# Use the following strategy to factor a difference or

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Chapter 1 / Exercise 41
College Algebra
Larson
Expert Verified
Use the following strategy to factor a difference or sum of two cubes. Factoring a Difference or Sum of Two Cubes a 3 b 3 ( a b ) ( a 2 ab b 2 ) a 3 b 3 ( a b ) ( a 2 ab b 2 ) Strategy for Factoring a 3 b 3 or a 3 b 3 1. The first factor is the original polynomial without the exponents, and the middle term in the second factor has the opposite sign from the first factor: a 3 b 3 ( a b )( a 2 ab b 2 ) a 3 b 3 ( a b )( a 2 ab b 2 ) opposite signs opposite signs 2. Recall the two perfect square trinomials a 2 2 ab b 2 and a 2 – 2 ab b 2 . The second factor is almost a perfect square trinomial. Just delete the 2.
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Chapter 1 / Exercise 41
College Algebra
Larson
Expert Verified
In Example 1, we used the first three perfect cubes, 1, 8, and 27. You should verify that 1, 8, 27, 64, 125, 216, 343, 512, 729, and 1000 are the first 10 perfect cubes. The polynomial ( a b ) 3 is not equivalent to a 3 b 3 because if a 2 and b 1, then ( a b ) 3 (2 1) 3 1 3 1 and a 3 b 3 2 3 1 3 8 1 7. Likewise, ( a b ) 3 is not equivalent to a 3 b 3 . U 2 V Factoring a Difference of Two Fourth Powers A difference of two fourth powers of the form a 4 b 4 is also a difference of two squares, ( a 2 ) 2 ( b 2 ) 2 . It can be factored by the rule for factoring a difference of two squares: a 4 b 4 ( a 2 ) 2 ( b 2 ) 2 Write as a difference of two squares. ( a 2 b 2 )( a 2 b 2 ) Difference of two squares ( a b )( a b ) ( a 2 b 2 ) Factor completely. Note that the sum of two squares a 2 b 2 is prime and cannot be factored. CAUTION 356 Chapter 5 Factoring 5-36 E X A M P L E 2 Factoring a difference of two fourth powers Factor each polynomial completely. a) x 4 16 b) 81 m 4 n 4 E X A M P L E 1 Factoring a difference or sum of two cubes Factor each polynomial. a) w 3 8 b) x 3 1 c) 8 y 3 27 Solution a) Because 8 2 3 , w 3 8 is a difference of two cubes. To factor w 3 8, let a w and b 2 in the formula a 3 b 3 ( a b ) ( a 2 ab b 2 ) : w 3 8 ( w 2) ( w 2 2 w 4 ) b) Because 1 1 3 , the binomial x 3 1 is a sum of two cubes. Let a x and b 1 in the formula a 3 b 3 ( a b )( a 2 ab b 2 ): x 3 1 ( x 1) ( x 2 x 1 ) c) 8 y 3 27 (2 y ) 3 3 3 This is a difference of two cubes. (2 y 3) ( 4 y 2 6 y 9 ) Let a 2 y and b 3 in the formula. Now do Exercises 1–16
We will use the factoring strategy in Example 3. 5-37 5.5 Difference and Sum of Cubes and a Strategy 357 Strategy for Factoring Polynomials Completely 1. Factor out the GCF (with a negative coefficient if necessary). 2. When factoring a binomial, check to see whether it is a difference of two squares, a difference of two cubes, or a sum of two cubes. A sum of two squares does not factor. 3. When factoring a trinomial, check to see whether it is a perfect square trinomial. 4. If the polynomial has four terms, try factoring by grouping. 5. When factoring a trinomial that is not a perfect square, use the ac method or the trial-and-error method. 6. Check to see whether any of the factors can be factored again. E X A M P L E 3 Factoring polynomials Factor each polynomial completely.