INSTRUCTOR GUIDANCE EXAMPLE: Week Five Discussion Factoring Since there are several different types of factoring problems assigned from pages 345346, four types will be demonstrated here to offer a selection, even though individual students will only be working two from these pages. #73. x3 - 2x2 - 9x + 18 x2(x - 2) - 9(x - 2) (x - 2)(x2 - 9) (x - 2)(x - 3)(x -+ 3) Four terms means start with grouping The common factor for each group is (x - 2) Notice the difference of squares in second group Now it is completely factored. #81. 6w2 - 12w - 18 6(w2 - 2w - 3) Every term has a GCF of 6 Common factor is removed, now have a trinomial Need two numbers that add to -2 but multiply to -3 Try with -3 and +1 This works, check by multiplying it back together 6(w - 3)(w + 1) #97. 8vw2 + 32vw + 32v 8v(w2 + 4w + 4) 8v(w + 2)(w + 2) 8v(w + 2)2 Every term has a GCF of 8v The trinomial is in the form of a perfect square Showing the squared binomial Writing the square appropriately #103. -3y3 + 6y2 - 3y -3y(y2 - 2y + 1) -3y(y - 1)(y - 1) -3y(y - 1)2 Every term has a GCF of -3y Another perfect square trinomial Showing the squared binomial Writing the square appropriately Here are two examples of problems similar to those assigned from page 353. 5b2 - 13b + 6 a = 5 and c = 6, so ac = 5(6) = 30. The factor pairs of 30 are 1, 30 2, 15 3, 10 5,6 5b2 - 3b - 10b + 6 b(5b - 3) - 2(5b - 3) (5b - 3)( b - 2) Now factor by grouping. The common binomial factor is (5b - 3). Check by multiplying it back together. -3(-10)=30 while -3+(-10)= -13 so replace -13b by -3b and -10b 3x2 + x - 14 3x2 - 6x + 7x - 14 3x(x - 2) + 7(x - 2) (x - 2)(3x + 7) a = 3 and c = -14, so ac =3(-14)= -42. The factor pairs of - 42 are 1, -42 -1, 42 3, -14 -3, 14 2, -21 -2, 21 6, -7 -6, 7 We see that -6(7) = -42 while -6 + 7 = 1 so replace x with -6x + 7x. Factor by grouping. The common binomial factor is (x - 2). Check by multiplying it back together. 6 Factoring completely Factor each polynomial completely. a) x3 6x2 16x b 4x3 + 4x2 + ) 4x Solution a) b First factor out 4x, the greatest common factor: ) To factor x2 + x + 1, we would need two integers with a product of 1 and a sum of 1. Because there are no such integers, x2 + x + 1 is prime, and the factorization is complete. Problem # 66 and b3 + 49b problem # 94 18a 2 + 3a 3 +36a 5 Factoring completely Factor each polynomial completely. a) 4x3 + 14x2 + 6x b 12x2y + 6xy + ) 6y Solution a) Check by multiplying. b 12x2y + 6xy + 6y = 6y(2x2 + x + 1) Factor out the GCF, 6y. ) To factor 2x2 + x + 1 by the ac method, we need two numbers with a product of 2 and a sum of 1. Because there are no such numbers, 2x2 + x + 1 is prime and the factorization is complete. Now do Exercises 75-84 Our first step in factoring is to factor out the greatest common factor (if it is not 1). If the first term of a polynomial has a negative coefficient, then it is better to factor out the opposite of the GCF so that the resulting polynomial will have a positive leading coefficient. Problem #78