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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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I review distribute aka f o i l

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


I review distribute aka f o i l

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


I review distribute aka f o i l

( x + 3 ) (x + 2 )


I review distribute aka f o i l

First

( x + 3 ) (x + 2 )

x2


I review distribute aka f o i l

First

Outside

( x + 3 ) (x + 2 )

x2+ 2x


I review distribute aka f o i l

First

Outside

( x + 3 ) (x + 2 )

Inside

x2+ 2x + 3x


I review distribute aka f o i l

First

Outside

( x + 3 ) (x + 2 )

Inside

Last

x2+ 2x + 3x + 6


I review distribute aka f o i l

First

Outside

( x + 3 ) (x + 2 )

Inside

Last

x2+ 2x + 3x + 6

x2+ 5x + 6


I review distribute aka f o i l

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


I review distribute aka f o i l

First

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

6x2


I review distribute aka f o i l

First

Outside

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

6x2+ 3x


I review distribute aka f o i l

First

Outside

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

Inside

6x2+ 3x + 4x


I review distribute aka f o i l

First

Outside

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

Inside

Last

6x2+ 3x + 4x + 2


I review distribute aka f o i l

First

Outside

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

Inside

Last

6x2+ 3x + 4x+ 2

6x2+ 7x + 2


Ii review trial and error

II. Review Trial and Error


I review distribute aka f o i l

Factoring Steps for Trial and Error

x2 + 5x + 6


I review distribute aka f o i l

Factoring Steps for Trial and Error

x2 + 5x + 6

1x2 + 5x + 6


I review distribute aka f o i l

Factoring Steps for Trial and Error

x2 + 5x + 6

1x2 + 5x + 6

(1x ____ ) (1x ____ )


I review distribute aka f o i l

Factoring Steps for Trial and Error

x2 + 5x + 6

1x2 + 5x + 6

(1x – 1) (1x + 6 )


I review distribute aka f o i l

Factoring Steps for Trial and Error

x2 + 5x + 6

1x2 + 5x + 6

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


I review distribute aka f o i l

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 – 6Error, Try Again


I review distribute aka f o i l

Factoring Steps for Trial and Error

x2 + 5x + 6


I review distribute aka f o i l

Factoring Steps for Trial and Error

x2 + 5x + 6

1x2 + 5x + 6


I review distribute aka f o i l

Factoring Steps for Trial and Error

x2 + 5x + 6

1x2 + 5x + 6

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


I review distribute aka f o i l

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!


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Factoring Steps for Trial and Error

3x2 – 4x – 4


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Factoring Steps for Trial and Error

3x2 – 4x – 4

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


I review distribute aka f o i l

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


I review distribute aka f o i l

Factoring Steps for Trial and Error

3x2 – 4x – 4


I review distribute aka f o i l

Factoring Steps for Trial and Error

3x2 – 4x – 4

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


I review distribute aka f o i l

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


I review distribute aka f o i l

Factoring Steps for Trial and Error

3x2 – 4x – 4


I review distribute aka f o i l

Factoring Steps for Trial and Error

3x2 – 4x – 4

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


I review distribute aka f o i l

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!


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Factoring Steps for Trial and Error

– 6x2 – x + 2


I review distribute aka f o i l

Factoring Steps for Trial and Error

– 6x2 – x + 2

Wow! How are we going to factor this??


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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.


Iii factoring by grouping

III. Factoring by Grouping


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Factoring Steps for Grouping

x2 + 5x + 6


I review distribute aka f o i l

Factoring Steps for Grouping

x2 + 5x + 6

x2 + 2x + 3x + 6


I review distribute aka f o i l

Factoring Steps for Grouping

x2 + 5x + 6

x2 + 2x + 3x + 6

x2 + 2x + 3x + 6


I review distribute aka f o i l

Factoring Steps for Grouping

x2 + 5x + 6

x2 + 2x + 3x + 6

x2 + 2x + 3x + 6

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


I review distribute aka f o i l

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)


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Factoring Steps for Grouping

3x2 – 4x – 4


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Factoring Steps for Grouping

3x2 – 4x – 4

3x2 – 6x + 2x – 4


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Factoring Steps for Grouping

3x2 – 4x – 4

3x2 – 6x + 2x – 4

3x2 – 6x + 2x – 4


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Factoring Steps for Grouping

3x2 – 4x – 4

3x2 – 6x + 2x – 4

3x2 – 6x + 2x – 4

3x(x – 2) 2(x – 2)


I review distribute aka f o i l

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)


I review distribute aka f o i l

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 . . .


