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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 is prime!

Don’t forget that 1 is a perfect square. We’ll need to use that fact in the next example.

Factor: 64 y 2 1 .

Solution

.
Is this a difference? Yes. .
Are the first and last terms perfect squares?
Yes - write them as squares. .
Factor as the product of conjugates. .
Check by multiplying.
( 8 y 1 ) ( 8 y + 1 )
64 y 2 1

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Factor: m 2 1 .

( m 1 ) ( m + 1 )

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Factor: 81 y 2 1 .

( 9 y 1 ) ( 9 y + 1 )

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Factor: 121 x 2 49 y 2 .

Solution

121 x 2 49 y 2 Is this a difference of squares? Yes. ( 11 x ) 2 ( 7 y ) 2 Factor as the product of conjugates. ( 11 x 7 y ) ( 11 x + 7 y ) Check by multiplying. ( 11 x 7 y ) ( 11 x + 7 y ) 121 x 2 49 y 2

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Factor: 196 m 2 25 n 2 .

( 16 m 5 n ) ( 16 m + 5 n )

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Factor: 144 p 2 9 q 2 .

( 12 p 3 q ) ( 12 p + 3 q )

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The binomial in the next example may look “backwards,” but it’s still the difference of squares.

Factor: 100 h 2 .

Solution

100 h 2 Is this a difference of squares? Yes. ( 10 ) 2 ( h ) 2 Factor as the product of conjugates. ( 10 h ) ( 10 + h ) Check by multiplying. ( 10 h ) ( 10 + h ) 100 h 2

Be careful not to rewrite the original expression as h 2 100 .

Factor h 2 100 on your own and then notice how the result differs from ( 10 h ) ( 10 + h ) .

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Factor: 144 x 2 .

( 12 x ) ( 12 + x )

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Factor: 169 p 2 .

( 13 p ) ( 13 + p )

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

Factor: x 4 y 4 .

Solution

x 4 y 4 Is this a difference of squares? Yes. ( x 2 ) 2 ( y 2 ) 2 Factor it as the product of conjugates. ( x 2 y 2 ) ( x 2 + y 2 ) Notice the first binomial is also a difference of squares! ( ( x ) 2 ( y ) 2 ) ( x 2 + y 2 ) Factor it as the product of conjugates. The last ( x y ) ( x + y ) ( x 2 + y 2 ) factor, the sum of squares, cannot be factored. Check by multiplying. ( x y ) ( x + y ) ( x 2 + y 2 ) [ ( x y ) ( x + y ) ] ( x 2 + y 2 ) ( x 2 y 2 ) ( x 2 + y 2 ) x 4 y 4

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Factor: a 4 b 4 .

( a 2 + b 2 ) ( a + b ) ( a b )

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Factor: x 4 16 .

( x 2 + 4 ) ( x + 2 ) ( x 2 )

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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 the GCF.

Factor: 8 x 2 y 18 y .

Solution

8 x 2 y 98 y Is there a GCF? Yes, 2 y —factor it out! 2 y ( 4 x 2 49 ) Is the binomial a difference of squares? Yes. 2 y ( ( 2 x ) 2 ( 7 ) 2 ) Factor as a product of conjugates. 2 y ( 2 x 7 ) ( 2 x + 7 ) Check by multiplying. 2 y ( 2 x 7 ) ( 2 x + 7 ) 2 y [ ( 2 x 7 ) ( 2 x + 7 ) ] 2 y ( 4 x 2 49 ) 8 x 2 y 98 y

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Factor: 7 x y 2 175 x .

7 x ( y 5 ) ( y + 5 )

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Factor: 45 a 2 b 80 b .

5 b ( 3 a 4 ) ( 3 a + 4 )

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Factor: 6 x 2 + 96 .

Solution

6 x 2 + 96 Is there a GCF? Yes, 6—factor it out! 6 ( x 2 + 16 ) Is the binomial a difference of squares? No, it is a sum of squares. Sums of squares do not factor! Check by multiplying. 6 ( x 2 + 16 ) 6 x 2 + 96

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Factor: 8 a 2 + 200 .

8 ( a 2 + 25 )

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Factor: 36 y 2 + 81 .

9 ( 4 y 2 + 9 )

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Factor sums and differences of cubes

There is another special pattern for factoring, one that we did not use when we multiplied polynomials. This is the pattern for the sum and difference of cubes. We will write these formulas first and then check them by multiplication.

a 3 + b 3 = ( a + b ) ( a 2 a b + b 2 ) a 3 b 3 = ( a b ) ( a 2 + a b + b 2 )

We’ll check the first pattern and leave the second to you.

.
Distribute. .
Multiply. a 3 a 2 b + a b 2 + a 2 b a b 2 + b 3
Combine like terms. a 3 + b 3

Practice Key Terms 3

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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