By reversing the rules for multiplication of binomials from Section 4.6, we get rules for factoring polynomials in certain forms.


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1 SECTION 5.4 Special Factoring Techniques Special Factoring Techniques OBJECTIVES 1 Factor a difference of squares. 2 Factor a perfect square trinomial. 3 Factor a difference of cubes. 4 Factor a sum of cubes. By reversing the rules for multiplication of binomials from Section 4.6, we get rules for factoring polynomials in certain forms. OBJECTIVE 1 Factor a difference of squares. The formula for the product of the sum and difference of the same two terms is Factoring a Difference of Squares 1x + y21x  y2 = x 2  y 2. Reversing this rule leads to the following special factoring rule. x 2 y 2 1x y21x y2 For example, The following conditions must be true for a binomial to be a difference of squares. 1. Both terms of the binomial must be squares, such as x 2, 9y 2 =13y2 2, m 216 = m =1m + 421m = 5 2, 1 = 1 2, m 4 =1m The terms of the binomial must have different signs (one positive and one negative). EXERCISE 1 Factor each binomial if possible. x x EXAMPLE 1 Factoring Differences of Squares Factor each binomial if possible. x 2  y 2 y21x a 249 = a =1a + 721a  72 = 1x + x 28 Because 8 is not the square of an integer, this binomial does not satisfy the conditions above. It is a prime polynomial. (d) p Since p is a sum of squares, it is not equal to 1 p p Also, we use FOIL and try the following. 1 p p p p Thus, p is a prime polynomial. y2 = p 28p + 16, not p = p 2 + 8p + 16, not p y 2  m 2 =1 y + m21 y  m2 ANSWERS 1. 1x x prime CAUTION As Example 1(d) suggests, after any common factor is removed, a sum of squares cannot be factored.
2 318 CHAPTER 5 Factoring and Applications EXERCISE 2 Factor each difference of squares. 9t a 249b 2 EXAMPLE 2 Factoring Differences of Squares Factor each difference of squares. 25m 216 =15m =15m m z 264t 2 x 2 =17z2 218t2 2  =17z + 8t217z  8t2 y 2 = 1x + y21x y2 Write each term as a square. Factor the difference of squares.  NOTE Always check a factored form by multiplying. EXERCISE 3 Factor completely. 16k 264 m v EXAMPLE 3 Factor completely. 81y 236 = 919y 242 = 9313y Factoring More Complex Differences of Squares = 913y y  22 Neither binomial can be factored further. Don t stop here. m 416 p 436 =1m =1 p =1 p p 262 =1m m 242 =1m m + 221m  22 Factor out the GCF, 9. Write each term as a square. Factor the difference of squares. Write each term as a square. Factor the difference of squares. Factor the difference of squares. Factor the difference of squares again. CAUTION Factor again when any of the factors is a difference of squares, as in Example 3. Check by multiplying. ANSWERS 2. 13t t a + 7b216a  7b k + 221k m m v v + 521v  52 OBJECTIVE 2 Factor a perfect square trinomial. The expressions 144, 4x 2, and 81m 6 are called perfect squares because 144 = 12 2, 4x 2 =12x2 2, and 81m 6 =19m A perfect square trinomial is a trinomial that is the square of a binomial. For example, x 2 + 8x + 16 is a perfect square trinomial because it is the square of the binomial x + 4. x 2 + 8x + 16 =1x + 421x + 42 =1x
3 SECTION 5.4 Special Factoring Techniques 319 On the one hand, a necessary condition for a trinomial to be a perfect square is that two of its terms be perfect squares. For this reason, 16x 2 + 4x + 15 is not a perfect square trinomial, because only the term 16x 2 is a perfect square. On the other hand, even if two of the terms are perfect squares, the trinomial may not be a perfect square trinomial. For example, x 2 + 6x + 36 has two perfect square terms, x 2 and 36, but it is not a perfect square trinomial. Factoring Perfect Square Trinomials x 2 2xy y 2 1x y2 2 x 2 2xy y 2 1x y2 2 The middle term of a perfect square trinomial is always twice the product of the two terms in the squared binomial (as shown in Section 4.6). Use this rule to check any attempt to factor a trinomial that appears to be a perfect square. EXERCISE 4 Factor y y EXAMPLE 4 Factoring a Perfect Square Trinomial Factor x x The x 2 term is a perfect square, and so is 25. Try to factor x x + 25 as 1x To check, take twice the product of the two terms in the squared binomial. 2 # x # 5 = 10x Middle term of x x + 25 Twice First term Last term of binomial of binomial Since 10x is the middle term of the trinomial, the trinomial is a perfect square. x x + 25 factors as 1x EXAMPLE 5 Factoring Perfect Square Trinomials Factor each trinomial. x 222x The first and last terms are perfect squares 1121 = 11 2 or Check to see whether the middle term of x 222x is twice the product of the first and last terms of the binomial x # x #1112 = 22x Middle term of x 222x Twice First Last term term Thus, x 222x is a perfect square trinomial. x 222x factors as 1x Same sign ANSWER 4. 1 y Notice that the sign of the second term in the squared binomial is the same as the sign of the middle term in the trinomial.
