How do you factor 6a^2+2a-1-8a^2+5?

Jun 22, 2016

$- 2 \left(\left(a - 2\right) \left(a + 1\right)\right)$

Explanation:

Group like terms
$= - 2 {a}^{2} + 2 a + 4$
Factor out the common term (-2)
$= - 2 \left({a}^{2} - a - 2\right)$
Now factor in form ${x}^{2} + b x + c$ into $\left(x + a\right) \left(x + b\right)$
Think what numbers add to -1 and multiply to -2
-2 and 1 fit.
Therefore, $= - 2 \left(\left(a - 2\right) \left(a + 1\right)\right)$

Let's verify our solution using $a = 5$

$- 2 \left(\left(5 - 2\right) \left(5 + 1\right)\right) = 6 {\left(5\right)}^{2} + 2 \left(5\right) - 1 - 8 {\left(5\right)}^{2} + 5$
$- 2 \left(\left(3\right) \left(6\right)\right) = 6 \left(25\right) + 10 - 1 - 8 \left(25\right) + 5$
$- 2 \left(18\right) = 150 + 10 - 1 - 200 + 5$
$- 36 = - 50 + 14$
$- 36 = - 36$

Try for any other $a$ and it will work.