How do you factor 2d^3-d^2-3+6d?

Apr 29, 2015

You may take a close look at the co-efficients.

Some rearranging:
$= \left(2 {d}^{3} - {d}^{2}\right) + \left(6 d - 3\right)$

Then we take common things out of the brackets:
$= {d}^{2} \left(2 d - 1\right) + 3 \left(2 d - 1\right)$

Since inside the brackets is now the same, we may go:
$= \left({d}^{2} + 3\right) \left(2 d - 1\right)$

${d}^{2} + 3$ cannot be factored further.