4 320 CHAPTER 5 Factoring and Applications EXERCISE 5 Factor each trinomial. t 218t p 228p x 2 + 6x + 4 (d) 80x x x 9m 224m + 16 =13m m =13m Twice First Last term term 25y y + 16 The first and last terms are perfect squares. and Twice the product of the first and last terms of the binomial 5y + 4 is which is not the middle term of This trinomial is not a perfect square. In fact, the trinomial cannot be factored even with the methods of the previous sections. It is a prime polynomial. (d) 12z z z = 3z14z z y 2 =15y2 2 2 # 5y # 4 = 40y, 25y y = 4 2 Factor out the common factor, 3z. = 3z312z z z2 + 20z + 25 is a perfect square trinomial. = 3z12z Factor. NOTE 1. The sign of the second term in the squared binomial is always the same as the sign of the middle term in the trinomial. 2. The first and last terms of a perfect square trinomial must be positive, because they are squares. For example, the polynomial x 22x  1 cannot be a perfect square, because the last term is negative. 3. Perfect square trinomials can also be factored by using grouping or the FOIL method, although using the method of this section is often easier. OBJECTIVE 3 Factor a difference of cubes. We can factor a difference of cubes by using the following pattern. Factoring a Difference of Cubes x 3 y 3 1x y21x 2 xy y 2 2 ANSWERS 5. 1t p prime (d) 5x14x This pattern for factoring a difference of cubes should be memorized. To see that the pattern is correct, multiply 1x  y21x 2 + xy + y 2 2. x 2 + xy + y 2 x  y Multiply vertically. (Section 4.5) y1x 2 + xy + y x 2 y  xy 2  y 3 x 3 + x 2 y + xy 2 x1x2 + xy + y 2 2 x 3  y 3 Add.
5 SECTION 5.4 Special Factoring Techniques 321 Notice the pattern of the terms in the factored form of x 3  y 3. x 3  y 3 = (a binomial factor)(a trinomial factor) The binomial factor has the difference of the cube roots of the given terms. The terms in the trinomial factor are all positive. The terms in the binomial factor help to determine the trinomial factor. x 3  y 3 =1x  y21 positive First term product of second term squared + the terms + squared x 2 + xy + y 2 2 CAUTION The polynomial x 3  y 3 is not equivalent to 1x  y2 3. x 3  y 3 1x  y2 3 =1x  y21x 2 + xy + y 2 2 =1x  y21x  y21x  y2 =1x  y21x 22xy + y 2 2 EXERCISE 6 Factor each polynomial. a t k (d) 125x 3343y 6 ANSWERS 6. 1a  321a 2 + 3a t t t k  421k 2 + 4k (d) 15x  7y 2 2 # 125x xy y 4 2 EXAMPLE 6 Factor each polynomial. (d) Factoring Differences of Cubes m Let x = m and y = 5 in the pattern for the difference of cubes. m = m =1m  521m 2 + 5m p 327 =12p x 3 y 3 =1m  521m 2 + 5m =12p p p =12p p 2 + 6p m 332 = 41m 382 = 41m t 3216s 6 =15t2 316s = 1x  y21x p2 2 = 2 2 p 2 = 4p 2, NOT 2p 2. = 41m  221m 2 + 2m + 42 =15t  6s t t16s s =15t  6s t ts s 4 2 xy + y 2 2 8p 3 =12p2 3 and 27 = 3 3. Let x = 2p, y = 3. Let x = m, y = = 25 Apply the exponents. Multiply. Factor out the common factor, 4. 8 = 2 3 Factor the difference of cubes. Write each term as a cube. Factor the difference of cubes. Apply the exponents. Multiply.
