Math 221 Intro to Algebra Week 5 DQ 1

INST RUCTOR GUIDANCE EXAMPLE: Week Five Discussion Factoring Since there are several different types of factoring problems assigned from pages 345 - 346, four types will be demonstrated here to offer a selection, even though individual students will only b e working two from these pages. #73. x 3 – 2x 2 – 9x + 18 Four terms means start with grouping x2(x – 2) – 9(x – 2) The common factor for each group is (x – 2) (x – 2)(x 2 – 9) Notice the difference of squares in second group (x – 2)(x – 3)(x -+ 3) Now it is completely factored. #81. 6w 2 – 12w – 18 Every term has a GCF of 6 6(w 2 – 2w – 3) Common factor is removed, now have a trinomial Need two numbers that add to -2 but multiply to -3 Try with -3 and +1 6(w – 3)(w + 1) This works , check by multiplying it back together #97. 8vw 2 + 32vw + 32v Every term has a GCF of 8 v 8v(w 2 + 4w + 4) The trinomial is in the form of a perfect square 8v(w + 2)(w + 2) Showing the squared binomial 8v (w + 2) 2 Writing the square appropriately #103. -3y 3 + 6y 2 – 3y Every term has a GCF of -3y -3y(y 2 – 2y + 1) Another perfect square trinomial -3y(y – 1)(y – 1) Showing the squared binomial -3y(y – 1)2 Writing the square appropriately Here are tw o examples of problems similar to those assigned from page 353. 5b 2 – 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 -3(-10 )=30 while -3+(-10 )= -13 so replace -13b by -3b and -10 b 5b 2 – 3b – 10 b + 6 Now facto r by grouping. b(5b – 3) – 2(5b – 3) The common binomial facto r is ( 5b – 3). (5b – 3)( b – 2) Check by multiplying it back together. 3x 2 + x – 14 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 . 3x 2 – 6x + 7x – 14 Factor by grouping . 3x( x – 2) + 7(x – 2) The common binomial factor is (x – 2). (x – 2)(3x + 7) Check by multiplying it b ack together.