How do you solve x^2-4x-5=0 by factoring?

Sep 29, 2015

The solutions are
 color(blue)(x=-1
 color(blue)(x=+5

Explanation:

x^2−4x−5=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:

${N}_{1} \cdot {N}_{2} = a \cdot c = 1 \cdot \left(- 5\right) = - 5$
AND
${N}_{1} + {N}_{2} = b = - 4$

After trying out a few numbers we get ${N}_{1} = - 5$ and ${N}_{2} = 1$

$\left(1\right) \cdot \left(- 5\right) = - 5$, and $1 + \left(- 5\right) = - 4$

x^2−color(blue)(4x)−5=x^2−color(blue)(5x+1x)−5

=x(x-5) +1(x−5)

=color(blue)((x+1)(x-5)  are the factors.

We can now obtain the solutions by equating the factors to zero.
=x+1 =0, color(blue)(x=-1
=x-5 =0, color(blue)(x=+5