Question

Factor completely.
6d2–4d–10

Answer 1

To factor the expression \(6d^2 - 4d - 10\), we can begin by looking for common factors in each term:

1. First, notice that each term in the expression has a common factor of 2. We can factor out the 2:

\[
2(3d^2 - 2d - 5)
\]

2. Next, we will focus on factoring the quadratic \(3d^2 - 2d - 5\). To factor this quadratic, we can look for two numbers that multiply to \(3 \times -5 = -15\) and add to \(-2\).

The numbers \(-5\) and \(3\) work, since:
\[
-5 + 3 = -2 \quad \text{and} \quad -5 \times 3 = -15
\]

3. Now, we can rewrite the middle term \(-2d\) using \(-5d\) and \(3d\):

\[
3d^2 - 5d + 3d - 5
\]

4. Next, we factor by grouping:

\[
(3d^2 - 5d) + (3d - 5)
\]

Factor out the common factors in each group:

\[
d(3d - 5) + 1(3d - 5)
\]

5. Now we can factor out the common binomial factor \(3d - 5\):

\[
(3d - 5)(d + 1)
\]

6. Finally, including the factor of 2 we factored out earlier, the complete factorization of the original expression \(6d^2 - 4d - 10\) is:

\[
2(3d - 5)(d + 1)
\]

So, the completely factored form is:

\[
\boxed{2(3d - 5)(d + 1)}
\]