5B. SPECIAL PRODUCTS 11 5b Special Products Special Forms In this section we revisit two special product forms that we learned in Chapter 5, the first of which was squaring a binomial. Squaring a binomial. Here are two earlier rules for squaring a binomial. 1. (a + b) 2 = a 2 +2ab + b 2 2. (a b) 2 = a 2 2ab + b 2 Now, because factoring is unmultiplying, it should be a simple matter to reverse the process of Example??. EXAMPLE 1. Factor each of the following trinomials: a) x 2 +10x +25 b)4y 2 12y +9 c)16a 2 24ab +9b 2 Solution: Because of the work already done in Example??, it is a simple task to factor each of these trinomials. a) x 2 +10x +25=(x +5) 2 b) 4y 2 12y +9=(2y 3) 2 c) 16a 2 24ab +9b 2 =(4a 3b) 2 EXAMPLE 2. Factor each of the following trinomials: a) 9x 2 42x +49 b)49a 2 +70ab +25b 2 c) 4x 2 37x +9 Solution: squares. Note that the first and last terms of each trinomial are perfect a) In the trinomial 9x 2 42x+49, note that (3x) 2 =9x 2 and 7 2 =49. Hence, the first and last terms are perfect squares. Taking the square roots, we suspect that 9x 2 42x + 49 factors as follows: 9x 2 42x +49=(3x 7) 2 However, we must check to see if the middle term is correct. Multiply 3x and 7, then double: 2(3x)(7) = 42x. Thus, the middle term is correct and we have the correct factorization of 9x 2 42x +49.
12 MODULE 5. FACTORING b) In the trinomial 49a 2 +70ab+25b 2,notethat(7a) 2 =49a 2 and (5b) 2 =25b 2. Hence, the first and last terms are perfect squares. Taking the square roots, we suspect that 49a 2 +70ab +25b 2 factors as follows: 49a 2 +70ab +25b 2 =(7a +5b) 2 However, we must check to see if the middle term is correct. Multiply 7a and 5b, then double: 2(7a)(5b) = 70ab. Thus, the middle term is correct and we have the correct factorization of 49a 2 +70ab +25b 2. c) In the trinomial 4x 2 37x +9, note that (2x) 2 =4x 2 and (3) 2 =9. Hence, the first and last terms are perfect squares. Taking the square roots, we suspect that 4x 2 37x + 9 factors as follows: 4x 2 37x +9=(2x 3) 2 However, we must check to see if the middle term is correct. Multiply 2x and 3, then double: 2(2x)(3) = 12x. However, this is not the middle term of 4x 2 37x + 9, so this factorization is incorrect! We must find another way to factor this trinomial. Comparing 4x 2 37x + 9 with ax 2 + bx + c, we need a pair of integers whose product is ac = 36 and whose sum is b = 37. The integer pair 1 and 36 comes to mind. Replace the middle term as a sum of like terms using this ordered pair. 4x 2 37x +9=4x 2 x 36x +9 37x = x 36x. = x(4x 1) 9(4x 1) Factor by grouping. =(x 9)(4x 1) Factor out 4x 1. This example clearly demonstrates how important it is to check your middle term. The first rule of factoring. problem is factor out the GCF. The first step to perform in any factoring EXAMPLE 3. Factor each of the following trinomials: a) 2x 3 y +12x 2 y 2 +18xy 3 b) 4x 5 +32x 4 64x 3 Solution: First factor out the GCF.
