Download Section 6.2

Chapter 6
Factoring
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CHAPTER
6
Factoring
6.1 Greatest Common Factor and
Factoring by Grouping
6.2 Factoring Trinomials
6.3 Factoring Special Products and
Factoring Strategies
6.4 Solving Equations by Factoring
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6.2
Factoring Trinomials
1. Factor trinomials of the form x2 + bx + c.
2. Factor trinomials of the form ax2 + bx + c,
where a  1, by trial.
3. Factor trinomials of the form ax2 + bx + c,
where a  1, by grouping.
4. Factor trinomials of the form ax2 + bx + c
using substitution.
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First we consider trinomials in the form x2 + bx + c in
which the coefficient of the squared term is 1. If the
trinomial is factorable, we have two binomial factors.
F O
I L
(x + 2)(x + 3) = x2 + 3x + 2x + 6
= x2 + 5x + 6
Sum of Product of
2 and 3
2 and 3
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Factoring x2 + bx + c
To factor a trinomial of the form x2 + bx + c,
1. Find two numbers with a product equal to c and a
sum equal to b.
2. The factored trinomial will have the form:
(x + ) (x + ), where the second terms are the
numbers found in Step 1.
Note: The signs of b and c may cause one or both
signs in the binomial factors to be minus signs.
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Example
Factor. x2 – 6x + 8
Solution We must find a pair of numbers whose
product is 8 and whose sum is –6. Note that if
two numbers have a positive product and
negative sum, they must both be negative. List
the possible products and sums.
Product
Sum
(–1)(–8) = 8
–1 + (–8) = –9
(–2)(–4) = 8
–2 + (–4) = –6
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This is the
correct
combination.
6
continued
Answer x2 – 6x + 8 = (x +(– 2))(x+( – 4))
= (x – 2)(x – 4)
Check Multiply the factors to verify that their
product is the original polynomial.
(x – 2)(x – 4) = x2 – 4x – 2x + 8 = x2 – 6x + 8
It checks.
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Example
Factor. x2 + 2x – 24
Solution: We must find a pair of numbers whose
product is –24 and whose sum is 2. Because the
product is negative, the two numbers have
different signs.
Product
Sum
(–2)(12) = 24
–2 + 12 = 10
(–4)(6) = 24
–4 + 6 = 2
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This is the
correct
combination.
8
continued
Answer x2 + 2x – 24 = (x – 4)(x + 6)
Check We can check by multiplying the binomial
factors to see if their product is the
original polynomial.
(x – 4)(x + 6) = x2 + 6x – 4x – 24
= x2 + 2x – 24
Multiply the factors using
FOIL.
The product is the original
polynomial.
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Example
Factor. 3x4 – 18x3 + 24x2
Solution
First, factor out the GCF, 3x2.
3x4 – 18x3 + 24x2 = 3x2 (x2 – 6x + 8)
= 3x2(x – 4)(x – 2)
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Example
Factor. x4 + 2x3 + 5x2
Solution
First, factor out the GCF, x2.
x2 ( x2 + 2x + 5)
Now try to factor the trinomial to two binomials.
We must find a pair of numbers whose product is
5 and whose sum is 2.
Product
Sum
(1)(5) = 5
1+5=6
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continued
Factor. x4 + 2x3 + 5x2
There is no combination of factors whose
product is 5 and sum is 2.
The trinomial x2 + 2x + 5 has no binomial factors
with integer terms, so x2 (x2 + 2x + 5) is the final
factored form.
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Example
Factor. 2 x3 y  2 x 2 y  84 xy
Solution
First, factor out the GCF, 2xy.
2 x3 y  2 x 2 y  84 xy  2 xy( x 2  x  42)
 2 xy ( x  6)( x  7)
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Example
Factor. a2 – ab – 20b2
Solution We must find a pair of terms whose product is
20b2 and whose sum is –1b.
Product
Sum
(–1b)(20b) = –20b2
–1b+ 20b= 19b
(1b)(–20b) = – 20b2
1b+ (–20b) = –19b
(2b)(–10b) = – 20b2
2 b+ (–10b) = –8b
(–2b)(10b) = – 20b2
–2b + 10b = 8b
(4b)(–5b) = – 20b2
4b + (–5b) = –1b
(–4b)(5b) = – 20b2
–4b + 5b = 1b
This is the
correct
combination.
Answer a2 – ab – 20b2 = (a – 5b)(a + 4b)
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Example
Factor. 4xy3 + 12xy2 – 72xy
Solution
This trinomial has a monomial GCF, 4xy.
4xy3 + 12xy2 – 72xy = 4xy(y2 + 3y – 18)
We must find a pair of numbers whose product is
–18 and whose sum is 3.
= 4xy(y – 3)(y + 6)
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Example
Factor. 6 x 2  13x  5
Solution
The first terms must multiply to equal 6x2.
These could be x and 6x, or 2x and 3x.
6 x 2  13x  5  
+



The last terms must multiply to equal –5.
Because –5 is negative, the last terms in the
binomials must have different signs. This factor
pair must be 1 and 5.
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continued
Now we multiply binomials with various
combinations of these first and last terms until we find
a combination whose inner and outer products
combine to equal 13x.
 x  5 6x 1  6x22  x  30x  5  6x22  29x  5
 x  1 6x  5  6x  5x  6x  5  6x  x  5
3x  1 2x  5  6x2 15x  2x  5  6x2 13x  5 Incorrect
combinations.
2
2
3
x

