Download Factoring out a greatest common factor As described in 6.1, the Distr

Math 65 Elementary Algebra
FACTORING
T. Henson
A. GCF: Factoring out a greatest common factor
As described in 6.1, the Distributive Property is used to factor out a greatest common
factor (GCF)
Example: For the polynomial 6 x3 − 15 x 2 + 30 x , the greatest common factor is 3x .
Factoring out 3x we have
6 x3 − 15 x 2 + 30 x = 3 x ( 2 x 2 − 5 x + 10 )
An unusual example of factoring out the GCF occurs in the following situation:
Example: In the expression 3 x ( x + 1) + 2 ( x + 1) there are TWO terms: 3 x ( x + 1) and
2 ( x + 1) . This may seem confusing, because in the expression x + 1 there are also two
terms. But we consider the expression 3 x ( x + 1) to be a single term because it represents
a multiplication: 3x and x + 1 are multiplied together (not added or subtracted).
Similarly, 2 ( x + 1) is a single term because 2 and x + 1 are multiplied together. When we
look at the expression 3 x ( x + 1) + 2 ( x + 1) in this way, then we can identify the quantity
( x + 1)
as a greatest common factor (also called a common grouped factor) and we can
factor it out. When we factor out ( x + 1) , the factors left behind are grouped together and
become the factor ( 3 x + 2 )
3 x ( x + 1) + 2 ( x + 1) = ( x + 1)( 3 x + 2 )
This type of factoring is the basis for Factoring by Grouping.
B. FACTORING BY GROUPING
This method of factoring is applied to polynomials consisting of four terms. The general
procedure is to look for the greatest common factor of the first two terms and the greatest
common factor of the last two terms and to factor each pair of terms independently. The
result should yield a common grouped factor. The next example illustrates this
procedure.
5
ax
+ 15
+
6
x + 2a
find the GCF
of these two
terms
find the GCF
of these two
terms
= 5
ax
+ 15
+
6
x + 2a
GCF=5 x
GCF= 2
= 5 x ( a + 3) + 2 ( a + 3)
= ( a + 3)( 5 x + 2 )
Factor out GCF in first two terms
and last two terms
Factor out common group factor of ( a + 3)
Math 65 Elementary Algebra
FACTORING
T. Henson
C. FACTORING TRINOMIALS OF THE FORM x 2 + bx + c
Whenever we factor trinomials, we know the factored form should be the product of two
binomial factors. The simplest type of trinomial is the trinomial of the form x 2 ± bx ± c .
This type of trinomial will have a factorization of the form ( x ± h )( x ± k ) where h and k
are numbers whose product is c and whose sum is b. The signs of b and c determine the
signs of h and k. Here are two examples.
Example 1: Factor x 2 − 12 x + 20 . To factor, we need to find two numbers which
multiplied together give us 20 but which when added together give us −12 . To get a
negative sum, at least one of the numbers must be negative and to get a positive product
both of the numbers must be negative (because a negative times a negative yields a
positive). So we try a few possibilities as shown in the table:
Two negative
factors of 20
Sum is −12
( −1)( −20 )
( −2 )( −10 )
( −1) + ( −20 ) = −21 (won't work)
( −2 ) + ( −10 ) = −12 (will work)
In the second row of the table we have found a winning combination: factors of 20
whose sum is −12 . We use the numbers −2 and −10 in the binomial factors. The
solution to the factoring problem is
x 2 − 12 x + 20 = ( x − 2 )( x − 10 )
You can always check by multiplying the two binomials together!