5B. SPECIAL PRODUCTS 13 a) In the trinomial 2x 3 y +12x 2 y 2 +18xy 3,wenotethattheGCFof2x 3 y, 12x 2 y 2,and18xy 3 is 2xy. Wefirstfactorout2xy. 2x 3 y +12x 2 y 2 +18xy 3 =2xy(x 2 +6xy +9y 2 ) We now note that the first and last terms of the resulting trinomial factor are perfect squares, so we take their square roots and try the following: =2xy(x +3y) 2 Of course, we must check that our middle term is correct. Because 2(x)(3y) = 6xy matches the middle term of x 2 +6xy +9y 2,wedohaveaperfectsquare trinomial and our result is correct. b) In the trinomial 4x 5 +32x 4 64x 3,wenotethattheGCFof4x 5,32x 4, and 64x 3 is 4x 3.Wefirstfactorout4x 3. 4x 5 +32x 4 64x 3 =4x 3 ( x 2 +8x 16) However, the first and third terms of x 2 +8x 16 are negative, and thus are not perfect squares. Let s begin again, this time factoring out 4x 3. 4x 5 +32x 4 64x 3 = 4x 3 (x 2 8x +16) This time the first and third terms of x 2 8x +16 are perfect squares. We take their square roots and write: = 4x 3 (x 4) 2 We must check that our middle term is correct. Because 2(x)(4) = 8x, we do have a perfect square trinomial and our result is correct. The Difference of Squares The second special product form we learned in Chapter 5 was the difference of squares. The difference of squares. Here is the difference of squares rule. (a + b)(a b) =a 2 b 2 Because factoring is unmultiplying, is can be a simple matter to reverse the process. EXAMPLE 4. Factor each of the following:
14 MODULE 5. FACTORING a) 9x 2 25 b) a 6 4b 6 Solution: Because of the work already done in previous examples, it is now a matter to factor (or unmultiply ) each of these problems. a) 9x 2 25 = (3x +5)(3x 5) b) a 6 4b 6 =(a 3 2b 3 )(a 3 +2b 3 ) In each case, note how we took the square roots of each term, then separated one set with a plus sign and the other with a minus sign. Because of the commutative property of multiplication, it does not matter which one you make plus and which one you make minus. Remember the first rule of factoring. The first rule of factoring. problem is factor out the GCF. The first step to perform in any factoring EXAMPLE 5. Factor: x 3 9x Solution: In x 3 9x, thegcfofx 3 and 9x is x. Factoroutx. x 3 9x = x(x 2 9) Note that x 2 9 is now the difference of two perfect squares. Take the square roots of x 2 and 9, which are x and 3, then separate one set with a plus sign and the other set with a minus sign. = x(x +3)(x 3) Factoring Completely Sometimes after one pass at factoring, factors remain that can be factored further. You must continue to factor in this case. EXAMPLE 6. Factor: x 4 16
5B. SPECIAL PRODUCTS 15 Solution: In x 4 16, we have the difference of two squares: (x 2 ) 2 = x 4 and 4 2 = 16. First, we take the square roots, then separate one set with a plus sign and the other set with a minus sign. x 4 16 = (x 2 +4)(x 2 4) Note that x 2 +4 is the sum of two squares and does not factor further. However, x 2 4 is the difference of two squares. Take the square roots, x and 2, then separate one set with a plus sign and the other set with a minus sign. =(x 2 +4)(x +2)(x 2) We cannot factor further. Revisiting Factoring by Grouping We usually factor a four-term expression by grouping. EXAMPLE 7. Factor: 2x 3 + x 2 50x 25 Solution: We factor by grouping. Factor an x 2 out of the first two terms and a 25 out of the second two terms. 2x 3 + x 2 50x 25 = x 2 (2x +1) 25(2x +1) Now we can factor out a 2x +1. =(x 2 25)(2x +1) We re still not done because x 2 25 is the difference of two squares and can be factored as follows: =(x +5)(x 5)(2x +1) Factoring Trinomials I In this section we concentrate on learning how to factor trinomials having the form ax 2 + bx + c which are not perfect squares. We must first be able to identify the coefficients a, b, and c.
