# Factoring Out Common Monomial Factors

Hello! Welcome to this video on **factoring polynomials using common monomial factors**. When factoring polynomials, this is always the first thing we want to look for, so let’s get some practice under our belts so we have a strong understanding of how this works.

Factor the binomial \(3x^{3}+6x\).

Start by finding a common factor between the two terms (\(3x^{3}\) and \(6x\)). These both have a common factor of \(3x\), which is the monomial we are factoring out. We are going to “undistribute” and pull \(3x\) out of both of these terms. Then we’ll write what’s left over in parentheses.

\(3x(\text{ })\)

To figure out what goes in the parentheses, divide each term by \(3x\).

\(\frac{3x^{3}}{3x} =\frac{x^{3}}{x} =x^{3-1}=x^{2}\)

and

\(\frac{6x}{3x} =\frac{2x}{x} =2\)

These terms added together are what go inside the parentheses. Therefore, we can write our factorization like this:

\(3x(x^{2}+2)\)

If we were to simplify this by distributing the \(3x\) into the set of parentheses, we would get the original binomial.

Let’s try another one!

Factor the polynomial \(21a^{4}b^{2}-14a^{3}b^{5}\).

Start by finding the greatest common factor between these two terms. Do this by finding the GCF of the constants (21 and 14), \(a\)-terms (\(a^{4}\) and \(a^{3}\)), and the \(b\)-terms (\(b^{2}\) and \(b^{5}\)) separately, and then multiply these together.

The GCF of 21 and 14 is 7.

The GCF of \(a^{4}\) and \(a^{3}\) is \(a^{3}\). The GCF of a variable raised to an exponent will always be that variable raised to the lowest exponent in the list. In this case, it’s 3.

The GCF of \(b^{2}\) and \(b^{5}\) is \(b^{2}\).

Therefore, the GCF of \(21a^{4}b^{2}\) and \(14a^{3}b^{5}\) is \(7a^{3}b^{2}\). Remember, we found the GCF of each of these sets of terms separately and then multiplied them together.

Now, factor out \(7a^{3}b^{2}\) from both terms.

\(7a^{3}b^{2}(\text{ })\)

Divide each term by \(7a^{3}b^{2}\).

\(\frac{21a^{4}b^{2}}{7a^{3}b^{2}}=\frac{3a^{4}b^{2}}{a^{3}b^{2}}=\frac{3ab^{2}}{b^{2}}=3a\)

\(\frac{-14a^{3}b^{5}}{7a^{3}b^{2}}=\frac{-2a^{3}b^{5}}{a^{3}b^{2}}=\frac{-2b^{5}}{b^{2}}=-2b^{3}\)

Add these terms together to get what goes inside the parentheses. This means our answer is:

\(7a^{3}b^{2}(3a-2b^{3})\)

Remember, if you distributed \(7a^{3}b^{2}\) into the parentheses, you would get the original expression, so that’s a good way to double check your answer to make sure you have the correct factored form.

Before we go, let’s try one last problem.

Factor the trinomial \(4x^{2}y+2xy^{2}-8x^{2}y^{2}\).

We need to start by finding a common factor between all three of these terms. We can do this by finding the greatest common factor (GCF) of the constants, \(x\)-terms, and \(y\)-terms, just like we did last time.

The GCF of 4, 2, and 8 is 2.

The GCF of \(x^{2}\), \(x\), and \(x^{2}\) is \(x\) because that’s the \(x\)-term with the lowest exponent. Remember, if there’s no exponent, it’s assumed to be 1.

The GCF of \(y\), \(y^{2}\), and \(y^{2}\) is \(y\).

This means that the GCF of \(4x^{2}y\), \(2xy^{2}\), and \(8x^{2}y^{2}\) is \(2xy\). So let’s factor \(2xy\) out of this trinomial.

\(2xy(\text{ })\)

Now, divide each term by \(2xy\).

\(\frac{4x^{2}y}{2xy}=2x\)

Notice that I did significantly fewer steps right here. I really followed all the same steps, but I didn’t write as much. So if you get more comfortable and don’t want to write out every intermediate step, that’s totally fine.

\(\frac{2xy^{2}}{2xy}=y\)

\(\frac{-8x^{2}y^{2}}{2xy} =-4xy\)

These terms added together are what go inside the parentheses. This means our answer is:

\(2xy(2x+y-4xy)\)

Remember, one way to check your work is by distributing your factored expressions. If you do, you should always get the original polynomial. If you don’t double check your work because you missed something along the way.

I hope this video on factoring out common monomial factors was helpful. Thanks for watching, and happy studying!