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I. Review Distribute (aka F.O.I.L.) II. Review Trial and Error III. Review Grouping III. Factoring Quadratics. I. Review Distribute (aka F.O.I.L.). ( x + 3 ) (x + 2 ). First. ( x + 3 ) (x + 2 ). x 2. First. Outside. ( x + 3 ) (x + 2 ). x 2 + 2x. First. Outside.

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slide1

I. Review Distribute (aka F.O.I.L.)II. Review Trial and ErrorIII. Review GroupingIII. Factoring Quadratics

slide4
First

( x + 3 ) (x + 2 )

x2

slide5
First

Outside

( x + 3 ) (x + 2 )

x2+ 2x

slide6
First

Outside

( x + 3 ) (x + 2 )

Inside

x2+ 2x + 3x

slide7
First

Outside

( x + 3 ) (x + 2 )

Inside

Last

x2+ 2x + 3x + 6

slide8
First

Outside

( x + 3 ) (x + 2 )

Inside

Last

x2+ 2x + 3x + 6

x2+ 5x + 6

slide10
First

( 3x + 2 ) (2x + 1 )

6x2

slide11
First

Outside

( 3x + 2 ) (2x + 1 )

6x2+ 3x

slide12
First

Outside

( 3x + 2 ) (2x + 1 )

Inside

6x2+ 3x + 4x

slide13
First

Outside

( 3x + 2 ) (2x + 1 )

Inside

Last

6x2+ 3x + 4x + 2

slide14
First

Outside

( 3x + 2 ) (2x + 1 )

Inside

Last

6x2+ 3x + 4x+ 2

6x2+ 7x + 2

slide17
Factoring Steps for Trial and Error

x2 + 5x + 6

1x2 + 5x + 6

slide18
Factoring Steps for Trial and Error

x2 + 5x + 6

1x2 + 5x + 6

(1x ____ ) (1x ____ )

slide19
Factoring Steps for Trial and Error

x2 + 5x + 6

1x2 + 5x + 6

(1x – 1) (1x + 6 )

slide20
Factoring Steps for Trial and Error

x2 + 5x + 6

1x2 + 5x + 6

(1x – 1) (1x + 6 ) F.O.I.L. to check answer

slide21
Factoring Steps for Trial and Error

x2 + 5x + 6

1x2 + 5x + 6

(1x – 1) (1x + 6 ) F.O.I.L. to check answer

(1x – 1) (1x + 6 ) = x2 + 5x – 6 Error, Try Again

slide23
Factoring Steps for Trial and Error

x2 + 5x + 6

1x2 + 5x + 6

slide24
Factoring Steps for Trial and Error

x2 + 5x + 6

1x2 + 5x + 6

(1x + 2) (1x + 3) F.O.I.L. to check answer

slide25
Factoring Steps for Trial and Error

x2 + 5x + 6

1x2 + 5x + 6

(1x + 2) (1x + 3) F.O.I.L. to check answer

(x + 2) (x + 3) = x2 + 5x + 6 Correct!

slide27
Factoring Steps for Trial and Error

3x2 – 4x – 4

(3x + 1) (1x – 4 ) F.O.I.L. to check answer

slide28
Factoring Steps for Trial and Error

3x2 – 4x – 4

(3x + 1) (1x – 4 ) F.O.I.L. to check answer

(3x + 1) (1x – 4) = 3x2 – 11x – 4 Error, Try Again

slide30
Factoring Steps for Trial and Error

3x2 – 4x – 4

(3x – 1) (1x + 4) F.O.I.L. to check answer

slide31
Factoring Steps for Trial and Error

3x2 – 4x – 4

(3x – 1) (1x + 4) F.O.I.L. to check answer

(3x – 1) (1x + 4) = 3x2 + 11x – 4 Error, Try Again

slide33
Factoring Steps for Trial and Error

3x2 – 4x – 4

(3x + 2) (1x – 2 ) F.O.I.L. to check answer

slide34
Factoring Steps for Trial and Error

3x2 – 4x – 4

(3x + 2) (1x – 2 ) F.O.I.L. to check answer

(3x + 2) (1x – 2) = 3x2 – 4x – 4 Correct!

slide36
Factoring Steps for Trial and Error

– 6x2 – x + 2

Wow! How are we going to factor this??

slide37
Factoring Steps for Trial and Error

– 6x2 – x + 2

Wow! How are we going to factor this??

Well, let’s try another method and see if that

will help with this problem.