16 MODULE 5. FACTORING EXAMPLE 8. Compare 3 7x +9x 2 with ax 2 + bx + c and identify the coefficients a, b, and c. Solution: Note that ax 2 + bx + c is arranged in descending powers of x, but 3 7x +9x 2 is arranged in descending powers of x. The first task is to arrange 3 7x +9x 2 in descending powers of x, then align it with ax 2 + bx + c for comparison. ax 2 + bx+c 9x 2 7x+3 Comparing, a =9,b = 7, and c =3. EXAMPLE 9. Compare 5 +x 2 +3x with ax 2 + bx + c and identify the coefficients a, b, and c. Solution: First, arrange 5+x 2 +3x in descending powers of x, then align it with ax 2 + bx + c for comparison. Note that the understood coefficient of x 2 is 1. ax 2 + bx+c 1x 2 +3x 5 Comparing, a =1,b =3,andc = 5. The ac-method We are now going to revie a technique called the ac-method (or ac-test) for factoring trinomials of the form ax 2 + bx + c. We begin with a simple example. EXAMPLE 10. Factor: x 2 +8x 48. Solution: We proceed as follows: 1. Compare x 2 +8x 48 with ax 2 + bx + c and identify a =1,b =8,and c = 48. Note that the leading coefficient is a =1. 2. Calculate ac. Notethatac =(1)( 48), so ac = 48.
5B. SPECIAL PRODUCTS 17 3. List all integer pairs whose product is ac = 48. 1, 48 1, 48 2, 24 2, 24 3, 16 3, 16 4, 12 4, 12 6, 8 6, 8 4. Circle the ordered pair whose sum is b =8. 5. Drop this ordered pair in place. 1, 48 1, 48 2, 24 2, 24 3, 16 3, 16 4, 12 4, 12 6, 8 6, 8 x 2 +8x 48 = (x 4)(x +12) 6. Use the FOIL shortcut to mentally check your answer. To multiply (x 4)(x +12),usethesesteps: Multiply the terms in the First positions: x 2. Multiply the terms in the Outer and Inner positions and combine the results mentally: 12x 4x =8x. Multiply the terms in the Last positions: 48. That is: F O I L (x 4)(x +12) = x 2 + 12x 4x 48 Combining like terms, (x 4)(x+12) = x 2 +8x 48, which is the original trinomial, so our solution checks. Speeding Things Up a Bit Some readers might already be asking Do I really have to list all of those ordered pairs if I already see the pair I need? The answer is No! If you see the pair you need, drop it in place and you are done. EXAMPLE 11. Factor: x 2 5x 24.
18 MODULE 5. FACTORING Solution: Compare x 2 5x 24 with ax 2 +bx+c and note that a =1,b = 5, and c = 24. Calculate ac =(1)( 24), so ac = 24. Now can you think of an integer pair whose product is ac = 24 and whose sum is b = 5? For some, the pair just pops into their head: 8 and 3. Drop the pair in place and you are done. x 2 5x 24 = (x 8)(x +3) Use the FOIL shortcut to check your answer. F O I L (x 8)(x +3) = x 2 + 3x 8x 24 Combining like terms, (x 8)(x+3) = x 2 5x 24, the original trinomial. Our solution checks. Note that if you combine the Outer and Inner products mentally, the check goes even faster. EXAMPLE 12. Factor: x 2 9x 36. Solution: Compare x 2 9x 36 with ax 2 +bx+c and note that a =1,b = 9, and c = 36. Calculate ac =(1)( 36), so ac = 36. Start listing the integer pairs whose product is ac = 36, but be mindful that you need an integer pair whose sum is b = 9. 1, 36 2, 18 3, 12 Note how we ceased listing ordered pairs the moment we found the pair we needed. Drop the circled pair in place. x 2 9x 36 = (x +3)(x 12) Use the FOIL shortcut to check your answer. F O I L (x +3)(x 12) = x 2 12x + 3x 36 Combining like terms, (x+3)(x 12) = x 2 9x 36, the original trinomial. Our solution checks. Note that if you combine the Outer and Inner products mentally, the check goes even faster.