5
2
x

1

6
x

3
x

10
x

5

6
x

7
x

5



2
2
2
x

1
3
x

5

6
x

10
x

3
x

5

6
x
 7x  5



2
2
2
x

5
3
x

1

6
x

2
x

15
x

5

6
x
 13x  5 Correct



combination.
Answer 6 x2  13x  5   2 x  53x 1
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Factoring by Trial and Error
To factor a trinomial of the form ax2 + bx + c, where
a ≠ 1, by trial and error:
1. Look for a monomial GCF in all of the terms. If
there is one, factor it out.
2. Write a pair of first terms whose product is ax2.
ax2





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3. Write a pair of last terms whose product is c.
c





4. Verify that the sum of the inner and outer products
is bx (the middle term of the trinomial).





Inner
+ Outer
bx
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If the sum of the inner and outer products is not
bx, then try the following:
a. Exchange the last terms of the binomials from
step 3; then repeat step 4.
b. For each additional pair of last terms, repeat
steps 3 and 4.
c. For each additional pair of first terms, repeat
steps 2 – 4.
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Example
2
6
x
x2
Factor.
Solution
There is no common monomial factor.
The first terms must multiply to equal 6x2.
These could be x and 6x, or 3x and 2x.
6x2  x  2 



+

The last terms must multiply to equal 2.
Because 2 is a prime number, its factors are 1
and 2.
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continued
Now find a combination whose inner and outer
products combine to equal –x.
2
2
6
x

1
x

2

6
x

12
x

x

2

6
x


 
   11x  2 
2
2
6
x

1
x

2

6
x

12
x

x

2

6
x


 
   11x  2 
 3x  2  2 x  1   6 x 2  3x  4 x  2    6 x 2  x  2 
2
2
3
x

2
2
x

1

6
x

3
x

4
x

2

6
x


 
   x  2
Correct combination.
Answer 6 x2  x  2  (3x  2)(2 x  1)
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Example
3
2
21
x

60
x
 9x
Factor.
Solution
2
21x3  60 x 2  9 x  3x  7 x  20 x  3
Factor out the
monomial GCF, 3x.
Now we factor the trinomial within the parentheses.
The first terms must multiply to equal 7x2.
These could be x and 7x.
3x  7 x 2  20 x  3  3x 


+

The last terms must multiply to equal 3.
Because 3 is a prime number, its factors are 1
and 3.
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continued
Now find a combination whose inner and outer
products combine to equal –20x.
3x  x  1 7 x  3  3x  7 x 2  3x  7 x  3  3x  7 x 2  4 x  3
3x  x  1 7 x  3  3x  7 x 2  3x  7 x  3  3x  7 x 2  4 x  3
3x  x  3 7 x  1  3x  7 x 2  x  21x  3  3x  7 x 2  20 x  3
3x  x  3 7 x  1  3x  7 x 2  x  21x  3  3x  7 x 2  20 x  3
Correct combination.
Answer 21x3  60x2  9x  3x  x  3 7 x  1
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Factoring ax2 + bx + c, Where a ≠ 1, by Grouping
To factor a trinomial of the form ax2 + bx + c, where
a ≠ 1, by grouping,
1. Look for a monomial GCF in all of the terms. If
there is one, factor it out.
2. Find two factors of the product ac whose sum is b.
3. Write a four-term polynomial in which bx is written
as the sum of two like terms whose coefficients are
the two factors you found in step 2.
4. Factor by grouping.
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Example
Factor. 2 x 2  15 x  7
Solution
For this trinomial, a = 2, b = –15, and c = 7, so
ac = (2)(7) = 14. We must find two factors of 14
whose sum is –15. Since 14 is positive and –15 is
negative, the two factors must be negative.
Factors of ac
Sum of Factors of ac
(–2)(–7) = 14
–2 + (–7) = –9
(–1)(–14) = 14
–1 + (– 14) = –15
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Correct
26
continued
2 x 2 –15x  7  2 x 2 –x – 14x  7
 x  2x 1  7  2x 1
  2 x 1 x  7 
Write -15x as –x – 14.
Factor x out 2x2 – x,
factor -7 out of -14x+7.
Factor out of 2x – 1.
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Example
Factor.12 x 4  14 x3  6 x 2
Solution
Notice that there is a GCF, 2x2, that we can factor out.
2x2(6x2 + 7x – 3)
Now we factor the trinomial within the parentheses.
a = 6, c = 3; ac = (6)(3) = 18
Find two factors of 18 whose sum is 7. Because the
product is negative, the two factors will have different
signs.
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continued
Factor.12 x 4  14 x3  6 x 2
Solution
Factors of ac
Sum of Factors of ac
(1)(18) = 18
1 + 18 = 17
(2)(9) = 18
2 + 9 = 7
(3)(6) = 18
3 + 6 = 3
Correct
2x2(6x2 + 7x – 3)
Now write 7x as 2x + 9x and then factor by
grouping.
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continued
2 x 2 (6 x 2 +7x  3  2 x 2 (6 x 2 2x + 9x  3)
 2 x2  2 x(3x 1)  3(3x 1)
 2 x2  2 x  3 3x 1
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Example
2
4(
y

5)
 10( y  5)  6
Factor using substitution.
Solution
If we substitute another variable, like u, for y – 5, we
can see that the polynomial is actually in the form
ax2 + bx + c.
4( y  5)2  10( y  5)  6
4u 2  10u  6
Substitute u for y – 5.
We can factor by grouping.
4u 2  10u  6  (2u  3)(2u  2)
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continued
Remember that u was substituted for y – 5, so we
must substitute y – 5 back for u to have the factored
form of our original polynomial.
 (2u  3)(2u  2)
  2  y  5   3  2  y  5   2
Substitute y – 5 for u.
This resulting factored form can be simplified.
  2 y 10  3 2 y 10  2
Distribute.
  2 y  7  2 y  8
Combine like terms.
 2  2 y  7  y  4
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