We can try Factoring by Grouping.

slide40
Factoring Steps for Grouping

x2 + 5x + 6

x2 + 2x + 3x + 6

slide41
Factoring Steps for Grouping

x2 + 5x + 6

x2 + 2x + 3x + 6

x2 + 2x + 3x + 6

slide42
Factoring Steps for Grouping

x2 + 5x + 6

x2 + 2x + 3x + 6

x2 + 2x + 3x + 6

x(x + 2) +3(x + 2)

slide43
Factoring Steps for Grouping

x2 + 5x + 6

x2 + 2x + 3x + 6

x2 + 2x + 3x + 6

x(x + 2) +3(x + 2)

(x + 2) (x + 3)

slide45
Factoring Steps for Grouping

3x2 – 4x – 4

3x2 – 6x + 2x – 4

slide46
Factoring Steps for Grouping

3x2 – 4x – 4

3x2 – 6x + 2x – 4

3x2 – 6x + 2x – 4

slide47
Factoring Steps for Grouping

3x2 – 4x – 4

3x2 – 6x + 2x – 4

3x2 – 6x + 2x – 4

3x(x – 2) 2(x – 2)

slide48
Factoring Steps for Grouping

3x2 – 4x – 4

3x2 – 6x + 2x – 4

3x2 – 6x + 2x – 4

3x(x – 2) + 2(x – 2)

(x – 2) (3x + 2)

slide49
Factoring Steps for Grouping

3x2 – 4x – 4

3x2 – 6x + 2x – 4

3x2 – 6x + 2x – 4

3x(x – 2) 2(x – 2)

(x – 2) (3x + 2)

Or it can be factored this way . . .

slide51
Factoring Steps for Grouping

3x2 – 4x – 4

3x2 + 2x – 6x – 4

slide52
Factoring Steps for Grouping

3x2 – 4x – 4

3x2 + 2x – 6x – 4

3x2 + 2x – 6x – 4

slide53
Factoring Steps for Grouping

3x2 – 4x – 4

3x2 + 2x – 6x – 4

3x2 + 2x – 6x – 4

x(3x + 2) – 2(3x + 2)

slide54
Factoring Steps for Grouping

3x2 – 4x – 4

3x2 + 2x – 6x – 4

3x2 + 2x – 6x – 4

x(3x + 2) – 2(3x + 2)

(3x + 2) (x – 2)

slide55
Factoring Steps for Grouping

– 6x2 – x + 2

slide56
Factoring Steps for Grouping

– 6x2 – x + 2

Oh, we are back to this problem again!

slide57
Factoring Steps for Grouping

– 6x2 – x + 2

Oh, we are back to this problem again!

Do you think Trial & Error or Grouping

will work easily on this problem?

slide58
Factoring Steps for Grouping

– 6x2 – x + 2

If you don’t think so, most people would agree with you.

What if I told you there was a different way that would work for the simple problems as well

as for problems like the one above?

slide59
Factoring Steps for Grouping

– 6x2 – x + 2

Well, let’s try the “Front times the Back” Method!!

slide61
Factoring Steps

1A. Factor out GCF, if necessary.

1B. Make x2 term positive, if necessary.

• Carry the GCF and / or – sign and put in front of ( ) ( ) in answer.

slide62
Factoring Steps

Do “Front times the Back” and get an answer.

“Front times the Back” means to multiply coefficient

of the x2 term (NUMBER ONLY NOT SIGN)

and the constant (NUMBER ONLY NOT SIGN).

slide63
Factoring Steps

3. Find the factors of the “Front times the Back’s”

answer that will add or subtract

to equal the middle term.

slide64
When these factors are multiplied,

the sign of that answer

must equal the

sign of the constant.

This step is very important because it

will eliminate the extraneous (extra) solutions.

slide65
5A. Separate the x2 into x • x.

5B. Write one x in the front of each ( )( ).

5C. Then write the factors in the back of each ( )( ).

example of answer:

(x + 3) (x + 5)

slide66
Divide only the numbers in the ( )( )

by the positive coefficient of x2.

Divide out common terms.

(aka: Simplify fractions.)

slide67
If there is a whole # left, that ( ) is finished.

If there is a fraction left, the denominator becomes the coefficient of x and the numerator is now the constant in the ( ).

Example: (x + 2) (x + ¾)

(x+2) ( 4x + 3) = answer

This is now in factored form.

slide68
TO SOLVE FOR X:

Set each ( ) equal to 0 and solve for x.