Example 2: Factor x 2 − 12 x − 45 . To factor, we need to find two numbers which
multiplied together give us −45 but which when added together give us −12 . Since the
only way to get a negative product is to multiply a positive times a negative, we know
that one of the factors will be negative, while the other will be positive. Again, we can
organize the search process in a table:
Two negative
Sum is −12
factors of −45
( −1)( 45)
(1)( −45)
( −3)(15)
( 3)( −15)
( −1) + 45 = 44 (won't work)
1 + ( −45 ) = −44 (won't work)
( −3) + 15 = 12 (won't work)
3 + ( −15 ) = −12 (will work)
The solution is: x 2 − 12 x − 45 = ( x + 3)( x − 15 ) . By the way, the order in which the
binomial factors are written does not matter; x 2 − 12 x − 45 = ( x − 15 )( x + 3) is also correct.
In this second example, it was pretty obvious that, whether positive or negative, 1 and 45
were not going to work. With a little practice and experience, the search process
illustrated by the table can be carried out in your head. However, if the constant term is
really large, the table can be a useful tool for organizing the search.
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Math 65 Elementary Algebra
FACTORING
T. Henson
D. FACTORING TRINOMIALS OF THE FORM ax 2 + bx + c : THE ac METHOD
What’s different here? The coefficient of the squared term is a number other than 1. For
example, suppose you had the trinomial 2 x 2 + 13x + 15 . The coefficient of the squared
term is 2. The method shown in part C will not work here. We will use a technique
called the “ac” method by many textbooks. The “a” and “c” refer to the coefficient a (the
coefficient of the squared term) and the constant term c. Before giving the full details of
the method, look at the following example. In this example, the trinomial 2 x 2 + 13x + 15
is re-written as a four-term polynomial and factored by grouping.
Example:
2 x 2 + 13x + 15
= 2 x 2 + 3 x + 10 x + 15
= 2
x 2
+
3 x + 10
+
15
x
GCF = x
GCF =5
= x ( 2 x + 3) + 5 ( 2 x + 3)
= ( 2 x + 3)( x + 5 )
Re-write 13x as 3x + 10 x
Find the GCF of the first two terms and GCF of
the last two terms
Factor out the GCF of the first two terms and the
GCF of the last two terms. A common group
factor of ( 2 x + 3) appears.
Factor out the common group factor and group
together the terms that remain behind.
The factoring process is complete and the solution is: 2 x 2 + 13x + 15 = ( 2 x + 3)( x + 5 ) .
Check by multiplying!
The “trick” was to split the middle term in the trinomial ( 13x in the example above) into
a sum of two terms so that the resulting 4-term polynomial could be factored by
grouping. But how did I know how to split up the middle term? The “ac” method will
tell us how.
The “ac” method for factoring the trinomial ax 2 + bx + c
Step 1: From the trinomial, identify a, b and c.
Step 2: Find the product ac
Step 3: Find two factors of the product ac whose sum is b. The signs of the factors will
be determined by the signs of ac and b. A table can help organize the work or you can
use the X method shown in class.
Here are two examples:
Example 1: Factor 6 x 2 − 19 x + 10 .
Step 1: From the trinomial, a = 6 , b = −19 and c = 10 .
Step 2: The product ac is 60
Step 3: Find two factors of 60 whose sum is b = −19 . Since ac is positive the
two factors must both be positive or they must both be negative. Since b is
negative, we will look for two negative numbers whose product is 60 and whose
sum is −19 .
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Math 65 Elementary Algebra
FACTORING
ac = 60
( −1)( −60 )
( −2 )( −30 )
( −3)( −20 )
( −4 )( −15)
T. Henson
Sum b = −19
( −1) + ( −60 ) = −61 Doesn’t work
( −2 ) + ( −30 ) = −32 Doesn’t work
( −3) + ( −20 ) = −23 Doesn’t work
( −4 ) + ( −15 ) = −19 ⇐ This is the required sum. When the required sum
has been found, the remaining possibilities need not be
checked
Now we know how to split up the middle term.
Re-write −19x as −4 x − 15 x
6 x 2 − 19 x + 10 = 6 x 2 − 4 x − 15 x + 10
2
Find the GCF of the first two terms and GCF of
= 6
x − 4
x − 15
+
10
x
GCF= 2 x
GCF= −5
the last two terms. Because the third term is
negative, we factor out −5
= 2 x ( 3x − 2 ) − 5 ( 3x − 2 ) Factor out the GCF of the first two terms and
the GCF of the last two terms. A common
group factor of ( 3x − 2 ) appears.
