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Objective The student will be able to: use grouping to factor polynomials with four terms. SOL: A.2c Designed by Skip Tyler, Varina High School 1. Factor rx + 2ry + kx + 2ky You have 4 terms - try factoring by grouping. (rx + 2ry) + (kx + 2ky) Find the GCF of each group. r(x + 2y) + k(x + 2y) The parentheses are the same! (r + k)(x + 2y) 2. Factor 2 2x - 3xz - 2xy + 3yz Check for a GCF: None Factor by grouping. Keep a + between the groups. (2x2 - 3xz) + (- 2xy + 3yz) Find the GCF of each group. x(2x – 3z) + y(- 2x + 3z) The signs are opposite in the parentheses! Keep-change-change! x(2x - 3x) - y(2x - 3z) (x - y)(2x - 3z) Objective The student will be able to: factor trinomials with grouping. SOL: A.2c Designed by Skip Tyler, Varina High School Here we go! 1) Factor x2 + 3x + 2 Use your factoring chart. Now we will learn Trinomials! You will set up a table with the following information. Product of the first and last coefficients Middle coefficient The goal is to find two factors in the first column that add up to the middle term in the second column. We’ll work it out in the next few slides. 1) Factor 2 x M A + 3x + 2 Create your MA table. Product of the first and last coefficients 1st coefficient is an unwritten “1” , the last coefficient is 2 Multiply +2 Add +3 Middle coefficient Here’s your task… What numbers multiply to +2 and add to +3? If you cannot figure it out right away, write the combinations. 1) Factor 2 x + 3x + 2 Place the factors in the table. Only 2 Factors are 2 and 1. Multiply +2 Add +3 +2, +1 +3, YES!! We are going to use these numbers in the next step! 1) Factor x2 + 3x + 2 Multiply +2 Add +3 +2, +1 +3, YES!! Hang with me now! Replace the middle number of the trinomial with our working numbers from the MA table x2 + 3x + 2 x2 + 2x + 1x + 2 Now, group the first two terms and the last two terms. We have two groups! (x2 + 2x)(+1x + 2) Almost done! Find the GCF of each group and factor it out. If things are done right, the parentheses x(x + 2) +1(x + 2) should be the same. Factor out the GCF’s. Write them in their own group. (x + 1)(x + 2) Tadaaa! There’s your answer…(x + 1)(x + 2) You can check it by multiplying. Piece of cake, huh? There is a shortcut for some problems too! (I’m not showing you that yet…) M A 2) Factor 5x2 - 17x + 14 Create your MA table. Product of the first and last coefficients Signs need to be the same as the middle sign since the product is positive. Multiply +70 -1, -70 -2, -35 -7, -10 Add -17 -71 -37 -17 Replace the middle term. 5x2 – 7x – 10x + 14 Group the terms. Middle coefficient (5x2 – 7x) (– 10x + 14) Factor out the GCF x(5x – 7) -2(5x – 7) The parentheses are the same! Weeedoggie! (x – 2)(5x – 7) Hopefully, these will continue to get easier the more you do them. Here are some hints to help you choose your factors in the MA table. 1) When the last term is positive, the factors will have the same sign as the middle term. 2) When the last term is negative, the factors will have different signs. Factor 1. 2. 3. 4. (2x + 10)(x + 1) (2x + 5)(x + 2) (2x + 2)(x + 5) (2x + 1)(x + 10) 2 2x + 9x + 10