1. Algebra I Notes Section 9.6 (A) Factoring ax 2 + bx + c With Leading Coefficient ≠ 1

2. In section 9.5, we learned how to factor quadratic polynomials whose leading coefficient = 1. We will now learn how to factor quadratic polynomials whose leading coefficient is ≠ 1. To factor quadratic polynomials when a ≠ 1, we need to find the factors of a and the factors of c such that when we add the Outer and Inner products of the FOIL method, their sum = b. Example – Factor the polynomialsWe must find the factors of a and c 1.3x 2 – 17x + 10- the factors of 3 (or a) are: ___________ - the factors of 10 (or c) are : ___________ We know that the binomials (without their signs) must be one of these four forms because of the factors: 1. 2. 3. 4. 1, 3 1, 10 & 2, 5 (x 1)(3x 10)(x 10)(3x 1)(x 2)(3x 5)(x 5)(3x 2)

3. If we look at the FOIL method for each of the possible factors, we get: F O I L 1. +++ 2. +++ 3. + ++ 4. +++ We know that in the original polynomial 3x 2 – 17x + 10, 10 (or c) must be positive, so the L multiplication of the FOIL method must yield a + 10 (for c). At the same time, the sum of the O and I multiplication must yield a - 17 (for b). With the correct choice of signs, which sum (of O and I) can possibly add up to -17? _______________ So we conclude: 3x 2 – 17x + 10 = ____________________ 3x 2 10x3x10 3x 2 x30x10 3x 2 5x6x10 3x 2 2x15x10 -2x & -15x (x – 5)(3x – 2)

4. 2. 3x 2 – 4x – 7- the factors of 3 (or a) are: _____________ - the factors of 7 (or c) are : _____________ We know that the binomials (without their signs) must be one of these two forms because of the factors: 1. 2. If we look at the FOIL method for each of the possible factors, we get: F O I L 1.++ + 2.++ + We know that in the original polynomial 3x 2 – 4x – 7, -7 (or c) must be negative, so the L multiplication of the FOIL method must yield a -7 (for c). At the same time, the sum of the O and I multiplication must yield a - 4 (for b). With the correct choice of signs, which sum (of O and I) can possibly add up to -4? ________________ So we conclude: 3x 2 – 4x – 7 = ________________ 1, 3 1, 7 (x 1)(3x 7)(x 7)(3x 1) 7x3x x21x -7x & 3x (x + 1)(3x – 7)

5. 3. 6x 2 – 2x – 8- begin by factoring out the common factor ___________ - the factors of 3 (or a) are: _______________ - the factors of 4 (or c) are : _______________ We know that the binomials (without their signs) must be one of these three forms because of the factors: 1. 2. 3. If we look at the FOIL method for each of the possible factors, we get: FOI L 1.+++ 2.+++ 3.+++ We know that in the FACTORED polynomial 3x 2 – x – 4, -4 (or c) must be negative, so the L multiplication of the FOIL method must yield a -4 (for c). At the same time, the sum of the O and I multiplication must yield a - 1 (for b). 2 1, 3 2(3x 2 – x – 4) 1, 4 & 2, 2 (x 1)(3x 4)(x 4)(3x 1)(x 2)(3x 2) 4x3x x12x 2x6x With the correct choice of signs, which sum (of O and I) can possibly add up to -1? ________________ So we conclude: 2(3x 2 – x – 4) = _______________________ -4x & 3x 2(x + 1)(3x – 4)

6. More Examples - Factor the polynomials. 1.3t 2 + 16t + 5 2. 6b 2 – 11b – 2 Factors of a : 1, 3 Factors of c : 1, 5 (t 1)(3t 5)(t 5)(3t 1) 5t, 3tt, 15t Must add to 16t : (t + 5)(3t + 1) Factors of a : 1, 6 2, 3 Factors of c : 1, 2 (2b 1)(3b 2)(2b 2)(3b 1) 4b, 3b2b, 6b (b 1)(6b 2)(b 2)(6b 1) 2b, 6bb, 12b Must add to -11b : (b – 2)(6b + 1)

7. 3. 5w 2 – 9w – 24. 4x 2 – 6x – 4 Factors of a : 1, 5 Factors of c : 1, 2 (w 1)(5w 2) (w 2)(5w 1) 2w, 5w w, 10w Must add up to -9w : (w - 2)(5w + 1) Factors of a : 1, 2 Factors of c : 1, 2 Factor out GCF : 2 2(2x 2 – 3x – 2) (x 1)(2x 2) (x 2)(2x 1) 2x, 2x x, 4x Must add up to -3x : 2(x – 2)(2x + 1)

8. 5. 12y 2 – 22y – 20 6. 9x 2 + 21x – 18 Factor out GCF : 2 2(6y 2 – 11y – 10) Factors of a : 1, 6 2, 3 Factors of c : 1, 10 2, 5 (2y 2)(3y 5) (2y 5)(3y 2) 10y, 6y 4y, 15y Must add up to -11y : 2(2y – 5)(3y + 2) Factor out GCF : 3 3(3x 2 + 7x – 6) Factors of a : 1, 3 Factors of c : 1, 6 2, 3 (x 2)(3x 3) (x 3)(3x 2) 3x, 6x 2x, 9x Must add up to 7x : 3(x + 3)(3x – 2)