Download M098 Carson Elementary and Intermediate Algebra 3e Section 6.3 Objectives

M098
Carson Elementary and Intermediate Algebra 3e
Section 6.3
Objectives
1.
2.
2
Factor trinomials of the form ax + bx + c, where, by trial.
2
Factor trinomials of the form ax + bx + c, where a ≠ 1, by grouping.
Vocabulary
Prior Knowledge
Factoring Sign Clues
Factoring by Grouping
New Concepts
2
1. Factoring trinomials of the ax + bx + c where a ≠ 1.
2
The general form for a trinomial is ax + bx + c. The coefficient of the squared term is referred to as a,
the coefficient of the middle term is b and the constant is c.
Some of the clues that were presented in the last section work only when a is 1. The problems become
more difficult when a is not 1. The book shows two methods for factoring these types of problems.
Objective 1 shows the “trial and error” or “guess and check” method. Trial and error is preferred by many
students and instructors because it is more of a mental process and is faster. Some people are gifted at
seeing these combinations of numbers and others struggle with them. The more practice you have, the
easier they become.
Objective 2 shows the factoring by grouping method. This method rewrites each trinomial as a 4-term
polynomial and then uses factoring by grouping.
There is a third method that was shared on the Internet a few years ago that also works well. It is related
to factoring by grouping. We call this Factoring the Illegal Way. Watch the animation to see how the
Illegal Factoring Method works.
The Illegal Method:
Factor: 6x 2  7x  3
This is difficult to factor because the coefficient of the squared term (a) is not 1. To remove the
6, multiply it with the c term (-3).
6x 2  7x  3
The new trinomial is
x 2  7x  18
This trinomial is easily factored as x  9x  2 .
An “illegal move” was committed so now it must be undone. Since 6 was multiplied in the
first step, to “undo” the illegal move, divide each constant in the binomials by 6.
9 
2

 x   x  
6 
6

This is factored form but integers are preferred in the binomials rather than fractions. To
remove the fractions, first reduce the fractions to lowest terms.
3 
1

 x   x  
2
3



V. Zabrocki 2011
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M098
Carson Elementary and Intermediate Algebra 3e
Section 6.3
Now take the denominator of each fraction and slide it in front of the x in the binomial, making
the denominator the coefficient of x.
3 
1

 x   x  
2 
3

The Illegal Move does
not work if a is a
negative number.
Factor out -1 and then
proceed.
2x  33x  1
Always foil to check your factors.
2x  33x  1  6x 2  7x  3
Do not forget to undo the illegal move. You must do both steps!!!
Example 1:
2
Factor 2x + 7x + 3
Factoring by Grouping
Multiply a · c and look for
factors of that number that
add (because of the +3) to
equal 7.
Illegal Method
Multiply a · c and rewrite the
trinomial.
2
Guess & Check
Since 2 and 3 are both prime
numbers, there is only one way
to factor each of them.
x + 7x + 6
Illegal Move - ·2
2·1=2
3·1=3
2·3=6
Factor using the sign clues:
6=1·6
6 = 2· 3
(x + 1)(x + 6)
Rewrite as a 4-term
polynomial replacing the
middle term (7x) with
(1x + 6x).
2
2x +1x + 6x + 3
Group the first and last terms.
2
Notice when this is foiled it
does NOT equal the original
trinomial. That’s because we
need to undo the Illegal Move.
The Illegal Move was to
multiply by 2 so to undo the
Illegal Move, divide the
constants by 2.
(2x + 1x) + (6x + 3)
Pull out the GCF from each
binomial.
x(2x + 1) + 3(2x + 1)
Factor out the binomial GCF.
(2x + 1)(x + 3)
Always foil to be sure the
factors are correct.
2
1 
6

 x   x  
2
2


Reduce the fractions if you
can and slide the remaining
denominator in front of the
variable.
1

 x  x  3
2

(2x + 1)(x + 3)
Start with the factors of 2 and
place them in the binomials.
(2x
)( x
)
The sign clues tell us the signs
are the same and both positive.
(2x + )( x + )
Now for the guess and check.
The last terms must be either 3
and 1 or 1 and 3. Try one
combination and foil to see if it
works.
(2x + 3)( x + 1)
2
2
2x + 2x + 3x + 3 = 2x + 5x + 3
The check didn’t work so switch
the 3 and 1 and check again by
foiling.
(2x + 1)( x + 3)
2
2
2x + 6x + 1x + 3 = 2x + 7x + 3
(2x + 1)(x + 3) = 2x + 7x + 3
Always foil to be sure the
factors are correct.
It works.
2
(2x + 1)(x + 3) = 2x + 7x + 3
V. Zabrocki 2011
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M098
Carson Elementary and Intermediate Algebra 3e
Section 6.3
Example 2: Do each of the following problems using all three methods until you decide
which one works the best for you.
2
3m – 10m + 3
(3m – 1)(m – 3)
2
(2y + 3)(2y + 1)
2
(4u + 3)(u – 5)
2
(3z + 5)(z – 6)
4y + 8y + 3
4u – 17u – 15
3z – 13z – 30
2
2
9x + 9xy – 10y
2
6u + 3u – 7
(3x – 2y)(3x + 5y)
prime
2. The first step in factoring should always be to remove the GCF.
Example 3: Factor. Remove the GCF first.
2
6a – 20a + 16
2
20xy – 50xy + 30x
V. Zabrocki 2011
2(3a – 4)(a – 2)
10x(y – 1)(2y – 3)
page 3