You already know how to use the distributive property of multiplication to multiply a monomial by a polynomial. For example, you can easily perform the following multiplication:

\(2x(3x + 4y) = 6x^2 + 8xy\)

But is there a way to begin with the polynomial, and break it down into its original factors? That is, could we look at a polynomial like \(6x^2 + 8xy\) and figure out that it is equal to \(2x(3x + 4y)\)?. The answer is yes. We will use a method called *factoring*.

In this article we will learn to *factor* a polynomial by searching for the *greatest common factor* of all of the terms in the polynomial. Let’s get right to it. Consider the polynomial

\(80v^2u - 8v^3 + 40v^2\)

We wish to find the *greatest common factor* of the three different terms above. Let’s consider them one at a time.

By breaking \(80v^2u\) into its *prime factorization*, we find that

\(80v^2u = 2\cdot 2\cdot 2\cdot 2\cdot 5\cdot v\cdot v\cdot u\)

Similarly, we look at \(-8v^3\) and we find

\(-8v^3 = -1\cdot 2\cdot 2\cdot 2\cdot v\cdot v\cdot v\)

Finally, we break down \(40v^2\) into

\(40v^2 = 2\cdot 2\cdot 2\cdot 5\cdot v\cdot v\)

Now we pick out the numbers that are common to every single term. We see that \(2\) appears in each term *three times*. So we will be able to *factor out \(2\cdot 2\cdot 2 = 8\) *from the polynomial. Also, we see that each and every term contains \(v\cdot v = v^2\). So we factor it out. Then

\(80v^2u - 8v^3 + 40v^3 = 8v^2\)(???)

Well, we have factored out the greatest common factor \(8v^2\). Now what is left of the polynomial inside the parentheses? Whatever is left from each term! Let’s work it out.

When we factor out \(8v^2\) from \(80v^2u\), we are left with \(2\cdot 5\cdot u = 10u\). Then \(10u\) will be inside the parentheses.

Similarly, when we factor out \(8v^2\) from \(-8v^3\), we are left with \(-v\).

Finally, when we factor out \(8v^2\) from \(40v^2\), we are left with \(5\).

Then our final *factorization* is \(80v^2u - 8v^3 + 40v^2 = 8v^2(10u - v + 5)\). Congratulation, you have just *factored!*

Let’s quickly try another example. Factor the polynomial \(-12x + 6xy^2 - 15x^3y^3\). To do this we consider each term separately and determine the greatest common factor. Can you figure out what it is? The greatest common factor is \(3x\)! ! If you cannot see this, repeat the procedure we did above for finding the greatest common factor. Well what happens when we factor out \(3x\) from each term? The first term is left with \(-4\), the second term is left with \(2y^2\), and the third term is left with \(-5x^2y^3\). Then our final factorization is

\(-12x + 6xy^2 - 15x^3y^3 = 3x(-4 + 2y^2 - 5x^2y^3)\)

Bellow you can **download** some** free** math worksheets and practice.

Factor the common factor out of each expression.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch bellow how to solve this example:**

Factor the common factor out of each expression.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch bellow how to solve this example:**

Factor the common factor out of each expression.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch bellow how to solve this example:**