6 322 CHAPTER 5 Factoring and Applications CAUTION A common error in factoring a difference of cubes, such as x 3  y 3 =1x  y21x 2 + xy + y 2 2, is to try to factor x 2 + xy + y 2. This is usually not possible. OBJECTIVE 4 Factor a sum of cubes. A sum of squares, such as cannot be factored by using real numbers, but a sum of cubes can. m , Factoring a Sum of Cubes x 3 y 3 1x y21x 2 xy y 2 2 Compare the pattern for the sum of cubes with that for the difference of cubes. Positive x 3  y 3 =1x  y21x 2 + xy + y 2 2 Difference of cubes Same sign Opposite sign Positive The only difference between the patterns is the positive and negative signs. x 3 + y 3 =1x + y21x 2  xy + y 2 2 Sum of cubes Same sign Opposite sign EXERCISE 7 Factor each polynomial. x a 3 + 8b 3 ANSWERS 7. 1x + 521x 25x a + 2b219a 26ab + 4b 2 2 EXAMPLE 7 Factoring Sums of Cubes Factor each polynomial. k = k = 3 3 =1k + 321k 23k Factor the sum of cubes. =1k + 321k 23k + 92 Apply the exponent. 8m n 3 =12m n2 3 8m 3 =12m2 3 and 125n 3 =15n2 3. =12m + 5n2312m2 22m15n2 +15n2 2 4 Factor the sum of cubes. =12m + 5n214m 210mn + 25n a b 3 =110a b2 3 Be careful: 12m2 2 = 2 2 m 2 and 15n2 2 = 5 2 n 2. =110a 2 + 3b23110a a 2 213b2 +13b2 2 4 Factor the sum of cubes. =110a 2 + 3b21100a 430a 2 b + 9b a = a = 100a 4
7 SECTION 5.4 Special Factoring Techniques 323 The methods of factoring discussed in this section are summarized here. Special Factorizations Difference of squares x 2 y 2 1x y21x y2 Perfect square trinomials x 2 2xy y 2 1x y2 2 x 2 2xy y 2 1x y2 2 Difference of cubes x 3 y 3 1x y21x 2 xy y 2 2 Sum of cubes x 3 y 3 1x y21x 2 xy y 2 2 The sum of squares can be factored only if the terms have a common factor. 5.4 EXERCISES Complete solution available on the Video Resources on DVD 1. Concept Check To help you factor the difference of squares, complete the following list of squares. 1 2 = 6 2 = 11 2 = 16 2 = 17 2 = 18 2 = 19 2 = 20 2 = 2. Concept Check The following powers of x are all perfect squares: x 2, x 4, x 6, x 8, x 10. On the basis of this observation, we may make a conjecture (an educated guess) that if the power of a variable is divisible by (with 0 remainder), then we have a perfect square. 3. Concept Check To help you factor the sum or difference of cubes, complete the following list of cubes. 1 3 = 2 2 = 7 2 = 12 2 = 2 3 = 3 2 = 8 2 = 13 2 = 3 3 = 4 2 = 9 2 = 14 2 = 4 3 = 5 2 = 10 2 = 15 2 = 5 3 = 6 3 = 7 3 = 8 3 = 9 3 = 10 3 = 4. Concept Check The following powers of x are all perfect cubes: x 3, x 6, x 9, x 12, x 15. On the basis of this observation, we may make a conjecture that if the power of a variable is divisible by (with 0 remainder), then we have a perfect cube. 5. Concept Check Identify each monomial as a perfect square, a perfect cube, both of these, or neither of these. 64x 6 y t 6 49x 12 (d) 81r Concept Check What must be true for x n to be both a perfect square and a perfect cube? Factor each binomial completely. If the binomial is prime, say so. Use your answers from Exercises 1 and 2 as necessary. See Examples y t x x m k m x r x x a 28
8 324 CHAPTER 5 Factoring and Applications p q r 225a m 2100p x w p r x y 410, p k Concept Check When a student was directed to factor k 481 from Exercise 30 completely, his teacher did not give him full credit for the answer 1k k The student argued that since his answer does indeed give k 481 when multiplied out, he should be given full credit. WHAT WENT WRONG? Give the correct factored form. 32. Concept Check The binomial 4x is a sum of squares that can be factored. How is this binomial factored? When can the sum of squares be factored? Concept Check Find the value of the indicated variable. 33. Find b so that x 2 + bx + 25 factors as 1x Find c so that 4m 212m + c factors as 12m Find a so that ay 212y + 4 factors as 13y Find b so that 100a 2 + ba + 9 factors as 110a Factor each trinomial completely. See Examples 4 and w 2 + 2w p 2 + 4p x 28x x 210x x x y y x 240x y 260y x 228xy + 4y z 212zw + 9w x xy + 9y t tr + 16r h 240hy + 8y x 248xy + 32y k 34k 2 + 9k 52. 9r 36r r z 4 + 5z 3 + z x 4 + 2x 3 + x 2 Factor each binomial completely. Use your answers from Exercises 3 and 4 as necessary. See Examples 6 and a m m b k p x y p y w 3216z x x y 38x x 316y w 3216z p q x y a b m 3 + 8p t 3 + 8s r s x 3125y t 364s m 6 + 8n r s x 9 + y x 9  y 9
