2 2 7 4 Factoring axbx bxcc Warm
2 2 7 -4 Factoring ax++bx bx++cc Warm Up Lesson Presentation Lesson Quiz Holt Algebra 1 Holt Mc. Dougal
7 -4 Factoring ax 2 + bx + c Warm Up Find each product. 1. (x – 2)(2 x + 7) Find each trinomial. 2. x 2 +4 x – 32 Holt Mc. Dougal Algebra 1
7 -4 Factoring ax 2 + bx + c ESSENTIAL QUESTION How do you factor quadratic trinomials of the form ax 2 + bx + c ? Holt Mc. Dougal Algebra 1
7 -4 Factoring ax 2 + bx + c When you multiply (3 x + 2)(2 x + 5), the coefficient of the x 2 -term is the product of the coefficients of the x-terms. Also, the constant term in the trinomial is the product of the constants in the binomials. (3 x + 2)(2 x + 5) = 6 x 2 + 19 x + 10 Holt Mc. Dougal Algebra 1
7 -4 Factoring ax 2 + bx + c Example 1: Factoring ax 2 + bx + c by Guess and Check Factor 6 x 2 + 11 x + 4 by guess and check. ( _ + _ ) ( x + _)( _ x + _) Write two sets of parentheses. The coefficient of the x 2 term is 6. The constant term in the trinomial is 4. (2 x + 4)(3 x + 1) = 6 x 2 + 14 x + 4 Try factors of 6 for the 2 (1 x + 4)(6 x + 1) = 6 x + 25 x + 4 coefficients and (1 x + 2)(6 x + 2) = 6 x 2 + 14 x + 4 factors of 4 for the (1 x + 1)(6 x + 4) = 6 x 2 + 10 x + 4 constant terms. (3 x + 4)(2 x + 1) = 6 x 2 + 11 x + 4 Holt Mc. Dougal Algebra 1
7 -4 Factoring ax 2 + bx + c Example 2 Factor each trinomial by guess and check. 6 x 2 + 11 x + 3 ( _ + _ ) ( x + _)( _ x + _) Write two sets of parentheses. The coefficient of the x 2 term is 6. The constant term in the trinomial is 3. (2 x + 1)(3 x + 3) = 6 x 2 + 9 x + 3 Try factors of 6 for the coefficients and (1 x + 3)(6 x + 1) = 6 x 2 + 19 x + 3 factors of 3 for the 2 (1 x + 1)(6 x + 3) = 6 x + 9 x + 3 constant terms. (3 x + 1)(2 x + 3) = 6 x 2 + 11 x + 3 Holt Mc. Dougal Algebra 1
7 -4 Factoring ax 2 + bx + c Example 3 Factor each trinomial by guess and check. 3 x 2 – 2 x – 8 ( _ + _ ) ( x + _)( _ x + _) Write two sets of parentheses. The coefficient of the x 2 term is 3. The constant term in the trinomial is – 8. (1 x – 1)(3 x + 8) = 3 x 2 + 5 x – 8 Try factors of 3 for the coefficients and 2 (1 x – 4)(3 x + 2) = 3 x – 10 x – 8 factors of 8 for the (1 x – 8)(3 x + 1) = 3 x 2 – 23 x – 8 constant terms. (1 x – 2)(3 x + 4) = 3 x 2 – 2 x – 8 Holt Mc. Dougal Algebra 1
7 -4 Factoring ax 2 + bx + c Example 4: Factoring ax 2 + bx + c When c is negative 3 x 2 – 16 x + 16 ( _ + _ ) ( x + _)( _ x + _) a = 3 and c = 16, Outer + Inner = – 16. Factors of 3 Factors of 16 Outer + Inner 1 and 3 – 1 and – 16 1(– 16) + 3(– 1) = – 19 1 and 3 – 2 and – 8 1( – 8) + 3(– 2) = – 14 – 4 and – 4 1( – 4) + 3(– 4)= – 16 1 and 3 (x – 4)(3 x – 4) Use the Foil method. Check (x – 4)(3 x – 4) = 3 x 2 – 4 x – 12 x + 16 = 3 x 2 – 16 x + 16 Holt Mc. Dougal Algebra 1
7 -4 Factoring ax 2 + bx + c 6 x 2 + 17 x + 5 Example 5 ( _ + _ ) ( x + _)( _ x + _) Factors of 6 Factors of 5 1 and 6 1 and 5 2 and 3 1 and 5 3 and 2 a = 6 and c = 5, Outer + Inner = 17. Outer + Inner 1(5) + 6(1) = 11 2(5) + 3(1) = 13 3(5) + 2(1) = 17 Use the Foil method. (3 x + 1)(2 x + 5) Check (3 x + 1)(2 x + 5) = 6 x 2 + 15 x + 2 x + 5 = 6 x 2 + 17 x + 5 Holt Mc. Dougal Algebra 1
