5.4 factoring ax 2 + bx +c 12/10/2012. in the previous section we learned to factor x 2 + bx + c...

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5.4 Factoring ax 2 + bx +c 12/10/2012

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Page 1: 5.4 Factoring ax 2 + bx +c 12/10/2012. In the previous section we learned to factor x 2 + bx + c where a = 1. In this section, we’re going to factor ax

5.4Factoring ax2 + bx +c

12/10/2012

Page 2: 5.4 Factoring ax 2 + bx +c 12/10/2012. In the previous section we learned to factor x 2 + bx + c where a = 1. In this section, we’re going to factor ax

In the previous section we learned to factor x2 + bx + c where a = 1.

In this section, we’re going to factor ax2 + bx + c where a ≠ 1.

Ex: Factor 3x2 + 7x +2

We’re still going to use the “Big X” Method.

Page 3: 5.4 Factoring ax 2 + bx +c 12/10/2012. In the previous section we learned to factor x 2 + bx + c where a = 1. In this section, we’re going to factor ax

The Big “X” method

a•c

b

Think of 2 numbers that Multiply to a•c and Add to b

#1 #2

add

multiply

Answer: Write the simplified answers in the 2 ( ). Top # is coefficient of x and bottom # is the 2nd term

Factor: ax2 + bx + c

a aSimplify like a fraction if needed

Simplify like a fraction if needed

Page 4: 5.4 Factoring ax 2 + bx +c 12/10/2012. In the previous section we learned to factor x 2 + bx + c where a = 1. In this section, we’re going to factor ax

3•2 = 6

7

Think of 2 numbers that Multiply to 6 and Add to 7

6 x 1 = 66 + 1 = 76 1

Answer: (1x + 2) (3x + 1) or (x + 2) (3x + 1)

Factor: 3x2 + 7x + 2

a•c

b

#1 #2

add

multiply

3 3Simplify like a fraction . ÷ by 3

2

1

a a

Page 5: 5.4 Factoring ax 2 + bx +c 12/10/2012. In the previous section we learned to factor x 2 + bx + c where a = 1. In this section, we’re going to factor ax

Checkpoint

1.

Factor the expression.

Factor when c is Positiveax 2 + bx + c

2x 2 + 11x + 5

2. 2y 2 + 9y + 7

3. 3r 2 + 8r + 5

ANSWER ( )1+2x ( )5+x

ANSWER ( )7+2y ( )1+y

ANSWER ( )5+3r ( )1+r

Page 6: 5.4 Factoring ax 2 + bx +c 12/10/2012. In the previous section we learned to factor x 2 + bx + c where a = 1. In this section, we’re going to factor ax

4(-9) = -36

-16

Think of 2 numbers that Multiply to -36 and Add to -16

-18 x 2 = -36 -18 + 2 = -16

-18 2

Answer: (2x - 9) (2x + 1)

Factor: 4x2 - 16x - 9

a•c

b

#1 #2

add

multiply

4 4Simplify like a fraction . ÷ by 2

-9

2

a a1

2 Simplify like a fraction . ÷ by 2

Page 7: 5.4 Factoring ax 2 + bx +c 12/10/2012. In the previous section we learned to factor x 2 + bx + c where a = 1. In this section, we’re going to factor ax

Do 6, 27 and -15 have any factors in common?Yes, 3. Factor 3 out.3(2x2 + 9x – 5). Then Factor what’s in the ( ).

2(-5) = -10

9

Think of 2 numbers that Multiply to -10 and Add to 9 -1 x 10 = -10 -1 + 10 = 9

-1 10

Answer: 3(2x - 1) (x + 5) (Don’t forget the 3!!!)

Factor: 6x2 + 27x - 15

a•c

b

#1 #2

add

multiply

2 2

a a5

1 Simplify like a fraction . ÷ by 2

Page 8: 5.4 Factoring ax 2 + bx +c 12/10/2012. In the previous section we learned to factor x 2 + bx + c where a = 1. In this section, we’re going to factor ax

Checkpoint

Factor the expression.

Factor ax 2 bx+ c+

6. 4w 2 6w 2+–

4. 6z 2 z+ 12– ANSWER ( )4 +3z ( )32z–

5. 11x 2 17x 6+ +

ANSWER ( )1–2w ( )1w –2

ANSWER ( )6+11x ( )1+x

Page 9: 5.4 Factoring ax 2 + bx +c 12/10/2012. In the previous section we learned to factor x 2 + bx + c where a = 1. In this section, we’re going to factor ax

Is the same as solving ax2+bx+c = 0

Graphically, finding the zeros of the quadratic function means finding the x-

intercepts of the parabola.

Finding the Zeros of the Function

Page 10: 5.4 Factoring ax 2 + bx +c 12/10/2012. In the previous section we learned to factor x 2 + bx + c where a = 1. In this section, we’re going to factor ax

Find the Zeros of a Quadratic FunctionExample 4

Find the zeros of x 23y 4.= – x –

x 230 4= – x – Let y 0.=

Factor the right side.( )3x 4–0 ( )x 1+=

3x 4– = 0 or x 1+ = 0 Use the zero product property.

Write original function.x 23 y 4= – x –

SOLUTION

To find the zeros of the function, let y = 0. Then solve for x.

3

4x = x = 1– Solve for x.

Page 11: 5.4 Factoring ax 2 + bx +c 12/10/2012. In the previous section we learned to factor x 2 + bx + c where a = 1. In this section, we’re going to factor ax

Find the Zeros of a Quadratic FunctionExample 4

ANSWER

The zeros of the function are3

4and 1.–

The zeros of a function are

also the x-intercepts of the

graph of the function. So,

the answer can be checked

by graphing

The x-intercepts of the

graph are and , so the

answer is correct.

x 23y 4.= – x –

3

4 1–

CHECK

Page 12: 5.4 Factoring ax 2 + bx +c 12/10/2012. In the previous section we learned to factor x 2 + bx + c where a = 1. In this section, we’re going to factor ax

Checkpoint Find the Zeros of a Quadratic Function

Find the zeros of the function.

ANSWER , 32

1

7. y = x 23 1– 2x – ANSWER , 13

1–

8. y = x 22 3– 7x +

ANSWER , 42

19. y = x 24 8– 18x +

Page 13: 5.4 Factoring ax 2 + bx +c 12/10/2012. In the previous section we learned to factor x 2 + bx + c where a = 1. In this section, we’re going to factor ax

Homework

5.4 p.244 #18-25, 46-48, 57-59