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Factoring Trinomials ax 2 + bx + c (a > 1) Model the following by drawing rectangles with the given areas. Label the sides. Factor the area/polynomial. (Hint: Since the sides multiply to equal the area, the sides are the factors.) (Hint: Since the sides multiply to equal the area, the sides are the factors.) a) 2x 2 + 7x + 6

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Factoring Trinomials ax 2 + bx + c (a > 1) Model the following by drawing rectangles with the given areas. Label the sides. Factor the area/polynomial. (Hint: Since the sides multiply to equal the area, the sides are the factors.) (Hint: Since the sides multiply to equal the area, the sides are the factors.) a) 2x 2 + 7x + 6 2x 3 2x 3 x 2x 2 + 7x + 6 x 2x 2 + 7x + 6 =(2x + 3)( x + 2) =(2x + 3)( x + 2) 2

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Factoring Trinomials ax 2 + bx + c (a > 1) b) 2x 2 + 9x + 4 c) 3x 2 + 11x + 6 d) 2x 2 + 3x – 9 e) 2x 2 – 7x + 3 f) 2x 2 + 5x – 12 g) 3x 2 - 16x + 16 h) 2x 2 – 7x – 4 i) 3x 2 - x - 4 j) 5x 2 + x – 18 k) 3x 2 - 4x - 15 l) 3x 2 + 4x + 1 m) 4x 2 + 4x – 15 n) 2x 2 – x – 1 o) 2x 2 + 5x + 2 p) 3x 2 - 5x – 2 q) 3x 2 - 4x + 1 r) Create your own trinomial of the form ax 2 + bx + c that can be factored. Factor.

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Factoring Trinomials ax 2 + bx + c (a > 1) - Conclusions Have any conclusions/rules been discovered about factoring ax 2 + bx + c other than drawing rectangles and determining the sides? Ex. 2x 2 + 5x – 12

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Factoring Trinomials ax 2 + bx + c (a > 1) - Conclusions Have any conclusions/rules been discovered about factoring ax 2 + bx + c other than drawing rectangles and determining the sides? Ex. 2x 2 + 5x – 12 2x -3 2x -3 x 2x 2 + 5x – 12 x 2x 2 + 5x – 12 = (2x - 3)( x + 4) = (2x - 3)( x + 4) 4 Have you noticed anything about the terms here? 4 Have you noticed anything about the terms here? x to = singles ( -3 x 4 = -12) x to = singles ( -3 x 4 = -12) and 2 x 4 + (-3) = 5 and 2 x 4 + (-3) = 5

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Factoring Trinomials ax 2 + bx + c (a > 1) - Conclusions Although the previous observations are true they may not help that much. If applied to 2x 2 + 9x + 4 it would require you to find two #’s that multiply to equal 4 while also finding a # that multiplies by one of the factors used above plus the other to equal the 9. ex. 2x 2 + 9x + 4 ex. 2x 2 + 9x + 4 (?x + ?)(x + ?) (?x + ?)(x + ?) x to equal 4 x to equal 4

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Factoring Trinomials ax 2 + bx + c (a > 1) - Conclusions Although the previous observations are true they may not help that much. If applied to 2x 2 + 9x + 4 it would require you to find two #’s that multiply to equal 4 while also finding a # that multiplies by one of the factors used above plus the other to equal the 9. ex. 2x 2 + 9x + 4 ex. 2x 2 + 9x + 4 (?x + ?)(x + ?) (?x + ?)(x + ?) x to equal 4 x to equal 4 and x + = 9 and x + = 9 For most of us, drawing or at least thinking about a rectangle may be the best option. For most of us, drawing or at least thinking about a rectangle may be the best option.

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8-4A Factoring Trinomials when the Leading Coefficient isn’t 1.

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