Intermediate Algebra
Answer/Discussion to Practice Problems
Tutorial 28: Factoring Trinomials
Answer/Discussion
to 1a

Note that this trinomial does not have a GCF.
So we go right into factoring the trinomial of the form 
Step 1: Set up a product of two ( ) where each will
hold two terms. 
It will look like this: (
)( ) 
Step 2: Find the factors that go in the first positions. 
Since we have x squared as our first term,
we will need the following:
(x
)(x
) 
Step 3: Find the factors that go in the last
positions. 
We need two numbers whose product is 14 and sum is 5. That
would have to be 7 and 2.
Putting that into our factors we get: 

*7 and 2 are two numbers whose prod. is 14 and sum is 5

Note that if we would multiply this out we would get the original trinomial. 
Answer/Discussion
to 1b


*Factor out the GCF of 3b

We are not finished, we can still factor the trinomial. It is
of the form . 
Step 1: Set up a product of two ( ) where each
will hold two terms. 
It will look like this: (
)( ) 
Step 2: Use trial and error to find the factors needed. 
In the first terms of the binomials, we need factors of 2 a squared. This would have to be 2a and a.
In the second terms of the binomials, we need factors of 1.
This would have to be 1 and 1.
Also, we need to make sure that we get the right combination of these
factors. 
Answer/Discussion
to 1c

Note that this trinomial does not have a GCF. 
Step 1: Substitute x for the noncoefficient
part of the middle term.
Since x is already being used in this problem,
let's use y for our substitution. 
Let

*Substitute y in
for x squared 
Step 2: If the trinomial is in the form OR ,
factor it accordingly. 
Now it is in a form that we do know how to factor. 
In the first terms of the binomials we need factors of 2 y squared.
This would have to be 2y and y.
In the second terms of the binomials we need factors of 20.
This would have to be 5 and 4, 5 and 4, 2 and 10, 2 and 10, 20 and
1, or 20 and 1 .
Also we need to make sure that we get the right combination of these
factors. 
Step 3: Substitute back in what you replaced x with in step 1. 

*Substitute x squared
back in for y 
Note that if we would multiply this out, we would get the original
trinomial. 
Answer/Discussion
to 1d

Note that this trinomial does not have a GCF. 
Step 1: Set up a product of two ( ) where each will
hold two terms. 
It will look like this: (
)( ) 
Step 2: Find the factors that go in the first positions. 
Since we have x squared as our first term,
we will need the following:
(x
)(x ) 
Step 3: Find the factors that go in the last
positions. 
We need two numbers whose product is 10 and sum is 4.
Can you think of any????
Since the product is a positive number and the sum is a negative
number, we only need to consider pairs of numbers where both signs are
negative.
One pair of factors of 10 is 10 and 1, which does not add up to
be 4.
Another pair of factors are 2 and 5, which also does not add up
to 4.
Since we have looked at ALL the possible factors, and none of them
worked, we can say that this polynomial is prime.
In other words, it does not factor. 
Last revised on July 15, 2011 by Kim Seward.
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