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216 Chapter 8 ◆ Factors and Factoring Exercise 4 ◆ Factoring by Grouping Factor completely. 1. 3. 5. 7. 8. 9. 10. 11. 12. a3 3a2 4a 12 x3 x2 x 1 x2 bx 3x 3b 3x 2y 6 xy x2y2 3x2 4y2 12 x2 y2 2xy 4 x2 6xy 9y2 a2 m2 n2 4 4n p2 r2 6pq 9q2 2. x3 x2 x 1 4. 2x3 x2 4x 2 6. ab a b 1 Expressions with Three or More Letters Factor completely. 13. ax bx 3a 3b 14. x2 xy xz yz 15. a2 ac ab bc 16. ay by ab y2 17. 2a bx2 2b ax2 18. 2xy wy wz 2xz 19. 6a2 2ab 3ac bc 20. x2 bx cx bc 21. b2 bc ab ac 22. bx cx bc x2 23. x2 a2 2ab b2 24. p2 y2 x2 2xy 8–5 The General Quadratic Trinomial When we multiply the two binomials (ax b) and (cx d), we get a trinomial with a leading coefficient of ac, a middle coefficient of (ad bc), and a constant term of bd. General Quadratic Trinomial (ax b)(cx d ) acx2 (ad bc)x bd 46 The general quadratic trinomial may be factored by trial and error or by the grouping method. Trial and Error To factor the general quadratic trinomial by trial and error, we look for four numbers, a, b, c, and d, such that ac the leading coefficient ad bc the middle coefficient bd the constant term Also, the signs in the factors are found in the same way as for trinomials with a leading coefficient of 1. Section 8–5 ◆◆◆ ◆ 217 The General Quadratic Trinomial Example 33: Factor 2x2 5x 3. Solution: The leading coefficient ac is 2, and the constant term bd is 3. Try a 1, c 2, b 3, and d1 Then ad bc 1(1) 3(2) 7. No good. It is supposed to be 5. We next try a 1, c 2, b 1, and d3 Then ad bc 1(3) 1(2) 5. This works. So 2x2 5x 3 (x 1)(2x 3) ◆◆◆ Grouping Method The grouping method eliminates the need for trial and error. ◆◆◆ Some people have a knack for factoring and can quickly factor a trinomial by trial and error. Other rely on the longer but surer grouping method. Example 34: Factor 3x2 16x 12. Solution: 1. Multiply the leading coefficient and the constant term. 3(12) 36 2. Find two numbers whose product equals 36 and whose sum equals the middle coefficient, 16. Two such numbers are 2 and 18. 3. Rewrite the trinomial, splitting the middle term according to the selected factors (16x 2x 18x). 3x2 2x 18x 12 Group the first two terms together and the last two terms together. (3x2 2x) (18x 12) 4. Remove common factors from each grouping. x(3x 2) 6(3x 2) 5. Remove the common factor (3x 2) from the entire expression: (3x 2)(x 6) which are the required factors. ◆◆◆ It is easy to make a mistake when factoring out a negative quantity. Thus when we factored out a 6 in going from Step 3 to Step 4 in Example 34, we got Common Error (18x 12) 6(3x 2) but not 6(3x 2) incorrect! Notice that if we had grouped the terms differently in Step 3 of Example 34, we would have arrived at the same factors as before. 3x2 18x 2x 12 3x(x 6) 2(x 6) (3x 2)(x 6) Sometimes an expression may be simplified before factoring, as we did in Sec. 8–3. 218 Chapter 8 ◆◆◆ ◆ Factors and Factoring Example 35: Factor 12(x y)6n (x y)3n z 6z2. Solution: If we substitute a (x y)3n, our expression becomes 12a2 az 6z2 Temporarily dropping the z’s gives 12a2 a 6 which factors into (4a 3)(3a 2) Replacing the z’s, we get 12a2 az 6z2 (4a 3z)(3a 2z) Finally, substituting back (x y)3n for a, we have 12(x y)6n (x y)3nz 6z2 [4(x y)3n 3z][4(x y)3n 2z] Exercise 5 ◆ ◆◆◆ The General Quadratic Trinomial Factor completely. 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 4x2 13x 3 5x2 11x 2 12b2 b 6 2a2 a 6 5x2 38x 21 3x2 6x 3 3x2 x 2 4x2 10x 6 4a2 4a 3 9a2 15a 14 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. 5a2 8a 3 7x2 23x 6 6x2 7x 2 2x2 3x 2 4x2 7x 15 2x2 11x 12 7x2 123x 54 3x2 11x 20 9x2 27x 18 16c2 48c 35 Expressions Reducible to Quadratic Trinomials Factor completely. 21. 49x6 14x3y 15y2 22. 18y2 42x2 24xy 23. 4x6 13x3 3 24. 5a2n 8an 3 25. 5x4n 11x2n 2 26. 12(a b)2 (a b) 6 27. 3(a x)2n 3(a x)n 6 Applications 28. An object is thrown upward with an initial velocity of 32 ft./s from a building 128 ft. above the ground. The height s of the object above the ground at any time t is given by s 128 32t 16t 2 Factor the right side of this equation. Section 8–6 ◆ 219 The Perfect Square Trinomial 29. An object is thrown into the air with an initial velocity of 23 m/s. To find the approximate time it takes for the object to reach a height of 12 m, we must solve the quadratic equation Width h 5t 2 23t 12 0 Factor the left side of this equation. 30. To find the depth of cut h needed to produce a flat of a certain width on a 1-cm-radius bar (Fig. 8–6), we must solve the equation 4h2 8h 3 0 1 cm Factor the left side of this equation. FIGURE 8–6 8–6 The Perfect Square Trinomial The Square of a Binomial In Chapter 2, we saw that the expression obtained when a binomial is squared is called a perfect square trinomial. ◆◆◆ Example 36: Square the binomial (2x 3). Solution: (2x 3)2 4x2 6x 6x 9 4x2 12x 9 ◆◆◆ Note that in the perfect square trinomial obtained in Example 36, the first and last terms are the squares of the first and last terms of the binomial. square (2x + 3)2 square 4x2 12x 9 The middle term is twice the product of the terms of the binomial. (2x 3)2 product 6x twice the product 4x2 12x 9 Also, the constant term is always positive. In general, the following equations apply: Perfect Square Trinomials (a b)2 a2 2ab b2 47 (a b)2 a2 2ab b2 48 Factoring a Perfect Square Trinomial We can factor a perfect square trinomial in the same way we factored the general quadratic trinomial in Sec. 8–5. However, the work is faster if we recognize that a trinomial is a perfect square. If it is, its factors will be the square of a binomial. The terms of that binomial are the square roots of the first and last terms of the trinomial. The sign in the binomial will be the same as the sign of the middle term of the trinomial. Any quadratic trinomial can be manipulated into the form of a perfect square trinomial by a procedure called completing the square. We will use that method in Sec. 14–2 to derive the quadratic formula.