Factoring Trinomials Whose Leading Coefficient is One

I. Procedure, Last sign positive
A. Factor out the GCF.
B. Set up two sets of parentheses.
C. Determine the factors of the First term.
D. Determine the signs in the parentheses:

1. If the last term is positive, the signs will be the same as the middle term:
a. ( + ) ( + )
b. ( - ) (- )
2. If the last term is negative, the signs will be positive and negative.

E. Determine the factors of the last term.
F. Check to ensure that you get the proper middle term.
G. This whole process is called the Trial and Error method.

II. Examples
A. Factor each of the following completely.
1. x² + 13x + 42
There is no GCF, so we set up parentheses immediately.
To get x², we have to multiply x times x.

(x )(x )
Since the last term is +, the signs will be the same as the middle term,
namely, +.

(x + )(x + )
Now, the factors of 42 are:
1 and 42, 2 and 21, 3 and 14, 6 and 7
Let’s try 1 and 42. We get:

(x + 1)(x + 42)
From the way we’ve set this up, we know that when we multiply it out, we will
get a first term of x² and a last term of 42. However, multiplying the outside
terms will give us 42x, multiplying the inside terms will give us 1x, and 42x +
1x is not 13x. So this is not the correct factorization.

We need to try all the other possibilities [(x + 2)(x + 21); (x + 3)(x + 14); and
(x + 6)(x + 7)] until we find the one that will give us the desired middle term.
In general, it is a good idea to start out with the factors that are closest
together and work your way out.

So it turns out that the answer in this case is:
Answer: (x + 6)(x + 7)

2. r² – 13r + 40
Again, no GCF, so we set up the parentheses.
To get r², we multiply r times r.

(r )(r )
Since the last sign is positive, the signs need to be - and - .
 

r - )(r - )
Finally, the factors of 40 are 1 and 40, 2 and 20, 4 and 10, 5 and 8. Let’s first
try 5 and 8 since they are closest together. We get:

(r - 5)(r - 8)
Checking to make sure that we get the correct middle term, we get:
r² - 8r - 5r + 40 = r² - 13r + 40
Which is what we want.
Answer: (r - 5)(r - 8)

3. Now you try one: r² – 9r + 20
Answer: (r – 5)(r – 4)

B. More Challenging Problems
1. 3x² – 24x + 36
First, we notice that there is a GCF, namely 3. Factoring that out, we get:

3(x² – 8x + 12)
The 3 is part of our answer, so we carry it along from now on. Setting up the
parentheses, observing that to get x² we have to multiply x times x, we get:

3(x )(x )
Since the last sign is positive, the signs must be the same as the middle term,
namely - and -.

3(x - )(x - )
The factors of 12 are 1 and 12, 2 and 6, 3 and 4. Let’s try 3 and 4 first:

3(x - 3)(x - 4)
Note that we just need to multiply (x - 3)(x - 4) to check.

x² - 3x - 4x + 12 = x² - 7x + 12
This is not the correct middle term, so let’s try 2 and 6 next:

3(x - 2)(x - 6)
Checking by multiplying, we get:

x² - 6x - 2x + 12 = x² - 8x + 12
This is what we want.
Answer: 3(x - 2)(x – 6)

2. Now you try one: 5r³ + 45r² + 100r
Answer: 5r(r + 4)(r + 5)

3. x² – 13xb + 36b²
This one is approached the same as the others, the only difference being that
there are two letters.

There is no GCF, setting up the parentheses, with x² being x times x, we get:

(x )(x )
The last sign is +, so the signs need to be the same as the middle term,
namely -.

(x - )(x - )
Now, the factors of 36b² are 1b and 36b, 2b and 18b, 3b and 12b, 4b and 9b,
6b and 6b. Let’s try 6b and 6b.

(x – 6b)(x – 6b)
Checking, we get -6xb - 6xb = -12xb, which is not what we want. So let’s try
4b and 9b.

(x – 4b)(x – 9b)
Checking, we get -9xb – 4xb = -13xb, which is what we want.
Answer: (x – 4b)(x – 9b)

4. Now you try one: x²+ 7xa + 10a²
Answer: (x + 5a)(x + 2a)

III. Procedure, last sign negative.
A. Factor out the GCF.
B. Set up two sets of parentheses.
C. Determine the factors of the First term.
D. Determine the signs in the parentheses:

1. If the last term is positive, the signs will be the same as the middle term:
a. ( + ) ( + )
b. (- ) (- )
2. If the last term is negative, the signs will be positive and negative.

E. Determine the factors of the last term.
F. Check to ensure that you get the proper middle term.
G. This whole process is called the Trial and Error method.

IV. Examples
A. Factor each of the following completely.
1. r² – r – 2
Again, no GCF, so we set up the parentheses.
To get r², we multiply r times r.

(r )(r )
Since the last sign is negative, the signs need to be + and - .
(r + )(r - )
Finally, the factors of 2 are 1 and 2.
(r + 1)(r – 2)
Checking, we see that we get the correct middle term, so the answer is:
Answer: (r + 1)(r – 2)

2. x² + 8x - 9
Again, there is no GCF, so we set up the parentheses. To get x² we need to
multiply x times x:
(x )(x )
Since the last term is negative, the signs need to be + and -:
(x + )(x - )
The factors of 9 and 1 and 9, 3 and 3. Let’s start with the factors that are
closest together, 3 and 3:
(x + 3)(x - 3)
Checking by multiplying, we get:
x² - 3x + 3x - 9 = x² - 9
This is not what we want, so we need to try 1 and 9:
(x + 1)(x - 9)
Checking by multiplying, we get:
x² - 9x + x - 9 = x² - 8x - 9
Our middle term has the correct number, wrong sign. So we switch the signs:
Answer: (x - 1)(x + 9)

3. Now you try one: r² – r - 20
Answer: (r – 5)(r + 4)

B. More Challenging Problems
1. 3x³ – 3x² – 18x
First, we notice that there is a GCF, 3x. Factoring that out, we get:
3x(x² – x – 6)
The 3x is part of our answer, so we carry it along from now on. Setting up the
parentheses, observing that to get x² we have to multiply x times x, we get:
3x(x )(x )
Since the last sign is negative, the signs must be + and -.
3x(x + )(x - )
The factors of 6 are 1 and 6, 2 and 3. Let’s try 2 and 3.
3x(x + 2)(x – 3)
Checking, we get –3x + 2x = -x, which is what we want.
Answer: 3x(x + 2)(x – 3)

2. Now you try one: 5r³ + 5r² - 100r
Answer: 5r(r - 4)(r + 5)

3. m³ n - 2m² n - 3mn
We begin by noticing that each term is divisible by mn, so we’ll factor out this
GCF to get:
mn(m² - 2m - 3)
Now we can concentrate on the trinomial in the parentheses. To get m², we
need to multiply m times m:
mn(m )(m )
Since the last sign is negative, our signs have to be + and -:
mn(m + )(m - )
To get 3, we need to multiply 3 times 1:
mn(m + 1)(m - 3)
Now we need to check by multiplying (m + 1) by (m - 3):
m² - 3m + m - 3 = m² - 2m - 3
This is what we want.
Answer: mn(m + 1)(m - 3)

4. Now you try one: z¹⁰ - 4z⁹ - 21z⁸
Answer: z⁸(z + 3)(z - 7)