Lesson Objectives

- Demonstrate an understanding of special products
- Learn how to factor a difference of squares
- Learn how to factor a perfect square trinomial
- Learn how to factor a sum or difference of cubes

## How to Factor Special Products

A few lessons ago, we studied special polynomial products. We learned that certain binomials can be multiplied together quickly using a pattern. In this lesson, we will reverse this process and show a quicker method to factor polynomials in certain forms.

(x + 5)(x - 5) = x

When we see a pattern that matches this format, we want to reverse the process.

x

Let's look at an example.

Example 1: Factor each.

x

We can think of this problem as:

x

Once we identify that we have the difference of squares, we can use the pattern to factor. Set up two sets of parentheses:

( )( )

Since x

(x )(x )

The signs need to alternate. One will be (+) and one will be (-):

(x + )(x - )

In our last position of each, we will place a 10:

x

Example 2: Factor each.

4x

We can think of this problem as:

(2x)

Once we identify that we have the difference of squares, we can use the pattern to factor. Set up two sets of parentheses:

( )( )

Since 4x

(2x )(2x )

The signs need to alternate. One will be (+) and one will be (-):

(2x + )(2x - )

In our last position of each, we will place a 5y:

4x

(x + y)

(x - y)

We obtain the first term squared, 2 times the first term and the second term, and the last term squared. The only difference between the two formulas is the sign of the middle term. If we have a perfect square trinomial and the middle term is positive, our trinomial will factor into a sum. If we have a perfect square trinomial and the middle term is negative, our trinomial will factor into a difference. Using this formula, it is easy to go backward. Let's look at an example.

Example 3: Factor each.

x

We could rewrite this as:

x

Once we know that we have a perfect square trinomial, we can then factor this as:

(x + 1)

Example 4: Factor each.

25x

We could rewrite this as:

(5x)

Once we know that we have a perfect square trinomial, we can then factor this as:

(5x - 2)

x

x

Let's take a look at some examples.

Example 5: Factor each.

27x

We could rewrite this as:

(3x)

Once we know that we have the sum of cubes, we follow the pattern:

(3x + 4)((3x)

(3x + 4)(9x

27x

Example 6: Factor each.

216x

We could rewrite this as:

(6x)

Once we know that we have the sum of cubes, we follow the pattern:

(6x - 5)((6x)

(6x - 5)(36x

216x

### Factoring the difference of squares

We previously learned that multiplying conjugates (the sum and difference of the same two terms) results in the first term squared minus the second term squared.(x + 5)(x - 5) = x

^{2}- 25When we see a pattern that matches this format, we want to reverse the process.

x

^{2}- y^{2}= (x + y)(x - y)Let's look at an example.

Example 1: Factor each.

x

^{2}- 100We can think of this problem as:

x

^{2}- 10^{2}Once we identify that we have the difference of squares, we can use the pattern to factor. Set up two sets of parentheses:

( )( )

Since x

^{2}is first, place an x in the first position of each set of parentheses:(x )(x )

The signs need to alternate. One will be (+) and one will be (-):

(x + )(x - )

In our last position of each, we will place a 10:

x

^{2}- 100 = (x + 10)(x - 10)Example 2: Factor each.

4x

^{2}- 25y^{2}We can think of this problem as:

(2x)

^{2}- (5y)^{2}Once we identify that we have the difference of squares, we can use the pattern to factor. Set up two sets of parentheses:

( )( )

Since 4x

^{2}is first, place a 2x in the first position of each set of parentheses:(2x )(2x )

The signs need to alternate. One will be (+) and one will be (-):

(2x + )(2x - )

In our last position of each, we will place a 5y:

4x

^{2}- 25y^{2}= (2x + 5y)(2x - 5y)### Factoring a Perfect Square Trinomials

When we square a binomial, we end up with a perfect square trinomial. Let's recall our formula:(x + y)

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

^{2}= x^{2}- 2xy + y^{2}We obtain the first term squared, 2 times the first term and the second term, and the last term squared. The only difference between the two formulas is the sign of the middle term. If we have a perfect square trinomial and the middle term is positive, our trinomial will factor into a sum. If we have a perfect square trinomial and the middle term is negative, our trinomial will factor into a difference. Using this formula, it is easy to go backward. Let's look at an example.

Example 3: Factor each.

x

^{2}+ 2x + 1We could rewrite this as:

x

^{2}+ 2 • x • 1 + 1^{2}Once we know that we have a perfect square trinomial, we can then factor this as:

(x + 1)

^{2}Example 4: Factor each.

25x

^{2}- 20x + 4We could rewrite this as:

(5x)

^{2}- 2 • (5x) • 2 + 2^{2}Once we know that we have a perfect square trinomial, we can then factor this as:

(5x - 2)

^{2}### Factoring the Sum or Difference of Cubes

When we come across the sum or difference of cubes, we can use a special factoring pattern:x

^{3}+ y^{3}= (x + y)(x^{2}- xy + y^{2})x

^{3}- y^{3}= (x - y)(x^{2}+ xy + y^{2})Let's take a look at some examples.

Example 5: Factor each.

27x

^{3}+ 64We could rewrite this as:

(3x)

^{3}+ 4^{3}Once we know that we have the sum of cubes, we follow the pattern:

(3x + 4)((3x)

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

^{2}- 12x + 16)27x

^{3}+ 64 = (3x + 4)(9x^{2}- 12x + 16)Example 6: Factor each.

216x

^{3}- 125We could rewrite this as:

(6x)

^{3}- (5)^{3}Once we know that we have the sum of cubes, we follow the pattern:

(6x - 5)((6x)

^{2}+ (6x)(5) + 5^{2}) =(6x - 5)(36x

^{2}+ 30x + 25)216x

^{3}- 125 = (6x - 5)(36x^{2}+ 30x + 25)#### Skills Check:

Example #1

Factor each. $$324x^{2}- 720x + 400$$

Please choose the best answer.

A

$$4(2x + 7)(2x - 7)$$

B

$$4(9x + 2)^{2}$$

C

$$4(9x - 10)^{2}$$

D

$$4(8x + 3)(8x - 3)$$

E

$$4(2x - 7)^{2}$$

Example #2

Factor each. $$125x^{2}- 320$$

Please choose the best answer.

A

$$5(-5x + 8)(5x - 8)$$

B

$$5(5x + 8)(5x - 8)$$

C

$$5(10x + 7)(10x - 7)$$

D

$$5(8x + 7)^{2}$$

E

$$5(8x - 1)^{2}$$

Example #3

Factor each. $$6x^{3}- 162y^{3}$$

Please choose the best answer.

A

$$3(x + 14)(x - 14)$$

B

$$14(x - 6)^{2}$$

C

$$3(x - 3y)(x^{2}+ 3xy + 9y^{2})$$

D

$$(x - 6y)(x^{2}+ 18xy + 27y^{2})$$

E

$$6(x - 3y)(x^{2}+ 3xy + 9y^{2})$$

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