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Algebra 1

6.6.3 Factoring the Difference of Squares

Algebra 16.6.3 Factoring the Difference of Squares

6.6.3 Factoring the Difference of Squares

6.6.3 • Factoring the Difference of Squares

Activity

Remember that ( a − b ) ( a − b ) and ( a + b ) ( a + b ) are a conjugate pair. When you multiply them together, you end up with the difference of squares, a 2 − b 2 a 2 − b 2 .

Let’s look at the pattern used to factor a difference of squares:

a 2 − b 2 = ( a − b ) ( a + b ) a 2 − b 2 = ( a − b ) ( a + b )

Work with a partner to factor using the difference of squares.

As you factor each binomial, take turns with your partner, alternating the completion of each step one at a time. If you do not know the next step, ask your partner for help.

1. 81 y 2 − 1 81 y 2 − 1

2. 196 m 2 − 25 n 2 196 m 2 − 25 n 2

3. 6 p 2 q 2 − 54 p 2 6 p 2 q 2 − 54 p 2

4. 7 a 4 c 2 − 7 b 4 c 2 7 a 4 c 2 − 7 b 4 c 2

5. a 2 + 6 a + 9 − 9 b 2 a 2 + 6 a + 9 − 9 b 2

6. x 2 + 6 x + 9 − 4 y 2 x 2 + 6 x + 9 − 4 y 2

7. Can a binomial with an addition operation be factored using the difference of squares? Explain.

Video: Factoring the Difference of Squares

Watch the following video to learn more about how to factor the difference of squares when there is also a greatest common factor.

Factoring the Difference of Squares

Self Check

Factor using the difference of squares: 121 p 2 − 9 q 2 .
  1. ( 11 p + 3 q ) 2
  2. ( 11 p − 3 q ) 2
  3. ( 11 q − 3 p ) ( 11 q + 3 p )
  4. ( 11 p − 3 q ) ( 11 p + 3 q )

Additional Resources

Factoring the Difference of Squares

The other special product you saw in the previous chapter was the Product of Conjugates pattern. You used this to multiply two binomials that were conjugates. Here’s an example:

( a − b ) ( a + b ) ( 3 x − 4 ) ( 3 x + 4 ) ( a − b ) ( a + b ) ( 3 x − 4 ) ( 3 x + 4 )

( a ) 2 − ( b ) 2 ( 3 x ) 2 − ( 4 ) 2 ( a ) 2 − ( b ) 2 ( 3 x ) 2 − ( 4 ) 2

9 x 2 − 16 9 x 2 − 16

A difference of squares pattern factors to a product of conjugates.

DIFFERENCE OF SQUARES PATTERN

If a a and b b are real numbers,

Remember, “difference” refers to subtraction. So, to use this pattern, you must make sure you have a binomial in which two squares are being subtracted.

How to factor the difference of squares:

Step 1 - Does the binomial fit the pattern?

Is this a difference?

Are the first and last terms perfect squares?

a 2 − b 2 _ _ _ − _ _ _ a 2 − b 2 _ _ _ − _ _ _

Step 2 - Write them as squares.

( a ) 2 − ( b ) 2 ( a ) 2 − ( b ) 2

Step 3 - Write the product of conjugates.

( a − b ) ( a + b ) ( a − b ) ( a + b )

Step 4 - Check by multiplying.

Example 1

Factor: 64 y 2 − 1 64 y 2 − 1 .

Step 1 - Does the binomial fit the pattern?

  • Is this a difference?
    Yes.
  • Are the first and last terms perfect squares?
    Yes.

Step 2 - Write them as the difference of squares.

a 2 − b 2 ( 8 y ) 2 − 1 2 a 2 − b 2 ( 8 y ) 2 − 1 2

Step 3 - Write the product of conjugates.

( a − b ) ( a + b ) ( 8 y − 1 ) ( 8 y + 1 ) ( a − b ) ( a + b ) ( 8 y − 1 ) ( 8 y + 1 )

Step 4 - Check by multiplying.

( 8 y − 1 ) ( 8 y + 1 ) ( 8 y − 1 ) ( 8 y + 1 )

64 y 2 − 1 ✓ 64 y 2 − 1 ✓

It is important to remember that sums of squares do not factor into a product of binomials. There are no binomial factors that multiply together to get a sum of squares. After removing any GCF, the expression a 2 + b 2 a 2 + b 2 is prime!

The next example shows variables in both terms.

Example 2

Factor: 144 x 2 − 49 y 2 144 x 2 − 49 y 2 .

