Question

Factor completely.
3j2+10j+8

Answer 1

To factor the quadratic expression \(3j^2 + 10j + 8\), we can start by identifying the coefficients:

- \(a = 3\) (coefficient of \(j^2\))
- \(b = 10\) (coefficient of \(j\))
- \(c = 8\) (constant term)

We are looking for two numbers that multiply to \(a \cdot c = 3 \cdot 8 = 24\) and add up to \(b = 10\).

The two numbers that satisfy these conditions are \(6\) and \(4\), since:

- \(6 + 4 = 10\)
- \(6 \cdot 4 = 24\)

Next, we can rewrite the middle term \(10j\) using \(6j\) and \(4j\):

\[
3j^2 + 6j + 4j + 8
\]

Now we group the terms:

\[
(3j^2 + 6j) + (4j + 8)
\]

Factor out the common terms in each group:

\[
3j(j + 2) + 4(j + 2)
\]

Now, notice that \((j + 2)\) is common in both terms:

\[
(3j + 4)(j + 2)
\]

The expression is now factored completely. Thus, the complete factorization of \(3j^2 + 10j + 8\) is:

\[
\boxed{(3j + 4)(j + 2)}
\]