Example: (x + 3) ( x + 5) = 0

x + 3 = 0 x + 5 = 0

x = 0 – 3 x = 0 – 5

x = – 3 x = – 5

If x equals – 3 or – 5, then at least one factor will be equal to 0 which will cause the left side to be 0. So 0 = 0 is true and the answer is x = {-3, -5}.

slide72
Actual Work Shown Mental Work

1x2+ 5x + 6

Step 1: No GCF

slide73
Actual Work Shown Mental Work

1x2+ 5x + 6

Step 1: No GCF

Step 2:

1 * 6 = 6

1 * 6

2 * 3

slide74
Actual Work Shown Mental Work

1x2+ 5x + 6

Step 1: No GCF

Step 2:Step 3:

1 * 6 = 6 + 6 +3

1 * 6 – 1+2

2 * 3 + 5 +5

slide75
Actual Work Shown Mental Work

1x2+ 5x + 6

Step 1: No GCF

Step 2:Step 3:

1 * 6 = 6 + 6 +3

1 * 6 – 1+2

2 * 3 + 5 +5

Even though there are two ways to get +5, only one set of factors will work.

slide76
Actual Work Shown Mental Work

1x2+ 5x + 6

Step 1: No GCF

Step 2:Step 3:

1 * 6 = 6 + 6 +3

1 * 6 – 1+2

2 * 3 + 5 +5

Even though there are two ways to get +5, only one set of factors will work.

Step 4:

(+6)(–1) = – #

(+3)(+2) = + #, and the last term is a + #, so only the +3 and the +2 will work.

slide77
Actual Work Shown Mental Work

1x2+ 5x + 6

Step 1: No GCF

Step 5: (x + 2) (x + 3)

Step 2:Step 3:

1 * 6 = 6 + 6 +3

1 * 6 – 1+2

2 * 3 + 5 +5

Even though there are two ways to get +5, only one set of factors will work.

Step 4:

(+6)(–1) = – #

(+3)(+2) = + #, and the last term is a + #, so only the +3 and the +2 will work.

slide78
Actual Work Shown Mental Work

1x2+ 5x + 6

Step 1: No GCF

Step 5: (x + 2) (x + 3)

Step 6: (x + 2) (x + 3)

1 1

Step 2:Step 3:

1 * 6 = 6 + 6 +3

1 * 6 – 1+2

2 * 3 + 5 +5

Even though there are two ways to get +5, only one set of factors will work.

Step 4:

(+6)(–1) = – #

(+3)(+2) = + #, and the last term is a + #, so only the +3 and the +2 will work.

slide79
Actual Work Shown Mental Work

1x2+ 5x + 6

Step 1: No GCF

Step 5: (x + 2) (x + 3)

Step 6: (x + 2) (x + 3)

1 1

Step 7-9: (x + 2) (x + 3)

Step 2:Step 3:

1 * 6 = 6 + 6 +3

1 * 6 – 1+2

2 * 3 + 5 +5

Even though there are two ways to get +5, only one set of factors will work.

Step 4:

(+6)(–1) = – #

(+3)(+2) = + #, and the last term is a + #, so only the +3 and the +2 will work.

slide80
To Solve for x:

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

x + 2 = 0 x + 3 = 0

x = 0 – 2 x = 0 – 3

x = – 2 x = – 3

x = {– 2, – 3}

slide83
Actual Work Shown Mental Work

1x2 – 5x – 6

Step 1: No GCF

slide84
Actual Work Shown Mental Work

1x2 – 5x – 6

Step 1: No GCF

Step 2:

1 * 6 = 6

1 * 6

2 * 3

slide85
Actual Work Shown Mental Work

1x2 – 5x – 6

Step 1: No GCF

Step 2:Step 3:

1 * 6 = 6– 6 – 3

1 * 6 + 1– 2

2 * 3 – 5 – 5

Even though there are two ways to get – 5, only one set of factors will work.

slide86
Actual Work Shown Mental Work

1x2 – 5x – 6

Step 1: No GCF

Step 2:Step 3:

1 * 6 = 6– 6 – 3

1 * 6 + 1– 2

2 * 3 – 5 – 5

Even though there are two ways to get – 5, only one set of factors will work.

Step 4:

(– 6)(+ 1) = – #

(– 3)(– 2) = + #, and the last term is a – #, so only the – 6 and the + 1 will work.

slide87
Actual Work Shown Mental Work

1x2 – 5x – 6

Step 1: No GCF

Step 5: (x – 6) (x + 1)

Step 2:Step 3:

1 * 6 = 6– 6 – 3

1 * 6 + 1– 2

2 * 3 – 5 – 5

Even though there are two ways to get – 5, only one set of factors will work.

Step 4:

(– 6)(+ 1) = – #

(– 3)(– 2) = + #, and the last term is a – #, so only the – 6 and the + 1 will work.

slide88
Actual Work Shown Mental Work

1x2 – 5x – 6

Step 1: No GCF

Step 5: (x – 6) (x + 1)

Step 6: (x – 6) (x + 1)

1 1

Step 2:Step 3:

1 * 6 = 6– 6 – 3

1 * 6 + 1– 2

2 * 3 – 5 – 5

Even though there are two ways to get – 5, only one set of factors will work.