5B. SPECIAL PRODUCTS 19 Factoring Trinomials II In this section we continue to factor trinomials of the form ax 2 + bx + c. In the last section, all of our examples had a = 1, and we were able to Drop in place our circled integer pair. However, in this section, a 1, and we ll soon see that we will not be able to use the Drop in place technique. However, readers will be pleased to learn that the ac-method will still apply, only with a slightly different twist. EXAMPLE 13. Factor: 2x 2 7x 15. Solution: We proceed as follows: 1. Compare 2x 2 7x 15 with ax 2 + bx + c and identify a =2,b = 7, and c = 15. Note that the leading coefficient is a = 2, so this case is different from all of the cases discussed in Section 5b. 2. Calculate ac. Notethatac =(2)( 15), so ac = 30. 3. List all integer pairs whose product is ac = 30. 1, 30 1, 30 2, 15 2, 15 3, 10 3, 10 5, 6 5, 6 4. Circle the ordered pair whose sum is b = 7. 1, 30 1, 30 2, 15 2, 15 3, 10 3, 10 5, 6 5, 6 5. Note that if we drop in place our circled ordered pair, (x+3)(x 10) 2x 2 7x 15. Right off the bat, the product of the terms in the First position does not equal 2x 2. Instead, we break up the middle term of 2x 2 7x 15 into a sum of like terms using our circled pair of integers 3and 10. 2x 2 7x 15 = 2x 2 +3x 10x 15 Now we factor by grouping. Factor an x out of the first two terms and a 5 outofthesecondtwoterms. Now we can factor out (2x +3). = x (2x +3) 5 (2x +3) =(x 5)(2x +3)
20 MODULE 5. FACTORING 6. Use the FOIL shortcut to mentally check your answer. To multiply (x 5)(2x +3),usethesesteps: Multiply the terms in the First positions: 2x 2. Multiply the terms in the Outer and Inner positions and combine the results mentally: 3x 10x = 7x. Multiply the terms in the Last positions: 15. That is: F O I L (x 5)(2x +3) = 2x 2 + 3x 10x 15 Combining like terms, (x 5)(2x +3) = 2x 2 7x 15, which is the original trinomial, so our solution checks. Speeding Things Up a Bit Some readers might already be asking Do I really have to list all of those ordered pairs if I already see the pair I need? The answer is No! If you see the pair you need, use it to break up the middle term of the trinomial as asum of like terms. EXAMPLE 14. Factor: 3x 2 7x 6. Solution: Compare 3x 2 7x 6 with ax 2 +bx+c and note that a =3,b = 7, and c = 6. Calculate ac =(3)( 6), so ac = 18. Now can you think of an integer pair whose product is ac = 18 and whose sum is b = 7? For some, the pair just pops into their head: 2 and 9. Break up the middle term into a sum of like terms using the pair 2 and 9. 3x 2 7x 6 =3x 2 +2x 9x 6 7x =2x 9x. = x (3x +2) 3 (3x +2) Factor by grouping. =(x 3)(3x +2) Factor out (3x +2). Use the FOIL shortcut to check your answer. F O I L (x 3)(3x +2) = 3x 2 + 2x 9x 6 Combining like terms, (x 3)(3x+2) = 3x 2 7x 6, the original trinomial. Our solution checks. Note that if you combine the Outer and Inner products mentally, the check goes even faster.
5B. SPECIAL PRODUCTS 21 First rule of factoring. The first step in factoring any polynomial is to factor out the greatest common factor. EXAMPLE 15. Factor: 30x 3 21x 2 18x. Solution: Note that the GCF of 30x 3,21x 2,and18x is 3x. Factor out this GCF. 30x 3 21x 2 18x = 3x 10x 2 3x 7x 3x 6 = 3x(10x 2 7x 6) Next, compare 10x 2 7x 6 with ax 2 + bx + c and note that a =10,b = 7, and c = 6. Start listing the integer pairs whose product is ac = 60, but be mindful that you need an integer pair whose sum is b = 7. 1, 60 2, 30 3, 20 4, 15 5, 12 Break up the middle term into a sum of like terms using our circled pair. 3x(10x 2 7x 6) =3x(10x 2 +5x 12x 18) 7x =5x 12x. =3x [ 5x(2x +1) 6(2x +1) ] Factor by grouping. =3x(5x 6)(2x +1) Factor out a 2x +1. Hence, 30x 3 21x 2 18x =3x(5x 6)(2x +1). Check: First, use the FOIL shortcut to multiply the two binomial factors, then distribute the monomial factor. 3x(5x 6)(2x +1)=3x(10x 2 7x 6) Apply the FOIL shortcut. =30x 3 21x 2 18x Distribute the 3x. Because this is the original polynomial, the solution checks.