Definition/Procedure Example Steps for Solving a Quadratic Equation by Factoring 1) Write the equation in the form ax2 bx c 0. 2) Factor the expression. 3) Set each factor equal to zero, and solve for the variable. 4) Check the answer(s). (p. 394) Solve 4m2 11 m2 2m 3. 3m2 2m 8 0 Standard form (3m 4)(m 2) 0 Factor. b R 3m 4 0 or m 2 0 3m 4 m 4 3 or m 2 The solution set is e 4 3 , 2f . Check the answers. 7.5 Applications of Quadratic Equations Pythagorean Theorem Given a right triangle with legs of length a and b and hypotenuse of length c, c b a the Pythagorean theorem states that a2 b2 c2 (p. 405) Find the length of side a. 5 a 4 Let b 4 and c 5 in a2 b2 c2. a2 (4)2 (5)2 a2 16 25 a2 9 0 (a 3)(a 3) 0 b R a 3 0 or a 3 0 a 3 or a 3 Reject 3 as a solution because the length of a side cannot be negative. Therefore, the length of side a is 3. (7.1) Find the greatest common factor of each group of terms. 1) 18, 27 2) 56, 80, 24 3) 33p5q3, 121p4q3, 44p7q4 4) 42r4s3, 35r2s6, 49r2s4 Factor out the greatest common factor. 5) 48y 84 6) 30a4 9a 7) 7n5 21n4 7n3 8) 72u3v3 42u3v2 24uv 9) a(b 6) 2(b 6) 10) u(13w 9) v(13w 9) Factor by grouping. 11) mn 2m 5n 10 12) jk 7j 5k 35 13) 5qr 10q 6r 12 14) cd2 5c d2 5 15) Factor out 4x from 8x3 12x2 4x. 16) Factor out 1 from r2 6r 2. (7.2) Factor completely. 17) p2 13p 40 18) f 2 17f 60 19) x2 xy 20y2 20) t2 2tu 63u2 21) 3c2 24c 36 22) 4m3n 8m2n2 60mn3 23) 5y2 11y 6 24) 3g2 g 44 25) 4m2 16m 15 26) 6t2 49t 8 27) 56a3 4a2 16a 28) 18n2 98n 40 29) 3s2 11st 4t2 30) 8f 2(g 11)3 6f(g 11)3 35(g 11)3 31) (3c 5)2 10(3c 5) 24 32) 2(k 1)2 15(k 1) 28 Chapter 7: Review Exercises 416 CHAPTER 7 Factoring Polynomials www.mhhe.com/messersmith
messersmith_power_intermediate_algebra_1e_ch4_7_10
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