Step 4:

(– 6)(+ 1) = – #

(– 3)(– 2) = + #, and the last term is a – #, so only the – 6 and the + 1 will work.

slide89
Actual Work Shown Mental Work

1x2 – 5x – 6

Step 1: No GCF

Step 5: (x – 6) (x + 1)

Step 6: (x – 6) (x + 1)

1 1

Step 7-9:(x – 6) (x + 1)

Step 2:Step 3:

1 * 6 = 6– 6 – 3

1 * 6 + 1– 2

2 * 3 – 5 – 5

Even though there are two ways to get – 5, only one set of factors will work.

Step 4:

(– 6)(+ 1) = – #

(– 3)(– 2) = + #, and the last term is a – #, so only the – 6 and the + 1 will work.

slide90
To Solve for x:

(x – 6) (x + 1) = 0

x – 6 = 0 x + 1 = 0

x = 0 + 6 x = 0 – 1

x = + 6 x = – 1

x = {+ 6, – 1}

slide92
Actual Work Shown Mental Work

2x2 + 7x – 4

Step 1: No GCF

slide93
Actual Work Shown Mental Work

2x2 + 7x – 4

Step 1: No GCF

Step 2:

2 * 4 = 8

1 * 8

2 * 4

slide94
Actual Work Shown Mental Work

2x2 + 7x – 4

Step 1: No GCF

Step 2:Step 3:

2 * 4 = 8 + 8

1 * 8 – 1

2 * 4 + 7

slide95
Actual Work Shown Mental Work

2x2 + 7x – 4

Step 1: No GCF

Step 2:Step 3:

2 * 4 = 8 + 8

1 * 8 – 1

2 * 4 + 7

Step 4:

Only the +8 added to the – 1 will equal +7.

And since (+ 8)(– 1) = – # , this verifies these are the two factors needed.

slide96
Actual Work Shown Mental Work

2x2 + 7x – 4

Step 1: No GCF

Step 5: (x + 8) (x – 1)

Step 2:Step 3:

2 * 4 = 8 + 8

1 * 8 – 1

2 * 4 + 7

Step 4:

Only the +8 added to the – 1 will equal +7.

And since (+ 8)(– 1) = – # , this verifies these are the two factors needed.

slide97
Actual Work Shown Mental Work

2x2 + 7x – 4

Step 1: No GCF

Step 5: (x + 8) (x – 1)

Step 6: (x + 8) (x – 1)

2 2

Step 2:Step 3:

2 * 4 = 8 + 8

1 * 8 – 1

2 * 4 + 7

Step 4:

Only the +8 added to the – 1 will equal +7.

And since (+ 8)(– 1) = – # , this verifies these are the two factors needed.

slide98
Actual Work Shown Mental Work

2x2 + 7x – 4

Step 1: No GCF

Step 5: (x + 8) (x – 1)

Step 6: (x + 8) (x – 1)

2 2

Step 7-9: (x + 4) (2x – 1)

Step 2:Step 3:

2 * 4 = 8 + 8

1 * 8 – 1

2 * 4 + 7

Step 4:

Only the +8 added to the – 1 will equal +7.

And since (+ 8)(– 1) = – # , this verifies these are the two factors needed.

slide99
To Solve for x:

(x + 4) (2x – 1)

x + 4 = 0 2x – 1= 0

x = 0 – 4 2x = 0 + 1

x = – 4 2x = 1

x = ½

x = { – 4, ½ }

slide100
Now here are some

practice problems. . .

slide102
x2 + 4x + 3

Step 1: No GCF

Step 2: “Front times the Back”

Step 3A: What factors of 3 add to = +4? AND

Step 3B: Do signs multiply to equal a + #?

slide103
x2 + 4x + 3

x2+3x +1x + 3

slide104
x2 + 4x + 3
  • x2+3x +1x + 3
  • Now separate the x2 into x • x
  • Put each x into ( ) ( )
  • Put the two red #s into ( ) ( )
slide105
1x2 + 4x + 3

x2+3x +1x + 3

( x +3) ( x +1)

Now, divide by the coefficient of x2.

slide106
1x2 + 4x + 3

x2+3x +1x + 3

( x +3) ( x +1)

1 1

slide107
1x2 + 4x + 3

x2+3x +1x + 3

( x +3) ( x +1)

1 1

Since dividing by 1 does not change the numbers, we have discovered that if the coefficient of x2 is 1, then we can eliminate the dividing by the x2 coefficient step.

