Question

Factor completely.
3p2+7p–6

Answer 1

To factor the quadratic expression \( 3p^2 + 7p - 6 \), we need to find two numbers that multiply to \( a \cdot c \) (the product of the coefficient of \( p^2 \) and the constant term) and add up to \( b \) (the coefficient of \( p \)).

Here, \( a = 3 \), \( b = 7 \), and \( c = -6 \).

First, compute \( a \cdot c = 3 \cdot (-6) = -18 \).

Now we are looking for two numbers that multiply to \( -18 \) and add up to \( 7 \). The numbers that satisfy this are \( 9 \) and \( -2 \):

- \( 9 \cdot -2 = -18 \)
- \( 9 + (-2) = 7 \)

Next, we can rewrite the middle term of the quadratic using these numbers:

\[
3p^2 + 9p - 2p - 6
\]

Now, we can group the terms:

\[
(3p^2 + 9p) + (-2p - 6)
\]

Factoring out the common factors in each group gives us:

\[
3p(p + 3) - 2(p + 3)
\]

Now we can factor out \( (p + 3) \):

\[
(3p - 2)(p + 3)
\]

Thus, the completely factored form of \( 3p^2 + 7p - 6 \) is:

\[
\boxed{(3p - 2)(p + 3)}
\]