadratic expression. The expression containing $x$ ‍ and the constant factored out in Step 5 combine to form the other factor of the quadratic expression.

Example: Factor

6 x^{2} - 7 x - 3

We're looking for two numbers,

a

and

b

, that meet the following criteria:

$a b$ ‍ is equal to the product of the coefficient of $x^{2}$ ‍ and the constant term, $(6) (- 3) = - 18$ ‍.
$a + b$ ‍ is equal to the coefficient of $x$ ‍, $- 7$ ‍.

a = - 9

and

b = 2

would work:

$(- 9) (2) = - 18$ ‍
$- 9 + 2 = - 7$ ‍

Now, we can use

a

and

b

to rewrite

6 x^{2} - 7 x - 3

and factor by grouping:

\begin{aligned} 6 x^{2} - 7 x - 3 & = 6 x^{2} - 9 x + 2 x - 3 \\ = 3 x (2 x - 3) + 1 (2 x - 3) \\ = (3 x + 1) (2 x - 3) \end{aligned}

Notice how the two groups,

6 x^{2} - 9 x

and

2 x - 3

, share the binomial factor

2 x - 3

. The other factors of the two groups,

3 x

and

1

, combine to form the other binomial factor

3 x + 1

6 x^{2} - 7 x - 3 = (3 x + 1) (2 x - 3)

Try it!

Try: factor a quadratic expression

4 x^{2} + 25 x + 6

To factor the quadratic equation above, first we need to find two numbers with a product equal to the product of the

x^{2}

-coefficient and the constant term,

4 \cdot 6 = 24

, and with a sum equal to the

x

-coefficient,

25

These two numbers are

Next, we use the two numbers to rewrite

4 x^{2} + 25 x + 6

4 x^{2} + 24 x + x + 6

The greatest common factor of

4 x^{2}

and

24 x

This means we can rewrite

4 x^{2} + 24 x + x + 6

as:

4 x (x + 6) + 1 (x + 6)

In the form of the quadratic expression above,

x + 6

is a common factor. This

x + 6

is a factor of

4 x^{2} + 25 + 6

, and the other factor is

How do I use special factoring?

Factor difference of squares: leading coefficient $\neq 1$ ‍

Khan Academy video wrapper

Factoring difference of squares: leading coefficient ≠ 1See video transcript

Special factoring

Special factoring rules are shortcuts for factoring polynomials with specific combinations of terms. Many students find that when they recognize opportunities to apply special factoring rules, they save valuable time on test day.

Anytime we see multiple perfect square terms in a polynomial expression, e.g.,

9

4 x^{2}

, check the other terms to see whether the expression satisfies the criteria for special factoring.

Square of sum:

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

Square of difference:

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

Difference of squares:

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

To use special factoring to factor a polynomial expression:

Factor out any common factors if possible.
Recognize that one or more terms in the expression are perfect squares.
Confirm that all of the terms in the expression satisfy the criteria for special factoring.
Apply the appropriate special factoring rule.

Let's look at some examples!

Factor

4 x^{2} + 12 x + 9

Notice that both

4 x^{2}

and

9

are perfect square terms:

4 x^{2} = (2 x)^{2}

, and

9 = 3^{2}

. The square of sum special factoring tells us that

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

. For

a = 2 x

and

b = 3

2 a b = 2 (2 x) (3) = 12 x

Since

2 a b = 12 x

, the

x

-term of the expression satisfies the criteria for special factoring, and we can use the square of sum special factoring to rewrite the expression:

4 x^{2} + 12 x + 9 = (2 x + 3)^{2}

We can check our work by squaring

2 x + 3

\begin{aligned} (2 x + 3)^{2} & = (2 x + 3) (2 x + 3) \\ = (2 x) (2 x) + (2 x) (3) + (3) (2 x) + (3) (3) \\ = 4 x^{2} + 6 x + 6 x + 9 \\ = 4 x^{2} + 12 x + 9 \end{aligned}

Notice how when we're FOILing, the products of the outer terms and the inner terms are both equal to

(2 x) (3)

, or

a b

. When adding the two together, we get

a b + a b = 2 a b

$4 x^{2} + 12 x + 9 = (2 x + 3)^{2}$ ‍

Factor

9 y^{4} - 25 x^{2}

Notice that both

9 y^{4}

and

25 x^{2}

are perfect square terms:

9 y^{4} = (3 y^{2})^{2}

, and

25 x^{2} = (5 x)^{2}

. Since the expression is the difference of two terms, we can apply the difference of squares special factoring. For

a = 3 y^{2}

and

b = 5 x

\begin{aligned} a^{2} - b^{2} & = (a + b) (a - b) \\ 9 y^{4} - 25 x^{2} & = (3 y^{2} + 5 x) (3 y^{2} - 5 x) \end{aligned}

We can check our work by multiplying

3 y^{2} + 5 x

and

3 y^{2} - 5 x

\begin{aligned} (3 y^{2} + 5 x) (3 y^{2} - 5 x) \\ = (3 y^{2}) (3 y^{2}) + (3 y^{2}) (- 5 x) + (5 x) (3 y^{2}) + (5 x) (- 5 x) \\ = 9 y^{4} + 15 x y^{2} - 15 x y^{2} - 25 x^{2} \\ = 9 y^{4} - 25 x^{2} \end{aligned}

Notice how when we're FOILing, the products of the outer terms and the inner terms are

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

and

(5 x) (3 y^{2})

, or

- a b

and

a b

. When adding the two together, we get

- a b + a b = 0

$9 y^{4} - 25 x^{2} = (3 y^{2} + 5 x) (3 y^{2} - 5 x)$ ‍

Try it!

Try: use special factoring to evaluate an expression

x + y = 8

and

x - y = 2

, what is the value of

(x^{2} - y^{2}) (x^{2} - 2 x y + y^{2})

x^{2} - y^{2}

is a difference of squares. This means we can factor the expression into

x^{2} - 2 x y + y^{2}

can be factored into a square of difference, or

Since

x + y = 8

and

x - y = 2

, we can plug in

8

for each

x + y

and

2

for each

x - y

. This means the value of

(x^{2} - y^{2}) (x^{2} - 2 x y + y^{2})

is equal to

, or

Your turn!

Practice: factor a quadratic expression

8 x^{2} - 5 x - 22 = (a x + 11) (b x - 2)

, where

a

and

b

are constants, what is the value of

a - b

Practice: use special factoring

Which of the following is equivalent to

4 x^{2} + 28 x + 49

Practice: use special factoring

Which of the following is equivalent to

2 x^{5} - 8 x

Things to remember

Square of sum:

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

Square of difference:

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

Difference of squares:

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

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Digital SAT Math

Course: Digital SAT Math > Unit 12

What are factoring quadratic and polynomial expressions questions?

How do I factor quadratic expressions?

Factoring quadratic expressions

Factoring quadratics by grouping

How do we factor by grouping?

Try it!

How do I use special factoring?

Factor difference of squares: leading coefficient ≠1‍

Special factoring

Let's look at some examples!

Try it!

Your turn!

Things to remember

Want to join the conversation?

Factor difference of squares: leading coefficient $\neq 1$ ‍