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9.4: FACTORING TO SOLVE ax 2 + bx + c = 0

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Factoring: A process used to break down any polynomial into simpler polynomials. 9.4: FACTORING TO SOLVE ax 2 + bx + c = 0. Zero-Product Property: For any real numbers a and b, If a b = 0 Then a = 0 or b = 0. Procedure:. 1) Always look for the GCF of all the terms. - PowerPoint PPT Presentation
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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.
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Page 1: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

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.

Page 2: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 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.

Page 3: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

GOAL:

Page 4: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

SOLVING BY FACTORING:

Ex: What are the solutions of:

x2+5x+6?

Page 5: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

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)

Page 6: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

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.

Page 7: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

SOLVING BY FACTORING:

Ex: What is the solutions of the equation:

5x2+11x+2?

Page 8: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

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

Page 9: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

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)

Page 10: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

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.

Page 11: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

YOU TRY IT:

Ex: What are the solutions of:

6x2+13x+5?

Page 12: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

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

Page 13: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

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)

Page 14: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

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.

Page 15: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

YOU TRY IT:

Ex: What are the solutions of:

3x2+4x = 15?

Page 16: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

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

Page 17: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

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)

Page 18: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

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.

Page 19: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

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?

Page 20: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

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

Page 21: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

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

Page 24: 9.4: FACTORING TO SOLVE ax 2  +  bx  + c = 0

CLASSWORK:

Page 508-509:

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


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