# How do you factor and solve x^2+2x-8=0?

Jun 10, 2016

The solutions for the equation are color(blue)(x=2 , color(blue)(x=-4

#### Explanation:

${x}^{2} + 2 x - 8 = 0$

We can Split the Middle Term of this expression to factorise it.

In this technique, if we have to factorise an expression like $a {x}^{2} + b x + c$, we need to think of 2 numbers such that:

${N}_{1} \cdot {N}_{2} = a \cdot c = 1 \cdot \left(- 8\right) = - 8$

AND

${N}_{1} + {N}_{2} = b = 2$

After trying out a few numbers we get ${N}_{1} = - 2$ and ${N}_{2} = 4$
$\left(- 2\right) \cdot 4 = - 8$, and $- 2 + 4 = 2$

${x}^{2} + \textcolor{b l u e}{2 x} - 8 = {x}^{2} + \textcolor{b l u e}{4 x - 2 x} - 8$

$= x \left(x + 4\right) - 2 \left(x + 4\right)$

$\textcolor{b l u e}{\left(x + 4\right)}$ is a common factor to each of the terms

 color(green)(( x - 2 )(x+4)  is the factorised form of the expression. Now we equate the factors to zero.

• color(blue)(x-2) =0 , color(blue)(x=2

• color(blue)(x + 4) =0 , color(blue)(x=-4

Jun 10, 2016

2 and -4

#### Explanation:

Solve $y = {x}^{2} + 2 x - 8.$
The 2 real roots have opposite signs because ac < 0
Find 2 real roots knowing sum (-b = -2) and product (c = -8).
Factor pairs of (c = -8) --> (-2, 4)(2, -4). This last sum is (-2 = -b).
Therefor, the 2 real roots are: 2 and -4.