9 Summary Exercises on Factoring 325 Although we usually factor polynomials using integers, we can apply the same concepts to factoring using fractions and decimals. z = z 2  a 3 4 b = A3 4B 2 = az baz b Factor the difference of squares. Apply the special factoring rules of this section to factor each binomial or trinomial. 83. p 84. q b x y t 90. m x 21.0x m t y 93. x y Brain Busters Factor each polynomial completely. x m + n2 21m  n a  b2 31a + b m 2  p 2 + 2m + 2p 98. 3r  3k + 3r 23k 2 36m PREVIEW EXERCISES Solve each equation. See Sections 2.1 and m  4 = t + 2 = t + 10 = x = 0 SUMMARY EXERCISES on Factoring As you factor a polynomial, ask yourself these questions to decide on a suitable factoring technique. Factoring a Polynomial 1. Is there a common factor? If so, factor it out. 2. How many terms are in the polynomial? Two terms: Check to see whether it is a difference of squares or a sum or difference of cubes. If so, factor as in Section 5.4. Three terms: Is it a perfect square trinomial? If the trinomial is not a perfect square, check to see whether the coefficient of the seconddegree term is 1. If so, use the method of Section 5.2. If the coefficient of the seconddegree term of the trinomial is not 1, use the general factoring methods of Section 5.3. Four terms: Try to factor the polynomial by grouping, as in Section Can any factors be factored further? If so, factor them. (continued)
10 326 CHAPTER 5 Factoring and Applications Match each polynomial in Column I with the best choice for factoring it in Column II. The choices in Column II may be used once, more than once, or not at all x x x 217x + 72 I 3. 16m 2 n + 24mn  40mn a 2121b p 260pq + 25q 2 6. z 24z r x 6 + 4x 43x w z 224z Factor each polynomial completely. II A. Factor out the GCF. No further factoring is possible. B. Factor a difference of squares. C. Factor a difference of cubes. D. Factor a sum of cubes. E. Factor a perfect square trinomial. F. Factor by grouping. G. Factor out the GCF. Then factor a trinomial by grouping or trial and error. H. Factor into two binomials by finding two integers whose product is the constant in the trinomial and whose sum is the coefficient of the middle term. I. The polynomial is prime. 11. a 24a a a y 26y y y 5168y a + 12b + 18c 16. m 23mn  4n p 217p z 26z + 7z z 27z m 210m x 3 y xy y a 58a 448a k 210k z 23za  10a z x 24x  5x n 2 r nr 350n 2 r 29. 6n 219n y y x m 2 + 2m y 25y m z z x x k 212k p p m 224z m 22m k 2 + 4k a 3 b 560a 4 b a 6 b k 3 + 7k 270k r  5s  rs 45. y y 530y m  16m k z y 2  y k p 1045p 9252p m m m m r rm + 9m z 212z h hg  14g z 345z z 59. k 211k p 2100m 2
11 Summary Exercises on Factoring k 312k 215k 62. y 24yk  12k p r m + 2p + mp 66. 2m 2 + 7mn  15n z 28z m 4400m 3 n + 195m 2 n m 236m a 281y x 2  xy + y y z z 216z m m m + 12n + 3mn q  6p + 3pq 77. 6a a y 642y 5120y a 3  b 3 + 2a  2b k 248k m 280mn + 25n y 3 z y 224y 4 z k 22kh  3h a 27a x a y 27yz  6z m 24m a ab  3b a RELATING CONCEPTS EXERCISES FOR INDIVIDUAL OR GROUP WORK A binomial may be both a difference of squares and a difference of cubes. One example of such a binomial is x 61. With the techniques of Section 5.4, one factoring method will give the completely factored form, while the other will not. Work Exercises in order to determine the method to use if you have to make such a decision. 91. Factor x 61 as the difference of squares. 92. The factored form obtained in Exercise 91 consists of a difference of cubes multiplied by a sum of cubes. Factor each binomial further. 93. Now start over and factor x 61 as the difference of cubes. 94. The factored form obtained in Exercise 93 consists of a binomial that is a difference of squares and a trinomial. Factor the binomial further. 95. Compare your results in Exercises 92 and 94. Which one of these is factored completely? 96. Verify that the trinomial in the factored form in Exercise 94 is the product of the two trinomials in the factored form in Exercise Use the results of Exercises to complete the following statement: In general, if I must choose between factoring first with the method for the difference of squares or the method for the difference of cubes, I should choose the method to eventually obtain the completely factored form. 98. Find the completely factored form of x by using the knowledge you gained in Exercises
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