slide108
Therefore,

the Factored Form of

x2 + 4x + 3

is (x+3) (x+1).

slide111
3x2 - 4x – 4

(x + 2) (x – 6)

slide112
3x2 - 4x – 4

(x + 2) (x – 6)

(x + 2) (x – 6)

3 3

slide113
3x2 - 4x – 4

(x + 2) (x – 6)

(x + 2) (x – 6)

3 3

(3x + 2) (x – 2)

slide114
3x2 - 4x – 4

(x + 2) (x – 6)

(x + 2) (x – 6)

3 3

(3x + 2) (x – 2)

You are now ready for

The Problem ! ! ! ! !

slide116
Actual Work Shown Mental Work

– 6x2 – x + 2

Step 1A: No GCF

slide117
Actual Work Shown Mental Work

– 6x2 – x + 2

Step 1A: No GCF

Step 1B: – 1(6x2 + x – 2)

slide118
Actual Work Shown Mental Work

– 6x2 – x + 2

Step 1A: No GCF

Step 1B: – 1(6x2 + x – 2)

Step 2:

6 * 2 = 12

1 * 12

2 * 6

  3 * 4

slide119
Actual Work Shown Mental Work

– 6x2 – x + 2

Step 1A: No GCF

Step 1B: – 1(6x2 + x – 2)

Step 2:Step 3:

6 * 2 = 12 + 4

1 * 12 – 3

2 * 6 + 1

  3 * 4

slide120
Actual Work Shown Mental Work

– 6x2 – x + 2

Step 1A: No GCF

Step 1B: – 1(6x2 + x – 2)

Step 2:Step 3:

6 * 2 = 12 + 4

1 * 12 – 3

2 * 6 + 1

  3 * 4

Step 4:

Only the +4 added to the – 3 will equal +1.

And since (+ 4)(– 3) = – # , this verifies these are the two factors needed.

slide121
Actual Work Shown Mental Work

– 6x2 – x + 2

Step 1A: No GCF

Step 1B: – 1(6x2 + x – 2)

Step 5: – 1 (x + 4) (x – 3)

Step 2:Step 3:

6 * 2 = 12 + 4

1 * 12 – 3

2 * 6 + 1

  3 * 4

Step 4:

Only the +4 added to the – 3 will equal +1.

And since (+ 4)(– 3) = – # , this verifies these are the two factors needed.

slide122
Actual Work Shown Mental Work

– 6x2 – x + 2

Step 1A: No GCF

Step 1B: – 1(6x2 + x – 2)

Step 5: – 1 (x + 4) (x – 3)

Step 6: – 1 (x + 4) (x – 3)

2 2

Step 2:Step 3:

6 * 2 = 12 + 4

1 * 12 – 3

2 * 6 + 1

  3 * 4

Step 4:

Only the +4 added to the – 3 will equal +1.

And since (+ 4)(– 3) = – # , this verifies these are the two factors needed.

slide123
Actual Work Shown Mental Work

– 6x2 – x + 2

Step 1A: No GCF

Step 1B: – 1(6x2 + x – 2)

Step 5: – 1 (x + 4) (x – 3)

Step 6: – 1 (x + 4) (x – 3)

6 6

Step 7-9: – 1(3x + 2) (2x – 1)

Step 2:Step 3:

6 * 2 = 12 + 4

1 * 12 – 3

2 * 6 + 1

  3 * 4

Step 4:

Only the +4 added to the – 3 will equal +1.

And since (+ 4)(– 3) = – # , this verifies these are the two factors needed.

slide124
Now try these problems . . .

(you may use GeoGebra).

x2 + 4x + 3

x2 + 7x + 6

x2 + 9x + 8

x2 + 10x + 9

x2 + 2x − 3

x2 + 5x − 6

x2 + 7x − 8

x2 + 8x − 9

slide125
The problems on this page and the next can be used as extra practice or as homework.

x2 − 2x − 3

x2 − 5x − 6

x2 − 7x − 8

x2 − 8x − 9

x2 − 4x + 3

x2 − 7x + 6

x2 − 9x + 8

x2 − 10x + 9

slide126
x2 + 4x + 3

x2 + 7x + 6

x2 + 9x + 8

x2 + 10x + 9

x2 + 2x − 3

x2 + 5x − 6

x2 + 7x − 8

x2 + 8x − 9

x2 − 2x − 3

x2 − 5x − 6

x2 − 7x − 8

x2 − 8x − 9

x2 − 4x + 3

x2 − 7x + 6

x2 − 9x + 8

x2 − 10x + 9

ad