c) Sum is 34. 7x2 34xy 5y 2 Product: 7 (5) 35 7x2 34xy 5y The integers whose product is 35 and whose sum is 34 are 35 and 1. Rewrite the middle term, 34xy, as 35xy xy. Factor by grouping. 2 7x2 35xy xy 5y 2 7x(x 5y) y(x 5y) Take out the common factor from each group. (x 5y)(7x y) Factor out (x 5y). Check: (x 5y)(7x y) 7x2 34xy 5y2 ✓ YOU TRY 1 EXAMPLE 2 In-Class Example 2 Factor 6p2 9p 42 completely. Answer: 3(2p 7)(p 2) Write down the steps on your own paper as you are reading the example. YOU TRY 2 Factor completely. a) 4p2 16p 15 b) 10y2 13y 4 c) 5a2 29ab 6b2 Factor 12n2 64n 48 completely. Solution It is tempting to jump right in and multiply 12 (48) 576 and try to think of two integers with a product of 576 and a sum of 64. However, fi rst ask yourself, “Can I factor out a GCF?” Yes! We can factor out 4. 12n2 64n 48 4(3n2 16n 12) Factor out 4. Product: 3 (12) 36 Now factor 3n2 16n 12 by fi nding two integers whose product is 36 and whose sum is 16. The numbers are 18 and 2. 4(3n2 18n 2n 12) 43n(n 6) 2(n 6) Take out the common factor from each group. 4(n 6)(3n 2) Factor out (n 6). Check by multiplying: 4(n 6)(3n 2) 4(3n2 16n 12) 12n2 64n 48 ✓ Factor completely. a) 24h2 54h 15 b) 20d 3 38d 2 12d 2 Factor ax2 bx c (a 1) by Trial and Error At the beginning of this section, we factored 2x2 11x 15 by grouping. Now we will factor it by trial and error, which is just reversing the process of FOIL. EXAMPLE 3 In-Class Example 3 Factor 5t 2 31t 6 completely. Answer: (5t 1)(t 6) Factor 2x2 11x 15 completely. Solution Can we factor out a GCF? No. So try to factor 2x2 11x 15 as the product of two binomials. Notice that all terms are positive, so all factors will be positive. SECTION 7.3 Factoring Trinomials of t www.mhhe.com/messersmith he Form ax2 bx c (a 1) 407
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