Greatest Common Factor Review

I. Remember
A. The Greatest Common Factor (GCF) of a group of numbers is the largest number
that will divide evenly into all the numbers.

B. For example, the GCF of 20 and 32 is 4 since

1. There is not a number larger than 4 that goes evenly into both, and
2. Four does go evenly into both.

C. Two numbers are said to be relatively prime if the GCF of the two numbers is 1.

II. GCF’s of Polynomials
A. The GCF for a polynomial is the monomial with the greatest coefficient and the
lowest power on the variable(s) that will divide evenly into each term of the
polynomial.

B. When we say factor the GCF of a polynomial, we mean to divide each term of the
polynomial by the GCF and write the resulting answer as a product ( times the GCF ).

C. Examples – Factor out the GCF .

1. 49xy – 21x²y² + 35x³y³
First, the largest number that will go evenly into all the coefficients is 7.
Next, each term contains an “x” and a “y”, with the lowest exponent on each
being 1.
So factoring out 7xy, we need to divide each term by that to get::
Answer: 7xy(7 – 3xy + 5x²y²)

2. 5ab² + 10a²b² + 15a³b²
First, the largest number that will go evenly into all the coefficients is 5.
Next, each term contains an “a” and a “b”; with the lowest exponent on “a”
being 1 and the lowest exponent on “b” being 2.
So factoring out 5ab2, we need to divide each term by that to get:
Answer: 5ab²(1 + 2a + 3a²)

3. Now you try one: 16x⁴ – 20x² – 16x
Answer: 4x(4x³ – 5x – 4)

4. 9x – 12y
First, the largest number that will go into both coefficients is 3.
Next, there are no common variables .
So factoring out 3, we will divide by that to get:
Answer: 3(3x – 4y)

5. Now you try one: 14x + 21
Answer: 7(2x + 3)

III. Factoring by Grouping
A. Procedure
1. Group terms that are similar together. Hint: We will usually group the terms
in half, i.e. – If there are four terms, we’ll group them 2 by 2, 6 terms, 3 by 3,
etc.
2. Factor out the GCF of each group.
3. Introduce a dummy variable (optional).
4. Factor out the GCF of the result.
5. Substitute back in for the dummy variable (optional, unless you did step 3,
then it’s required).
6. Check by multiplying.

B. Examples – Factor by Grouping
1. x² – ax + bx – ab
First, let’s group these 2 by 2 to get:
(x² – ax) + (bx – ab)
Notice that in the first group, the GCF is x. In the second group, the GCF is b.
Factor out both to get:
x(x – a) + b(x – a)
What makes this difficult to work with now is the parenthesis. So we are
going to introduce a dummy variable to take the place of the parenthesis and
make what looks difficult into something much simpler .
Let u = (x – a)
Substituting, we get:
xu + bu
We can now see that “u” is the GCF, so we will factor it out to get:
u(x + b)
But “u” was not in the original problem, so it can’t be in the answer. But what
is “u” equal to? Let’s substitute that into this expression to get :
Answer: (x – a)(x + b)

2. 3ax + 21x – a – 7
Let’s first group these 2 by 2 to get:
(3ax + 21x) – (a + 7)
Notice that we changed the sign on both the a AND the 7 in the second
group. The reason we do this is because if we decided to do something
rather stupid and say multiply out what we just did, we want to make sure that
we get back what we started with.
So now, we notice that the GCF of the first group is 3x. We also notice that
the second group does not have a GCF (other than 1). Factoring out 3x, we
get:
3x(a + 7) – (a + 7)
Let’s introduce a dummy variable:
Let u = (a + 7)
We now substitute for the parenthesis to get:
3xu – u
We immediately notice that “u” is the GCF. Factoring it out, we get:
u(3x – 1)
Now, substitute back in for “u” to get:
Answer: (a + 7)(3x – 1)

3. Now you try one: xy + 2x + 4y + 8
Answer: (y + 2)(x + 4)