For example, the greatest common factor in \(3x^2+9x\) is \(3x\) since you can evenly divide each term of \(3x^2+9x\) by both a 3 and \(x\), but cannot divide each term by higher powers of \(3\) or \(x\), or any other factor for that matter.
Essentially, what the above entails is determining everything that a collection of terms has in common, and then systematically “undistributing” those factors by pulling them out in front of the quantity.
In this expression, the Greatest Common Factor is \(2x\) because \(2\) and \(x\) are the only factors that both evenly divide \(4x^2\) and \(6x\). So we drop some parentheses around the whole quantity, “undistribute” this \(2x\), writing \(2x\) out in front of the quantity, and remove a factor of \(2\) and \(x\) from both terms (dividing each term by \(2x\)). This gives you the following:
$$\begin{align}4x^2+2x&=(4x^2+2x)\\&=2x(2x+3)\end{align}$$
Note that if you were distribute the \(2x\) back into the quantity, you would get what you started with before factoring. This is a useful way of making sure you factored correctly.
Effectively, factoring “undoes” the process of FOILing or distribution.
So, if you want to make sure you did your factoring correctly (either here or in any factoring problem), just expand/distribute/FOIL your answer and see if it gives you the expression you started with. If so, you did it right!
Therefore, if you see factoring problems on an exam or quiz or homework, it is always a good idea to check your work in this way if you have time. If you don’t check yourself, you are leaving your grade up to chance unnecessarily… and that can be very deadly on an exam.
\(3(x^2+3)\)
The greatest common factor of the terms \(3x^2\) and \(9\) is 3; i.e. 3 is the “largest” factor of both of these. So plop the \(3\) out in front of the whole quantity, then divide the numbers you see inside the quantity by \(3\).
\(2x(x+4)\)
\(2x(6x^2-2x+1)\)
\(x(13x-24)\)
\(-2x^2y^4(2x-y)\)
\(4k^2m^3(1+2k^2-3m)\)
\((x+y)(5+6x)\)
\((2x+1)(4x-2)\)