# How do you factor completely 2x^2+6x-80?

May 12, 2016

y = 2(x - 5)( x + 8)

#### Explanation:

$y = 2 \left({x}^{2} + 3 x - 40\right)$
Factor the trinomial in parentheses.
Find 2 numbers knowing sum (b = 3) and product (c = -40).
They are (-5, 8).
y = 2(x - 5)(x + 8)

May 12, 2016

$2 {x}^{2} + 6 x - 80 = \textcolor{b l u e}{2 \left(x - 5\right) \left(x + 8\right)}$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} 2 {x}^{2} + 6 x - 80$

First observe that all the terms have a constant factor of $2$;
so we can write:
$\textcolor{w h i t e}{\text{XXX}} 2 \left({x}^{2} + 3 x - 40\right)$

Next lets think about factors of $40$ whose difference is $3$

$40 = 2 \times 20$ ...not a difference of $3$
$40 = 4 \times 10$ ...not a difference of $3$
$40 = 5 \times 8$ ...looks as if we've found it

In $\left({x}^{2} + 3 x - 40\right)$ the last term $\left(- 40\right)$ is negative so one of $5$ or $8$ must be negative;
and the middle term $\left(+ 3 x\right)$ is positive so the larger of $5$ and $8$ must be positive.

Therefore our complete factors are:
$\textcolor{w h i t e}{\text{XXX}} 2 \left(x - 5\right) \left(x + 8\right)$