7 -4 Factoring ax 2 + bx + c 9 x 2 – 15 x + 4 ( _ + _ )( _ +_) ( x + _)( _ x + _) Example 6 a = 9 and c = 4, Outer + Inner = – 15. Factors of 9 Factors of 4 Outer + Inner 3 and 3 – 1 and – 4 3(– 4) + 3(– 1) = – 15 3 and 3 – 2 and – 2 3(– 2) + 3(– 2) = – 12 – 4 and – 1 3(– 1) + 3(– 4)= – 15 3 and 3 (3 x – 4)(3 x – 1) Use the Foil method. Check (3 x – 4)(3 x – 1) = 9 x 2 – 3 x – 12 x + 4 = 9 x 2 – 15 x + 4 Holt Mc. Dougal Algebra 1
7 -4 Factoring ax 2 + bx + c Example 7 3 x 2 + 13 x + 12 ( _ + _ ) ( x + _)( _ x + _) Factors of 3 Factors of 12 1 and 3 1 and 12 2 and 6 1 and 3 3 and 4 1 and 3 a = 3 and c = 12, Outer + Inner = 13. Outer + Inner 1(12) + 3(1) = 15 1(6) + 3(2) = 12 1(4) + 3(3) = 13 Use the Foil method. (x + 3)(3 x + 4) Check (x + 3)(3 x + 4) = 3 x 2 + 4 x + 9 x + 12 = 3 x 2 + 13 x + 12 Holt Mc. Dougal Algebra 1
7 -4 Factoring ax 2 + bx + c Example: Factoring ax 2 + bx + c 3 n 2 + 11 n – 4 ( _ + _ ) ( x + _)( _ x + _) When c is Negative Factors of 3 Factors of – 4 1 and 3 – 1 and 4 1 and 3 – 2 and 2 – 4 and 1 1 and 3 4 and – 1 1 and 3 a = 3 and c = – 4, Outer + Inner = 11. Outer + Inner 1(4) + 3(– 1) = 1 1(2) + 3(– 2) = – 4 1(1) + 3(– 4) = – 11 1(– 1) + 3(4) = 11 (n + 4)(3 n – 1) Use the Foil method. Check (n + 4)(3 n – 1) = 3 n 2 – n + 12 n – 4 = 3 n 2 + 11 n – 4 Holt Mc. Dougal Algebra 1
7 -4 Factoring ax 2 + bx + c Example: Factoring ax 2 + bx + c When b & c are Negative 4 x 2 – 15 x – 4 ( _ + _ ) ( x + _)( _ x + _) Factors of 4 Factors of – 4 1 and 4 – 2 and 2 – 4 and 1 1 and 4 (x – 4)(4 x + 1) a = 4 and c = – 4, Outer + Inner = – 15. Outer + Inner 1(4) + 4(– 1) = 0 1(2) + 4(– 2) = – 6 1(1) + 4(– 4) = – 15 Use the Foil method. Check (x – 4)(4 x + 1) = 4 x 2 + x – 16 x – 4 = 4 x 2 – 15 x – 4 Holt Mc. Dougal Algebra 1
7 -4 Factoring ax 2 + bx + c Example 4 n 2 – n – 3 ( _ + _ ) ( x + _)( _ x + _) a = 4 and c = – 3, Outer + Inner = – 1. Factors of 4 Factors of – 3 Outer + Inner 1 and 4 1(– 3) + 4(1) = 1 1 and – 3 1(3) – 4(1) = – 1 1 and 4 – 1 and 3 (4 n + 3)(n – 1) Use the Foil method. Check (4 n + 3)(n – 1) = 4 n 2 – 4 n + 3 n – 3 = 4 n 2 – n – 3 Holt Mc. Dougal Algebra 1
7 -4 Factoring ax 2 + bx + c Example: Factoring ax 2 + bx + c – 2 x 2 + 7 x - 3. When a is Negative (_ x + )(_ x+ ) Now: a = 2 and c = -3; Outer + Inner = 5 Factors of -2 Factors of -3 1 and -2 3 and -1 1 and -2 -3 and 1 Outer + Inner 7 -1(1) + 3(-2) = -7 1(1) + -2(-3) = (x - 3)(-2 x + 1) ANS: (x - 3)(-2 x + 1) Holt Mc. Dougal Algebra 1
7 -4 Factoring ax 2 + bx + c Example – 17 x – 12 ( _ + _ ) -1( x + _)( _ x + _) – 6 x 2 – 1(6 x 2 + 17 x + 12) Factor out – 1. a = 6 and c = 12; Everything positive! Factors of 6 Factors of 12 Outer + Inner 17 2 and 3 4 and 3 2(3) + 3(4) = 18 2 and 3 3 and 4 2(4) + 3(3) = (2 x + 3)(3 x + 4) – 1(2 x + 3)(3 x + 4) Holt Mc. Dougal Algebra 1
7 -4 Factoring ax 2 + bx + c Lesson Quiz Factor each trinomial. Check your answer. 1. 5 x 2 + 17 x + 6 (5 x + 2)(x + 3) 2. 2 x 2 + 5 x – 12 (2 x– 3)(x + 4) 3. 6 x 2 – 23 x + 7 (3 x – 1)(2 x – 7) 4. – 4 x 2 + 11 x + 20 (–x + 4)(4 x + 5) Holt Mc. Dougal Algebra 1
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