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Homework Help: Mathematics: Algebra: Trinomials
This lesson explains how to factor trinomials. The process presented is
essentially the opposite of the
FOIL Method,
which is a process used to multiply two binomials. Make sure you understand the
FOIL Method lesson first.
Examine the following expression which consists of one binomial in
parentheses multiplying another binomial in parentheses.
(2k + 7)(3k - 10)
Using the FOIL Method, we can simplify the expression, obtaining the work
below and a final result which is a trinomial.
6k2 - 20k + 21k - 70
6k2 + k - 70
After the two sets of parentheses were multiplied, like terms were combined.
Notice that a trinomial, which consists of three terms, remains. This lesson
explains how to factor a trinomial like this into two binomials (which we had
before the FOIL Method was applied).
The first problem we will examine is below.
-16 - m2 -10m
In this problem, all of the terms are negative. Therefore, our first step
is to factor out the Greatest Common Factor which is a -1 or simply a minus
sign.
-(16 + m2 + 10m)
To make the factoring process a little more consistent and easier, it is a
good idea to keep the terms in order by the variable's exponent. Since m is the
only variable letter in this expression we will order the terms from the highest
power of m to the lowest power of m.
-(m2 + 10m + 16)
Before attempting to factor any more, there are a few simple questions you
can ask to make sure that the expression is factorable as a trinomial.
| Question |
Answer and Reason |
| After the GCF has already been factored out, are all
variables in the first (m2) and last (16) terms to an even power?
|
Yes. (m2
is the only variable in the first and last terms, and its exponent, 2, is an
even number.)
|
| Are all of the middle term's (10m) exponents to half of the power of one
of the exponents on the outside terms?
|
Yes. (In 10m, the middle term, m
is the only variable. Its exponent is not shown and therefore is 1 which is
half of 2, the exponent of m in the first term.)
|
The next step is to carry down the minus sign and write sets of open
parentheses, side by side.
-( )( )
Handling Variables
On the left side of each set of parentheses, write all of the variables from the
first term of the trinomial with half their exponents. The first
term is m2, thus after the exponent is cut in half, m is placed inside
each set of parentheses.
-( m )( m )
Since all of the terms in the original set of parentheses is positive, two
plus signs are placed below.
-( m + )( m + )
Handling Coefficients
Now identify the coefficient of the first and last terms in the original
set of parentheses, m2 + 10m + 16. The coefficients are 1 and 16.
Now write all pairs of factors of 1 in a vertical column and then write all
pairs of factors of 16 in another vertical column.
Factors of 1
1 * 1
|
Factors of 16
1 * 16
2 * 8
4 * 4
|
Choosing Factor Pairs
Now we must choose a pair of factors of 1 and a pair of factors of 16 to
insert into the pairs of parentheses. But how is this done? For the number 1,
there is only 1 pair of factors, eliminating any choice. Thus, the 1s can be
inserted on the left side of each set of parentheses.
-(1m + )(1m + )
Finally, a pair of factors of 16 must be chosen. This is a matter of trial
and error. To make sure all pairs are considered, start using factors at the
top of the list then check each pair below it, in order.
We start by choosing the first pair, 1 * 16. We place the factors on the right
side of each set of parentheses.
-(1m + 1 )(1m + 16)
Now test whether this is the correct pair. Multiply the expression using the
FOIL Method:
-(1m2 + 16m + m + 16)
-(1m2 + 17m + 16)
Note that the resulting trinomial is not the same as the one we started with,
so this is the incorrect pair of factors of 16. Try the next pair:
-(1m + 2 )(1m + 8 )
-(1m2 + 8m + 2m + 16)
-(1m2 + 10m + 16)
Since the result here is eqivalent to the original, the correct pair of factors
of 16 has been chosen. Additionally, the answer to the problem is.
-(1m + 2)(1m + 8)
Factoring A Trinomial with a Negative Sign
Examine this expression.
s2 - 5s + 6
Only one of the three terms is negative. As a result, a minus sign should
not be factored out as in the previous example. This expression will be
factored much like the expression on the previous page, but we will need
to work with the minus sign as we build two sets of parentheses.
To start factoring, first write out two empty sets of parentheses.
( )( )
Handling Variables
Now on the left of each set of parentheses, write all of the variables from
the first term with half the exponent. The first term is s2, after
dividing its exponent by two, we place s on the left of each set of parentheses.
( s )( s )
The last term in the trinomial does not have any variables, so we do not need
to carry any variables into the parentheses as a result.
Handling Coefficients
Now identify the coefficient of the first and last terms in the original
set of parentheses, s2 - 5s + 6. The coefficients are 1 and 6.
