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# I. Review Distribute (aka F.O.I.L.) - PowerPoint PPT Presentation

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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### I. Review Distribute (aka F.O.I.L.)

( x + 3 ) (x + 2 )

x2

Outside

( x + 3 ) (x + 2 )

x2+ 2x

Outside

( x + 3 ) (x + 2 )

Inside

x2+ 2x + 3x

Outside

( x + 3 ) (x + 2 )

Inside

Last

x2+ 2x + 3x + 6

Outside

( x + 3 ) (x + 2 )

Inside

Last

x2+ 2x + 3x + 6

x2+ 5x + 6

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

6x2

Outside

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

6x2+ 3x

Outside

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

Inside

6x2+ 3x + 4x

Outside

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

Inside

Last

6x2+ 3x + 4x + 2

Outside

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

Inside

Last

6x2+ 3x + 4x+ 2

6x2+ 7x + 2

### II. Review Trial and Error

x2 + 5x + 6

x2 + 5x + 6

1x2 + 5x + 6

x2 + 5x + 6

1x2 + 5x + 6

(1x ____ ) (1x ____ )

x2 + 5x + 6

1x2 + 5x + 6

(1x – 1) (1x + 6 )

x2 + 5x + 6

1x2 + 5x + 6

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

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

x2 + 5x + 6

x2 + 5x + 6

1x2 + 5x + 6

x2 + 5x + 6

1x2 + 5x + 6

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

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!

3x2 – 4x – 4

3x2 – 4x – 4

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

3x2 – 4x – 4

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

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

3x2 – 4x – 4

3x2 – 4x – 4

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

3x2 – 4x – 4

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

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

3x2 – 4x – 4

3x2 – 4x – 4

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

3x2 – 4x – 4

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

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

– 6x2 – x + 2

– 6x2 – x + 2

Wow! How are we going to factor this??

– 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.

### III. Factoring by Grouping

x2 + 5x + 6

x2 + 5x + 6

x2 + 2x + 3x + 6

x2 + 5x + 6

x2 + 2x + 3x + 6

x2 + 2x + 3x + 6

x2 + 5x + 6

x2 + 2x + 3x + 6

x2 + 2x + 3x + 6

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

x2 + 5x + 6

x2 + 2x + 3x + 6

x2 + 2x + 3x + 6

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

(x + 2) (x + 3)

3x2 – 4x – 4

3x2 – 4x – 4

3x2 – 6x + 2x – 4

3x2 – 4x – 4

3x2 – 6x + 2x – 4

3x2 – 6x + 2x – 4

3x2 – 4x – 4

3x2 – 6x + 2x – 4

3x2 – 6x + 2x – 4

3x(x – 2) 2(x – 2)

3x2 – 4x – 4

3x2 – 6x + 2x – 4

3x2 – 6x + 2x – 4

3x(x – 2) + 2(x – 2)

(x – 2) (3x + 2)

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 . . .

3x2 – 4x – 4

3x2 – 4x – 4

3x2 + 2x – 6x – 4

3x2 – 4x – 4

3x2 + 2x – 6x – 4

3x2 + 2x – 6x – 4

3x2 – 4x – 4

3x2 + 2x – 6x – 4

3x2 + 2x – 6x – 4

x(3x + 2) – 2(3x + 2)

3x2 – 4x – 4

3x2 + 2x – 6x – 4

3x2 + 2x – 6x – 4

x(3x + 2) – 2(3x + 2)

(3x + 2) (x – 2)

– 6x2 – x + 2

– 6x2 – x + 2

Oh, we are back to this problem again!

– 6x2 – x + 2

Oh, we are back to this problem again!

Do you think Trial & Error or Grouping

will work easily on this problem?

– 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?

– 6x2 – x + 2

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

### IV. Factoring Quadratics

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.

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).

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

to equal the middle term.

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.

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 ( )( ).

(x + 3) (x + 5)

Divide only the numbers in the ( )( )

by the positive coefficient of x2.

Divide out common terms.

(aka: Simplify fractions.)

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.

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}.

x2+ 5x + 6

1x2+ 5x + 6

1x2+ 5x + 6

Step 1: No GCF

1x2+ 5x + 6

Step 1: No GCF

Step 2:

1 * 6 = 6

1 * 6

2 * 3

1x2+ 5x + 6

Step 1: No GCF

Step 2:Step 3:

1 * 6 = 6 + 6 +3

1 * 6 – 1+2

2 * 3 + 5 +5

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.

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.

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.

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.

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.

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

x + 2 = 0 x + 3 = 0

x = 0 – 2 x = 0 – 3

x = – 2 x = – 3

x = {– 2, – 3}

x2 – 5x – 6

1x2 – 5x – 6

1x2 – 5x – 6

Step 1: No GCF

1x2 – 5x – 6

Step 1: No GCF

Step 2:

1 * 6 = 6

1 * 6

2 * 3

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.

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.

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.

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.

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.

(x – 6) (x + 1) = 0

x – 6 = 0 x + 1 = 0

x = 0 + 6 x = 0 – 1

x = + 6 x = – 1

x = {+ 6, – 1}

2x2 + 7x – 4

2x2 + 7x – 4

Step 1: No GCF

2x2 + 7x – 4

Step 1: No GCF

Step 2:

2 * 4 = 8

1 * 8

2 * 4

2x2 + 7x – 4

Step 1: No GCF

Step 2:Step 3:

2 * 4 = 8 + 8

1 * 8 – 1

2 * 4 + 7

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.

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.

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.

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.

(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 }

practice problems. . .

x2 + 4x + 3

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 + #?

x2 + 4x + 3

x2+3x +1x + 3

• x2 + 4x + 3

• x2+3x +1x + 3

• Now separate the x2 into x • x

• Put each x into ( ) ( )

• Put the two red #s into ( ) ( )

1x2 + 4x + 3

x2+3x +1x + 3

( x +3) ( x +1)

Now, divide by the coefficient of x2.

1x2 + 4x + 3

x2+3x +1x + 3

( x +3) ( x +1)

1 1

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.

the Factored Form of

x2 + 4x + 3

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

3x2 - 4x – 4

3x2 - 4x – 4

(x + 2) (x – 6)

3x2 - 4x – 4

(x + 2) (x – 6)

(x + 2) (x – 6)

3 3

3x2 - 4x – 4

(x + 2) (x – 6)

(x + 2) (x – 6)

3 3

(3x + 2) (x – 2)

3x2 - 4x – 4

(x + 2) (x – 6)

(x + 2) (x – 6)

3 3

(3x + 2) (x – 2)

You are now ready for

The Problem ! ! ! ! !

– 6x2 – x + 2

– 6x2 – x + 2

Step 1A: No GCF

– 6x2 – x + 2

Step 1A: No GCF

Step 1B: – 1(6x2 + x – 2)

– 6x2 – x + 2

Step 1A: No GCF

Step 1B: – 1(6x2 + x – 2)

Step 2:

6 * 2 = 12

1 * 12

2 * 6

3 * 4

– 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

– 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.

– 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.

– 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.

– 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.

(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

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

x practice or as homework.2 + 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