Factor out the common group factor and group
= ( 3x − 2 )( 2 x − 5 )
together the terms that remain behind.
The factoring process is complete and the solution is: 6 x 2 − 19 x + 10 = ( 3 x − 2 )( 2 x − 5 ) .
Check by multiplying!
As discussed earlier in part C, certain combinations of factors are obviously not going to
work (such as −1 and −60 in this last example) and need not be tried. However,
sometimes the choice is not obvious and many possibilities may need to be tested.
By the way, when you split up the middle term, the order in which you write the terms
does not matter. Here is the same example as above, but with the middle term written as
−15 x − 4 x :
6 x 2 − 19 x + 10
= 6 x 2 − 15 x − 4 x + 10
2
= 6x
− 15
x + 10
x − 4
GCF =3 x
GCF =−2
= 3x ( 2 x − 5 ) − 2 ( 2 x − 5)
= ( 2 x − 5 )( 3 x − 2 )
Re-write −19x as −15 x − 4 x
Find the GCF of the first two terms and GCF of
the last two terms. Because the third term is
negative, we factor out −2
Factor out the GCF of the first two terms and
the GCF of the last two terms. A common
group factor of ( 2 x − 5 ) appears.
Factor out the common group factor and group
together the terms that remain behind.
Because multiplication is commutative, this answer is the same as before.
Here is another example.
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Math 65 Elementary Algebra
FACTORING
T. Henson
Example 2: Factor 6 x 2 − 11x − 10 .
Step 1: From the trinomial, a = 6 , b = −11 and c = −10 .
Step 2: The product ac is −60
Step 3: Find two factors of −60 whose sum is b = −11 . Since ac is negative, one
of the factors will be negative and the other positive. We will look for two
numbers whose product is −60 and whose sum is −11 .
ac = −60
( −1)( 60 )
(1)( −60 )
( −2 )( 30 )
( 2 )( −30 )
( −3)( 20 )
( 3)( −20 )
( −4 )(15)
( 4 )( −15)
Sum b = −11
( −1) + 60 = 59
1 + ( −60 ) = −59
( −2 ) + 30 = 28
2 + ( −30 ) = −28
( −3) + 20 = 17
3 + ( −20 ) = −17
( −4 ) + 15 = 11
4 + ( −15 ) = −11
Doesn’t work
Doesn’t work
Doesn’t work
Doesn’t work
Doesn’t work
Doesn’t work
Doesn’t work
⇐ This is the required sum. When the required sum has
been found, the remaining possibilities need not be
checked
Now we know how to split up the middle term.
Re-write −11x as −15 x + 4 x
6 x 2 − 11x − 10
= 6 x 2 − 15 x + 4 x − 10
2
Find the GCF of the first two terms and GCF of
= 6x
− 15
x − 10
x + 4
GCF =3 x
GCF = 2
the last two terms.
= 3x ( 2 x − 5 ) + 2 ( 2 x − 5 ) Factor out the GCF of the first two terms and
the GCF of the last two terms. A common
group factor of ( 2 x − 5 ) appears.
Factor out the common group factor and group
= ( 2 x − 5 )( 3 x + 2 )
together the terms that remain behind.
The factoring process is complete and the solution is: 6 x 2 − 11x − 10 = ( 2 x − 5 )( 3x + 2 ) .
Check by multiplying!
A couple of notes about this last example:
1. When looking for factors of −60 that add up to −11 , it is quite obvious that
combinations involving 1 and 60 or even 2 and 30 will not work, so in practice
you would not test those combinations.
2. Also, when splitting the middle I chose to put the negative term first – to write
−15 x + 4 x - so that I would not have to factor out a negative during the factoring
by grouping step.
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