9.4: FACTORING TO SOLVE ax2 + 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 = 0Then a = 0 or b = 0.

FACTORING ax2 + 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:

x2+5x+6?

FACTORING: To factor a quadratic trinomial with a coefficient of 1 in the x2, we must look at the b and c coefficients:

x2+5x+6 = 0 x2+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 2x3 since 2+3 = 5= bThus 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 = 0X = -2

x+3 = 0X = -3

The solutions are x = -3, and -2.

SOLVING BY FACTORING:

Ex: What is the solutions of the equation:

5x2+11x+2?

FACTORING TO SOLVE: To factor a quadratic trinomial with a coefficient ≠ 1 in the x2, we must look at the b and ac coefficients:

5x2+11x+2 = 0 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

Re-write using factors of ac that = b.

5x2+11x+2 5x2 + 1x + 10x + 2 Look at the GCF of the first two terms:

Thus the factored form is: (5x+1) (x+2)

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)

FACTORING: To find the solutions (x-intercepts) we go one step further:

(5x+1) (x+2) = 0Using the Zero-Product property

(5x+1) (x+2) = 0(5x+1) = 0 (x+2) = 0

5x = -1X = -1/5

x+2 = 0X = -2

The solutions are x = -2, and -1/5.

YOU TRY IT:

Ex: What are the solutions of:

6x2+13x+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 = 0 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

Re-write using factors of ac that = b.

6x2+13x+5 6x2 + 3x + 10x + 5 Look at the GCF of the first two terms:

Thus the factored form is:(3x+5)(2x+1)

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)

FACTORING: To find the solutions (x-intercepts) we go one step further:

(2x+1) (3x+5) = 0Using the Zero-Product property

(2x+1) (3x+5) = 0(2x+1) = 0 (3x+5) = 0

2x = -1X = -1/2

3x = -5X = -5/3

The solutions are x = -1/2, and -5/3.

YOU TRY IT:

Ex: What are the solutions of:

3x2+4x = 15?

FACTORING: Since a ≠ 1, we still look at the b and ac coefficients: 3x2+4x-15 = 0 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

Re-write: using factors of ac that = b.

3x2+4x-15 3x2 -5x + 9x -15 Look at the GCF of the first two terms:

Thus the factored form is: (3x-5) (x+3)

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)

FACTORING: To find the solutions (x-intercepts) we go one step further:

(3x-5) (x+3) = 0Using the Zero-Product property

(3x-5) (x+3) = 0(3x-5) = 0 (x+3) = 0

3x = 5X = 5/3

x = -3x = -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 in2. What should the outer dimensions of the frame be?

SOLUTION:

2x +11

2x +17A = b h

A = (2x +11)(2x +17)315 = (2x +11)(2x +17)

Adding an x to both sides of the picture we get:

315 = 4x2 + 56x + 187 FOIL

SOLUTION:

4(x+16)(x-2) = 0

To solve we now put it in the ax2+bx+c = 0 form:

315 = 4x2 + 56x + 1874x2 + 56x + 187-315 = 04x2 + 56x -128 = 04(x2 + 14x -32) = 0

(x+16)= 0 (x-2) = 0x = -16 x = 2

VIDEOS: SOLVING BY

FACTORING Solving by factoring:

http://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_formula_tutorial/v/solving-quadratic-equations-by-completing-the-square

http://www.khanacademy.org/math/algebra/quadratics/factoring_quadratics/v/Example%201:%20Solving%20a%20quadratic%20equation%20by%20factoring

VIDEOS: SOLVING BY

FACTORING Solving by factoring:

http://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_formula_tutorial/v/solving-quadratic-equations-by-completing-the-square

http://www.khanacademy.org/math/algebra/quadratics/factoring_quadratics/v/Example%202:%20Solving%20a%20quadratic%20equation%20by%20factoring

CLASSWORK:

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Problems: 4, 7, 15, 21, 25, 36, 37.