Step 1 - Is this a difference of squares?

144 x 2 − 49 y 2 144 x 2 − 49 y 2

Yes.

( 12 x ) 2 − ( 7 y ) 2 ( 12 x ) 2 − ( 7 y ) 2

Step 2 - Factor as the product of conjugates.

( 12 x − 7 y ) ( 12 x + 7 y ) ( 12 x − 7 y ) ( 12 x + 7 y )

Step 3 - Check by multiplying.

( 12 x − 7 y ) ( 12 x + 7 y ) ( 12 x − 7 y ) ( 12 x + 7 y )

44 x 2 − 49 y 2 ✓ 44 x 2 − 49 y 2 ✓

As always, you should look for a common factor first whenever you have an expression to factor. Sometimes a common factor may “disguise” the difference of squares, and you won’t recognize the perfect squares until you factor out the GCF.

Also, to completely factor the binomial in the next example, we’ll factor a difference of squares twice!

Example 3

Factor: 48 x 4 y 2 − 243 y 2 48 x 4 y 2 − 243 y 2 .

Step 1 - Is there a GCF?

48 x 4 y 2 − 243 y 2 48 x 4 y 2 − 243 y 2

Yes, 3 y 2 3 y 2 —factor it out!

3 y 2 ( 16 x 4 − 81 ) 3 y 2 ( 16 x 4 − 81 )

Step 2 - Is the binomial a difference of squares? Yes.

3 y 2 [ ( 4 x 2 ) 2 − ( 9 ) 2 ] 3 y 2 [ ( 4 x 2 ) 2 − ( 9 ) 2 ]

Step 3 - Factor as a product of conjugates.

3 y 2 ( 4 x 2 − 9 ) ( 4 x 2 + 9 ) 3 y 2 ( 4 x 2 − 9 ) ( 4 x 2 + 9 )

Step 4 - Notice the first binomial is also a difference of squares!

3 y 2 [ ( 2 x ) 2 − ( 3 ) 2 ] ( 4 x 2 + 9 ) 3 y 2 [ ( 2 x ) 2 − ( 3 ) 2 ] ( 4 x 2 + 9 )

Step 5 - Factor it as the product of conjugates.

3 y 2 ( 2 x − 3 ) ( 2 x + 3 ) ( 4 x 2 + 9 ) 3 y 2 ( 2 x − 3 ) ( 2 x + 3 ) ( 4 x 2 + 9 )

The last factor, the sum of squares, cannot be factored.

Step 6 - Check by multiplying.

3 y 2 ( 2 x − 3 ) ( 2 x + 3 ) ( 4 x 2 + 9 ) 3 y 2 ( 2 x − 3 ) ( 2 x + 3 ) ( 4 x 2 + 9 )

3 y 2 ( 4 x 2 − 9 ) ( 4 x 2 + 9 ) 3 y 2 ( 4 x 2 − 9 ) ( 4 x 2 + 9 )

3 y 2 ( 16 x 4 − 81 ) 3 y 2 ( 16 x 4 − 81 )

48 x 4 y 2 − 243 y 2 ✓ 48 x 4 y 2 − 243 y 2 ✓

Example 4

Factor: x 2 − 6 x + 9 − y 2 x 2 − 6 x + 9 − y 2 .

Step 1 - Factor by grouping the first three terms.

Step 2 - Use the perfect square trinomial pattern.

( x − 3 ) 2 − y 2 ( x − 3 ) 2 − y 2

Step 3 - Is there a difference of squares?

Yes—write them as squares.

a 2 − b 2 ( x − 3 ) 2 − y 2 a 2 − b 2 ( x − 3 ) 2 − y 2

Step 4 - Factor as the product of conjugates.

( a − b ) ( a + b ) ( ( x − 3 ) − y ) ( ( x − 3 ) + y ) ( a − b ) ( a + b ) ( ( x − 3 ) − y ) ( ( x − 3 ) + y )

( x − 3 − y ) ( x − 3 + y ) ( x − 3 − y ) ( x − 3 + y )

You may want to rewrite the solution as ( x − y − 3 ) ( x + y − 3 ) ( x − y − 3 ) ( x + y − 3 ) .

Try it

Factoring the Difference of Squares

Factor using the difference of squares.

1. 121 m 2 − 1 121 m 2 − 1

2. 2 x 4 y 2 − 32 y 2 2 x 4 y 2 − 32 y 2

3. x 2 − 10 x + 25 − y 2 x 2 − 10 x + 25 − y 2

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