Question

Factor completely.
2g2–3g–2

Answer 1

To factor the quadratic expression \(2g^2 - 3g - 2\), we'll start by looking for two numbers that multiply to \(a \cdot c\) (in this case, \(2 \cdot -2 = -4\)) and add to \(b\) (which is \(-3\)).

The two numbers that meet these criteria are \(-4\) and \(1\) because:
- \(-4 \cdot 1 = -4\)
- \(-4 + 1 = -3\)

Now we can rewrite the middle term, \(-3g\), using \(-4g\) and \(1g\):

\[
2g^2 - 4g + 1g - 2
\]

Next, we group the terms:

\[
(2g^2 - 4g) + (1g - 2)
\]

Now we factor out the greatest common factor from each group:

\[
2g(g - 2) + 1(g - 2)
\]

Now we can factor out the common binomial factor \((g - 2)\):

\[
(2g + 1)(g - 2)
\]

Therefore, the completely factored form of the expression \(2g^2 - 3g - 2\) is:

\[
\boxed{(2g + 1)(g - 2)}
\]