Factor 4n3 12n2 40n completely. Solution Ask yourself, “Can I factor out a GCF?” Yes. The GCF is 4n. 4n3 12n2 40n 4n(n2 3n 10) Look at the trinomial and ask yourself, “Can I factor again?” Yes. The integers whose product is 10 and whose sum is 3 are 5 and 2. Therefore, 4n3 12n2 40n 4n(n2 3n 10) 4n(n 5)(n 2) Ask yourself, “Can I factor again?” No. Therefore, 4n3 12n2 40n 4n(n 5)(n 2). Check: 4n(n 5)(n 2) 4n(n2 2n 5n 10) 4n(n2 3n 10) 4n3 12n2 40n ✓ YOU TRY 3 Factor completely. a) 7p4 42p3 56p2 b) 3a2b 33ab 90b 4 Factor a Trinomial Containing Two Variables If a trinomial contains two variables and we cannot take out a GCF, the trinomial may still be factored according to the method outlined in this section. Factor x2 12xy 32y2 completely. Solution Ask yourself, “Can I factor out a GCF?” No. Notice that the fi rst term is x2. Let’s rewrite the trinomial as so that we can think of 12y as the coeffi cient of x. Find two expressions whose product is 32y2 and whose sum is 12y. They are 4y and 8y since 4y 8y 32y2 and 4y 8y 12y. We cannot factor (x 4y)(x 8y) any more, so this is the complete factorization. The check is left to the student. YOU TRY 4 Factor completely. x2 12yx 32y2 x2 12xy 32y2 (x 4y)(x 8y) a) m2 10mn 16n2 b) 5a3 40a2b 45ab2 ANSWERS TO YOU TRY EXERCISES 1) a) 3, 7 b) 6, 3 c) 2, 10 2) a) (m 4)(m 7) b) (c 12)(c 4) c) prime d) (r 9)(r 5) e) (r 8)(r 3) f ) (h 6)(h 6) or (h 6)2 3) a) 7p2(p 4)(p 2) b) 3b(a 5)(a 6) 4) a) (m 2n)(m 8n) b) 5a(a b)(a 9b) EXAMPLE 3 In-Class Example 3 Factor 2z3 16z2 18z completely. Answer: 2z(z 9)(z 1) Add “Can I factor out a GCF?” as the first step in this section’s procedure. EXAMPLE 4 In-Class Example 4 Factor a2 7ab 10b2 completely. Answer: (a 2b)(a 5b) SECTION 7.2 Factoring Tr www.mhhe.com/messersmith inomials of the Form x2 bx c 403
messersmith_power_introductory_algebra_1e_ch4_7_10
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