9 4 FACTORING TO SOLVE ax 2 bx

9. 4: FACTORING TO SOLVE ax 2 + bx + c = 0 Factoring: A process used to break down any polynomial into simpler polynomials. Zero-Product Property: For any real numbers a and b, If ab = 0 Then a = 0 or b = 0.

FACTORING ax 2 + bx + c = 0 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:

SOLVING BY FACTORING: Ex: What are the solutions of: 2 x +5 x+6?

FACTORING: To factor a quadratic trinomial with a coefficient of 1 in the x 2, we must look at the b and c coefficients: 2 x +5 x+6 =0 2 x +bx+c b= +5 c = +6 Look at the factors of C: c = +6 : (1)(6), (2)(3) Take the pair that equals to b when adding the two integers. In our case it is 2 x 3 since 2+3 = 5= b Thus the factored form is: (x+2)(x+3)

FACTORING: To find the solutions (x-intercepts) we go one step further: (x+2)(x+3) = 0 Using the Zero-Product property (x+2)(x+3) = 0 (x+2) = 0 (x+3) = 0 x+2 = 0 x+3 = 0 X = -2 X = -3 The solutions are x = -3, and -2.

SOLVING BY FACTORING: Ex: What is the solutions of the equation: 2 5 x +11 x+2?

FACTORING TO SOLVE: To factor a quadratic trinomial with a coefficient ≠ 1 in the x 2, we must look at the b and ac coefficients: 2 5 x +11 x+2 =0 2 ax +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 1 x 10 since 1+10 =11= b

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

FACTORING: To find the solutions (x-intercepts) we go one step further: (5 x+1) (x+2) = 0 Using the Zero-Product property (5 x+1) (x+2) = 0 (5 x+1) = 0 (x+2) = 0 5 x = -1 x+2 = 0 X = -1/5 X = -2 The solutions are x = -2, and -1/5.

YOU TRY IT: Ex: What are the solutions of: 2 6 x +13 x+5?

SOLUTION: To factor a quadratic trinomial with a coefficient ≠ 1 in the x 2, we must look at the b and ac coefficients: 2 6 x +13 x+5 =0 2 ax +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 3 x 10 since 3+10 =13= b

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

FACTORING: To find the solutions (x-intercepts) we go one step further: (2 x+1) (3 x+5) = 0 Using the Zero-Product property (2 x+1) (3 x+5) = 0 (2 x+1) = 0 (3 x+5) = 0 2 x = -1 3 x = -5 X = -1/2 X = -5/3 The solutions are x = -1/2, and -5/3.

YOU TRY IT: Ex: What are the solutions of: 2 3 x +4 x = 15?

FACTORING: Since a ≠ 1, we still look at the b and ac coefficients: 3 x 2+4 x-15 = 0 ax 2+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

Re-write: using factors of ac that = b. 2 3 x +4 x-15 3 x 2 -5 x + 9 x -15 Look at the GCF of the first two terms: 3 x 2 - 5 x x(3 x - 5) Look at the GCF of the last two terms: 9 x -15 3(3 x -5) Look at the GCF of both: x(3 x - 5) + 3(3 x - 5) Thus the factored form is: (3 x-5) (x+3)

FACTORING: To find the solutions (x-intercepts) we go one step further: (3 x-5) (x+3) = 0 Using the Zero-Product property (3 x-5) (x+3) = 0 (3 x-5) = 0 (x+3) = 0 3 x = 5 x = -3 X = 5/3 x = -3 The solutions are x = -3 and 5/3.

REAL-WORLD: You want to make a frame for the photo. You want the frame to be the same width all the way around and the total area to be 315 in 2. What should the outer dimensions of the frame be?

SOLUTION: Adding an x to both sides of the picture 2 x +11 we get: A=bh 2 x +17 A = (2 x +11)(2 x +17) 315 = (2 x +11)(2 x +17) FOIL 2 315 = 4 x + 56 x + 187

SOLUTION: To solve we now put it in 2 the ax +bx+c = 0 form: 315 = 4 x 2 + 56 x + 187 -315 = 0 2 4 x + 56 x -128 = 0 2 4(x + 14 x -32) = 0 4(x+16)(x-2) = 0 (x+16)= 0 (x-2) = 0 x = -16 x = 2

VIDEOS: SOLVING BY FACTORING Solving by factoring: http: //www. khanacademy. org/math/algebra/quadratics /factoring_quadratics/v/Example%201: %20 Solving%20 a %20 quadratic%20 equation%20 by%20 factoring http: //www. khanacademy. org/math/trigonometry/polynomial_and_ratio nal/quad_formula_tutorial/v/solving-quadratic-equations-by-completingthe-square

VIDEOS: SOLVING BY FACTORING Solving by factoring: http: //www. khanacademy. org/math/algebra/quadratics/f actoring_quadratics/v/Example%202: %20 Solving%20 a%20 quadratic%20 equation%20 by%20 factoring http: //www. khanacademy. org/math/trigonometry/polynomial_and_ratio nal/quad_formula_tutorial/v/solving-quadratic-equations-by-completingthe-square

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