8.6: FACTORING ax2 + bx + c where a ≠ 1 Factoring: A process used to break down any polynomial into simpler polynomials.

FACTORING ax2 + bx + c Procedure: 1) Always look for the GCF of all the terms 2) Factor the remaining terms – pay close attention to the value of coefficient a and follow the proper steps. 3) Re-write the original polynomial as a product of the polynomials that cannot be factored any further.

GOAL:

5x2+11x+2? FACTORING: When AC is POSSIBLE. Ex: What is the FACTORED form of: 5x2+11x+2?

5x2+11x+2 ax2+bx+c  b= +11  ac =(5)(2)  ac = +10 : (1)(10), (2)(5) FACTORING: To factor a quadratic trinomial with a coefficient ≠ 1 in the x2, we must look at the b and ac coefficients: 5x2+11x+2 ax2+bx+c  b= +11  ac =(5)(2) Look at the factors of ac:  ac = +10 : (1)(10), (2)(5) Take the pair that equals to b when adding the two integers. In our case it is 1x10 since 1+10 =11= b

5x2+11x+2 5x2 + 1x + 10x + 2 5x2 + 1x  x(5x + 1) 10x + 2  2(5x + 1) Re-write using factors of ac that = b. 5x2+11x+2 5x2 + 1x + 10x + 2 Look at the GCF of the first two terms: 5x2 + 1x  x(5x + 1) Look at the GCF of the last two terms: 10x + 2  2(5x + 1) Look at the GCF of both:  x(5x + 1) + 2(5x + 1) Thus the factored form is: (5x+1) (x+2)

YOU TRY IT: Ex: What is the FACTORED form of: 6x2+13x+5?

6x2+13x+5 ax2+bx+c  b= +13  ac =(6)(5) SOLUTION: To factor a quadratic trinomial with a coefficient ≠ 1 in the x2, we must look at the b and ac coefficients: 6x2+13x+5 ax2+bx+c  b= +13  ac =(6)(5) Look at the factors of C: ac = +30 :(1)(30), (2)(15), (3)(10) Take the pair that equals to b when adding the two integers. In our case it is 3x10 since 3+10 =13= b

6x2+13x+5  6x2 + 3x + 10x + 5 6x2 + 3x  3x(2x + 1) 10x + 5 Re-write using factors of ac that = b. 6x2+13x+5  6x2 + 3x + 10x + 5 Look at the GCF of the first two terms: 6x2 + 3x  3x(2x + 1) Look at the GCF of the last two terms: 10x + 5  5(2x + 1) Look at the GCF of both:  3x(2x + 1)+ 5(2x + 1) Thus the factored form is:(3x+5)(2x+1)

CLASSWORK: Page 508-509: Problems: 1, 2, 5, 8, 9, 11, 20.

3x2+4x-15? FACTORING: When ac is Negative. Ex: What is the FACTORED form of: 3x2+4x-15?

 ac = -45 : (-1)(45), (1)(-45) (-3)(15), (3)(-15) (-5)(9), (5)(-9) FACTORING: Since a ≠ 1, we still look at the b and ac coefficients: 3x2+4x-15 ax2+bx+c  b= +4  ac =(3)(-15) Look at the factors of ac:  ac = -45 : (-1)(45), (1)(-45) (-3)(15), (3)(-15) (-5)(9), (5)(-9) Take the pair that equals to b when adding the two integers. In our case it is (-5)(9)since -5+9 =+4 =b

3x2+4x-15 3x2 -5x + 9x -15 3x2 - 5x  x(3x - 5) 9x -15  3(3x -5) Re-write: using factors of ac that = b. 3x2+4x-15 3x2 -5x + 9x -15 Look at the GCF of the first two terms: 3x2 - 5x  x(3x - 5) Look at the GCF of the last two terms: 9x -15  3(3x -5) Look at the GCF of both:  x(3x - 5) + 3(3x - 5) Thus the factored form is: (3x-5) (x+3)

YOU TRY IT: Ex: What is the FACTORED form of: 10x2+31x-14?

10x2+31x-14 ax2+bx+c  b= +31  ac =(10)(-14) FACTORING: Since a ≠ 1, we still look at the b and ac coefficients: 10x2+31x-14 ax2+bx+c  b= +31  ac =(10)(-14) Look at the factors of ac:  ac = -140 : (-1)(140), (1)(-140) (-2)(70), (2)(-70) (-4)(35), (4)(-35) Take the pair that equals to b when adding the two integers. This time it is(-4)(35)since -4+35=+31=b

10x2+31x-14  10x2-4x + 35x -14 10x2 - 4x  2x(5x - 2) 35x -14 Re-write using factors of ac that = b. 10x2+31x-14  10x2-4x + 35x -14 Look at the GCF of the first two terms: 10x2 - 4x  2x(5x - 2) Look at the GCF of the last two terms: 35x -14  7(5x - 2) Look at the GCF of both:  2x(5x - 2)+ 7(5x - 2) Thus the factored form is:(2x+7)(5x-2)

CLASSWORK: Page 508-509: Problems: 3, 6, 14, 16, 17, 18, 34.

REAL-WORLD: The area of a rectangular knitted blanket is 15x2-14x-8. What are the possible dimensions of the blanket?

15x2-14x-8 ax2+bx+c  b= -14  ac =(15)(-8) SOLUTION: Since a ≠ 1, we still look at the b and ac coefficients: 15x2-14x-8 ax2+bx+c  b= -14  ac =(15)(-8) Look at the factors of ac:  ac = -120 : (-1)(120), (1)(-120) (-2)(60), (2)(-60),(-3)(40), (3)(-40) (-4)(30), (4)(-30), (-5)(24), (5)(-24) (-6)(20), (6)(-20), (-8)(15), (8)(-15) This time it is(6)(-20)since 6-20=-14=b

15x2-14x-8  15x2 + 6x - 20x -8 15x2 + 6x  3x(5x + 2) -20x -8 Re-write using factors of ac that = b. 15x2-14x-8  15x2 + 6x - 20x -8 Look at the GCF of the first two terms: 15x2 + 6x  3x(5x + 2) Look at the GCF of the last two terms: -20x -8 -4 (5x + 2) Look at the GCF of both:  3x(5x + 2)- 4(5x + 2) Thus the factored form is:(3x-4)(5x+2)

VIDEOS: Factoring Quadratics Factoring when a ≠ 1: http://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_factoring/v/factoring-trinomials-by-grouping-5 http://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_factoring/v/factoring-trinomials-by-grouping-6

CLASSWORK: Page 508-509: Problems: 4, 7, 15, 21, 25, 36, 37.