Question

factor completely.
$g^{2}f + gf + 3g^{2} + 3g$

Explanation:

Step1: Group the terms

We can group the first two terms and the last two terms together:
\((g^{2}f + gf) + (3g^{2} + 3g)\)

Step2: Factor out the GCF from each group

For the first group \((g^{2}f + gf)\), the greatest common factor (GCF) is \(gf\). Factoring it out, we get:
\(gf(g + 1)\)

For the second group \((3g^{2} + 3g)\), the GCF is \(3g\). Factoring it out, we get:
\(3g(g + 1)\)

So now the expression becomes:
\(gf(g + 1) + 3g(g + 1)\)

Step3: Factor out the common binomial factor

Now we can see that both terms have a common factor of \((g + 1)\) and also a common factor of \(g\) (from \(gf\) and \(3g\)). Wait, actually, the common binomial factor is \((g + 1)\), and the other common factor is \(g\) (since \(gf\) and \(3g\) both have \(g\)). Wait, let's re - examine:

\(gf(g + 1)+3g(g + 1)=g(g + 1)(f + 3)\)

We factor out \(g\) from \(gf\) and \(3g\) first (wait, actually, the common factor between \(gf(g + 1)\) and \(3g(g + 1)\) is \(g(g + 1)\)? No, let's do it step by step.

The two terms are \(gf(g + 1)\) and \(3g(g + 1)\). The greatest common factor of these two terms is \(g(g + 1)\)? Wait, no. Let's factor out \(g(g + 1)\):

\(gf(g + 1)+3g(g + 1)=g(g + 1)(f + 3)\)

Answer:

\(g(g + 1)(f + 3)\)