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Factoring Steps for Grouping

3x2 – 4x – 4


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Factoring Steps for Grouping

3x2 – 4x – 4

3x2 + 2x – 6x – 4


I review distribute aka f o i l

Factoring Steps for Grouping

3x2 – 4x – 4

3x2 + 2x – 6x – 4

3x2 + 2x – 6x – 4


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Factoring Steps for Grouping

3x2 – 4x – 4

3x2 + 2x – 6x – 4

3x2 + 2x – 6x – 4

x(3x + 2) – 2(3x + 2)


I review distribute aka f o i l

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)


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Factoring Steps for Grouping

– 6x2 – x + 2


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Factoring Steps for Grouping

– 6x2 – x + 2

Oh, we are back to this problem again!


I review distribute aka f o i l

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?


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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?


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Factoring Steps for Grouping

– 6x2 – x + 2

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


Iv factoring quadratics

IV. Factoring Quadratics


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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.


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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).


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Factoring Steps

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

answer that will add or subtract

to equal the middle term.


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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.


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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)


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Divide only the numbers in the ( )( )

by the positive coefficient of x2.

Divide out common terms.

(aka: Simplify fractions.)


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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.


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TO SOLVE FOR X:

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

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

x + 3 = 0x + 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}.


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Now let’s try some . . .


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Actual Work ShownMental Work

x2+ 5x + 6


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Actual Work ShownMental Work

1x2+ 5x + 6


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Actual Work ShownMental Work

1x2+ 5x + 6

Step 1: No GCF


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Actual Work ShownMental Work

1x2+ 5x + 6

Step 1: No GCF

Step 2:

1 * 6 = 6

1 * 6

2 * 3


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Actual Work ShownMental 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


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Actual Work ShownMental 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.


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Actual Work ShownMental 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.


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Actual Work ShownMental 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.


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Actual Work ShownMental 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.


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Actual Work ShownMental 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.


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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}


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Actual Work ShownMental Work

x2 – 5x – 6


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Actual Work ShownMental Work

1x2 – 5x – 6


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Actual Work ShownMental Work

1x2 – 5x – 6

Step 1: No GCF


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Actual Work ShownMental Work

1x2 – 5x – 6

Step 1: No GCF

Step 2:

1 * 6 = 6

1 * 6

2 * 3


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Actual Work ShownMental 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.


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Actual Work ShownMental 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.


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Actual Work ShownMental 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.


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Actual Work ShownMental 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.


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Actual Work ShownMental 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.


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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}


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Actual Work ShownMental Work

2x2 + 7x – 4


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Actual Work ShownMental Work

2x2 + 7x – 4

Step 1: No GCF


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Actual Work ShownMental Work

2x2 + 7x – 4

Step 1: No GCF

Step 2:

2 * 4 = 8

1 * 8

2 * 4


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Actual Work ShownMental Work

2x2 + 7x – 4

Step 1: No GCF

Step 2:Step 3:

2 * 4 = 8 + 8

1 * 8 – 1

2 * 4 + 7


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Actual Work ShownMental 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.


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Actual Work ShownMental 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.


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Actual Work ShownMental 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.


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Actual Work ShownMental 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.


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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 }


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Now here are some

practice problems. . .


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x2 + 4x + 3


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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 + #?


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x2 + 4x + 3

x2+3x +1x + 3


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  • x2 + 4x + 3

  • x2+3x +1x + 3

  • Now separate the x2 into x • x

  • Put each x into ( ) ( )

  • Put the two red #s into ( ) ( )


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1x2 + 4x + 3

x2+3x +1x + 3

( x +3) ( x +1)

Now, divide by the coefficient of x2.


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1x2 + 4x + 3

x2+3x +1x + 3

( x +3) ( x +1)

1 1


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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.


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Therefore,

the Factored Form of

x2 + 4x + 3

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


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Now, let’s try another problem.


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3x2 - 4x – 4


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3x2 - 4x – 4

(x + 2) (x – 6)


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3x2 - 4x – 4

(x + 2) (x – 6)

(x + 2) (x – 6)

3 3


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3x2 - 4x – 4

(x + 2) (x – 6)

(x + 2) (x – 6)

3 3

(3x + 2) (x – 2)


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3x2 - 4x – 4

(x + 2) (x – 6)

(x + 2) (x – 6)

3 3

(3x + 2) (x – 2)

You are now ready for

The Problem ! ! ! ! !


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Actual Work ShownMental Work

– 6x2 – x + 2


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Actual Work ShownMental Work

– 6x2 – x + 2

Step 1A: No GCF


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Actual Work ShownMental Work

– 6x2 – x + 2

Step 1A: No GCF

Step 1B: – 1(6x2 + x – 2)


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Actual Work ShownMental 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


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Actual Work ShownMental 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


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Actual Work ShownMental 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.


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Actual Work ShownMental 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.


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Actual Work ShownMental 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.


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Actual Work ShownMental 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.


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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


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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


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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


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