Lane ORCCA (2020–2021): Open Resources for Community College Algebra

Example 7.1.1. Finding the Greatest Common Factor.

What is the common factor between \(6x^2\) and \(70x\text{?}\) Break down each of these into its prime factors:

With \(2\) and \(x\) in common, the greatest common factor is \(2x\text{.}\)

Example 7.1.5.

Factor the polynomial \(3x^3+3x^2-9\text{.}\)

1. We identify the greatest common factor as \(3\text{,}\) because \(3\) is the only common factor between \(3x^3\text{,}\) \(3x^2\) and \(9\text{.}\)

2. We write:

   \begin{equation*} 3x^3+3x^2-9=3(\phantom{x^2}+\phantom{x^2}-\phantom{3})\text{.} \end{equation*}

3. We ask the question “\(3\) times what gives \(3x^3\text{?}\)” The answer is \(x^3\text{.}\) Now we have:

   \begin{equation*} 3x^3+3x^2-9=3(x^3+\phantom{x^2}-\phantom{3})\text{.} \end{equation*}

   We ask the question “\(3\) times what gives \(3x^2\text{?}\)” The answer is \(x^2\text{.}\) Now we have:

   \begin{equation*} 3x^3+3x^2-9=3(x^3+x^2-\phantom{3})\text{.} \end{equation*}

   We ask the question “\(3\) times what gives \(9\text{?}\)” The answer is \(3\text{.}\) Now we have:

   \begin{equation*} 3x^3+3x^2-9=3(x^3+x^2-3)\text{.} \end{equation*}

To check that this is correct, multiplying through \(3(x^3+x^2-3)\) should give the original expression \(3x^3+3x^2-9\text{.}\) We check this, and it does.

Example 7.1.9.

Factor \(-35m^5+5m^4-10m^3\text{.}\)

First, we identify the greatest common factor. The number \(5\) is the greatest common factor of the three coefficients (which were \(-35\text{,}\) \(5\text{,}\) and \(-10\)) and also \(m^3\) is the largest expression that divides \(m^5\text{,}\) \(m^4\text{,}\) and \(m^3\text{.}\) Therefore, the greatest common factor is \(5m^3\text{.}\)

In this example, the leading term is a negative number. When this happens, we will make it common practice to take that negative as part of the greatest common factor. So we will proceed by factoring out \(-5m^3\text{.}\) Note the sign changes.

Example 7.1.10.

Factor \(14-7n^2+28n^4-21n\text{.}\)

Notice that the terms are not in a standard order, with powers of \(n\) decreasing as you read left to right. It is usually a best practice to rearrange the terms into the standard order first.

The number \(7\) divides all of the numerical coefficients. Separately, no power of \(n\) is part of the greatest common factor because the \(14\) term has no \(n\) factors. So the greatest common factor is just \(7\text{.}\) We proceed by factoring that out:

Example 7.1.11.

Factor \(24ab^2+16a^2b^3-12a^3b^2\text{.}\)

There are two variables in this polynomial, but that does not change the factoring strategy. The greatest numerical factor between the three terms is \(4\text{.}\) The variable \(a\) divides all three terms, and \(b^2\) divides all three terms. So we have:

Example 7.1.12.

Factor \(4m^2n-3xy\text{.}\)

There are no common factors in those two terms (unless you want to count \(1\) or \(-1\text{,}\) but we do not count these for the purposes of identifying a greatest common factor). In this situation we can say the polynomial is prime, and leave it as it is.

Example 7.1.13.

Factor \(-x^3+2x+18\text{.}\)

There are no common factors in those three terms, and it would be correct to state that this polynomial is prime. However, since its leading coefficient is negative, it may be wise in some situations to factor out a negative sign. So, it could be factored as \(-\mathopen{}\left(x^3-2x-18\right)\mathclose{}\text{.}\) Note that every term is negated as the leading negative sign is extracted.