Sign of

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Section 6.3

Factoring Trinomials



Factoring Trinomials of the Form ax 2  bx  c :

Characteristics:

1. Coefficient of x 2 is something other than 1 or 0

2. b and c may be any real number, , , 1, or

otherwise, but not 0

From FOIL we get:

2x  x Factors of a (1st terms)

2 x 2  7 x  6  2 x  3 x  2

3 2 Factors of c (Last terms)



2 x 2  7 x  6  2 x  3 x  2

3x Product of Inner terms

+4x Product of Outer terms

+7x Sum of Inner and Outer

Products

In words,

1. The first terms of each binomial are factors of a.

2. The last terms of each binomial are factors of c.

3. b is the sum of the inner and outer products.

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Factoring Trinomials of the Form ax 2  bx  c :

1. Write all pairs of factors of a (the x 2 term).

2. Write all pairs of factors of c.

3. Try various combinations of these factors until

the correct middle term is found.

4. If no combination is found, the polynomial is

prime.

Factor 3x 2  5 x  2

1. Factors of 3(the a term): 1, 3

2. Factors of 2 (the c term): 1, 2

Try  x  23x  1  3x 2  7 x  2

The a and c terms are correct but the b term is

wrong. So these are not the correct factors.

Next try the other combination. Simply switch the 1

and 2 numbers of c.

Try  x  13x  2  3x 2  5 x  2



This is the trinomial we are trying to factor so it is

correct.

Ex. Factor: 5 x 2  7 x  2

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In the previous examples, all the coefficients were +

Next, we consider negative coefficients

Important Patterns

Sign of Sign of Action to

___c___ Factors determine b

+ Both + or Both  Sum



 Opposite Signs Difference







Summary of signs:

Sign of Sign of Signs in

___c___ ____b___ Binomial

+ + Both +

 Both 



 + Larger product +

 Larger product 



Ex. Factor: a. 2 x 2  x  3 b. 3x 2  2 x  5

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Lastly, note that if a polynomial has no GCF, then

none of its factors can have can have a GCF.

Ex. Factor 2 x 2  5 x  2

Factors of a: 1, 2 Factors of c: 1, 2

Using these combinations would result in the

following factored forms:

A:  x  22 x  1 B:  x  12 x  2

Note that the factor 2 x  2 in B. has a GCF of 2.

In order for this to be true, the original trinomial

would have a GCF of 2. It does not, therefore B is

not a factored form. So, A is the factored form.



Ex. Factor 12 x 2  7 x  12

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Perfect Square Trinomials: A trinomial that is a

perfect square.

a  b 2  a 2  2ab  b 2

 x  42  x 2  8 x  16

Characteristics:

1. The 1st and 3rd terms are both squares

2. The middle term is twice the product of a and b

or its opposite.  2ab



Ex. Factor 9 x 2  6 x  1

9 x 2  3x  3x 1  1 1 23x 1  6x opposite

of  6 x

Answer: 3x  12

Perfect Square Numbers:

12  1 5 2  25

22  4 6 2  36

32  9 7 2  49

4 2  16 8 2  64



Ex. Factor 4 x 2  12 x  9

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Factoring ax2  bx  c by Grouping: Use a technique

similar to the one used to factor x 2  bx  c but with

one small difference.

1. Find 2 numbers whose product is ac and whose

sum is b.

2. Write bx as the sum of the factors found in step 1.

3. Factor by grouping.



Ex. Factor a. 3x 2  14 x  8. b. 6 x 2  7 x  5

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Summary

A Factoring Strategy



I. GCF



II. Recognize situation, ID number of terms



1. Two Terms—Difference of 2 squares



2. Four Terms—Grouping



3. Three Terms—ID value of a

Situation Action

a. a 1 PST, middle term = 2 a c

OR

Try Combinations or Grouping



Note: Sum of two squares can not be factored.

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Section 6.3 Example solutions

Factor: 5 x 2  7 x  2

Factors of a: 1, 5 Factors of c: 1, 2

Combinations: There are 2

A B

1 5 1 5

1 2 2 1

1 10 = 11 2 5=7

B is the correct one

Answer:  x  15x  2

Factor 2 x 2  x  3

Factors of a: 1, 2 Factors of c: 1, 3

c is negative so DIFF

Combinations: There are 2

1 2 1 2

1 3 3 1

1 6=5 3 2=1

B is correct. Want a +1 so the 3 is + and 1 is 

Answer:  x  12 x  3

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Section 6.3 Example solutions continued

Factor 3x 2  2 x  5 c is  so want DIFF

1 3 1 3

1 5 5 1

1 15 = 14 5 3=2

2 is negative so 5 is negative

Answer:  x  13x  5

Factor 12 x 2  7 x  12 NO GCF

a: 1, 12; 2, 6; 3, 4 c: 1, 12; 2, 6; 3, 4 DIFF

18 combinations

1 12 1 12 1 12 1 12 1 12 1 12

1 12 12 1 2 6 6 2 3 4 4 3

1 144 No No No No No

= 143

2 6 2 6 2 6 2 6 2 6 2 6

1 12 12 1 2 6 6 2 3 4 4 3

No No No No No No

3 4 3 4 3 4 3 4 3 4 3 4

1 12 12 1 2 6 6 2 3 4 4 3

No No No No 9 16 No

=7

7 is neg so 16 is neg so the 4 is neg

Answer: 3x  44 x  3

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Section 6.3 Example solutions continued

Factor 4 x 2  12 x  9

Both a and c are perfect squares and c is +

So try PST

4 x2  2 x  2 x 9  3 3 22 x 3  12x

Opp of  12x

So trinomial is a PST

Answer: 2 x  32

a. Factor 3x 2  14 x  8. ac = 24

c is + so signs same and use sum

Prod = 24 Sum = 14 Factors are 12 and 2

Write as 3x 2  12 x  2 x  8

3x x  4  2 x  4

x  43x  2

b. Factor 6 x 2  7 x  5 ac = 30

c is  so signs are diff and use diff

Prod = 30 Sum =  7 Factors are  10 and 3

Write as 6 x 2  10 x  3x  5

2 x3x  5  3x  5

3x  52 x  1