Now write all pairs of factors of 1 in a vertical column and write all pairs
of factors of 6 in another vertical column.
Since this problem involves a negative term, we also include the negative
factors of 6.
Factors of 1
1 * 1
|
Factors of 6
1 * 6
-1 * -6
2 * 3
-2 * -3
|
Choosing Factor Pairs
Now we must choose a pair of factors of 1 and a pair of factors of 16 to
insert into the pairs of parentheses. For the number 1, there is only one pair
of factors, eliminating any choice. Thus the 1s can be inserted on the left side
of each set of parentheses.
(1s )(1s )
Now a pair of factors of 6 must be chosen. This is a matter of trial and error.
To make sure all pairs are considered, check each pair of factors starting with
the pair on the top of the list and working toward the bottom of the list.
A summary of the trial and error process using the FOIL Method is below.
(1s + 1)(1s + 6)
s2 + 6s + s + 6
s2 + 7s + 6
(1s + -1)(1s - 6)
s2 - 6s - s + 6
s2 -7s + 6
(1s + 2)(1s + 3)
s2 + 3s + 2s + 6
s2 + 5s + 6
(1s - 2)(1s - 3)
s2 - 3s - 2s + 6
s2 - 5s + 6
The expression s2 - 5s + 6 is equivialent to the original
expression, therefore (1s - 2)(1s - 3) is the
final answer.
An Example With Two Negative Signs
The problem below is similar to the last problem, but it has two negative
signs in the expression.
p2 - 20p - 21
First, write out two sets of empty parentheses.
( )( )
Write all of the variables from the first term with half the exponent
in the front of each set of parentheses. Again, since there are no variables
in the last term, nothing is written on the right side of each set of
parentheses.
( p )( p )
Write out the factors of the coefficients of the first and last terms.
The previous problems have led to the observation that the first term's
coefficient of 1 results in 1 * 1 as the pair of factors chosen.
(1p )(1p )
Now the pairs of factors of 21 (both positive and negative) are computed.
Factors of -21
1 * -21
-1 * 21
3 * -7
-3 * 7
Use trial and error to find which pair of factors to use.
(1p + 1)(1p - 21)
p2 - 21p + 1p - 21
p2 - 20p - 21
The expression p2 - 20p - 21 is equivalent to the original
expression. Therefore, (1p + 1)(1p - 21) is the answer.
Factoring A Trinomial with a Negative Sign
Examine the expression below.
6c2 + c - 12
The expression is a trinomial, like the expressions which we factored
on previous pages. There is a small variation in this problem however -- the
first term has a coefficient that is not 1.
This expression will be factored much like the others, the main difference
is that we will need to find the correct pair of factors for the first term's
coefficient (6) in addition to the correct pair of factors for
the last term's coefficient (-12).
Start out by writing two sets of empty parentheses.
( )( )
Write each variable of the first and last terms in their respective
positions, but with only half the exponent. The first term has the variable
c2, with half its exponent it becomes c which is written on the left
of each set of parentheses. The last term does not have any variables to write.
( c )( c )
Because the last term is negative, write a plus sign in the middle of the
first set of parentheses and a - sign in the middle of the second.
(If the last term were positive you would write a + sign in each.)
( c + )( c - )
Now write out the positive factors of the coefficient of first and last.
Factors of 6
1 * 6
2 * 3
|
Factors of -12
1 * -12
-1 * 12
2 * -6
-2 * 6
3 * -4
-3 * 4
|
To find the right combination of factors, we will use Trial and Error as in
the last two lessons, but we must try different factors for both the first and
last terms. Each pair of factors must be tried in two arrangements. That is the
pair "1 * 6" must be tried as both "1 * 6" and "6 * 1". The work below will
explain this a little better.
After you write out a possible combination, FOIL the factored polynomial.
If the result is eqivalent to the original expression, you have found the
answer. All of the possible combinations for the above problem are shown below.
(1c + 1)(6c - 12)
(1c + 12)(6c - 1)
(1c + 2)(6c - 6)
(1c + 6)(6c - 2)
(1c + 3)(6c - 4)
(1c + 4)(6c - 3)
(2c + 1)(3c - 12)
(2c + 12)(3c - 1)
(2c + 2)(3c - 6)
(2c + 6)(3c - 2)
(2c + 3)(3c - 4)
(2c + 4)(3c - 3)
After using the FOIL Method to check each possible solution above, you should
find that (2c + 3)(3c - 4) is the correct factorization.
Homework Help: Mathematics: Algebra
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