# How do you solve -16p^2-64p-64=0 by factoring?

Aug 20, 2015

The solution is
color(green)(p=-2

#### Explanation:

$- 16 {p}^{2} - 64 p - 64 = 0$

We can Split the Middle Term of this expression to factorise it and thereby find solutions.

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

${N}_{1} \cdot {N}_{2} = a \cdot c = - 16 \cdot - 64 = 1024$
and
${N}_{1} + {N}_{2} = b = - 64$

After trying out a few numbers we get ${N}_{1} = - 32$ and ${N}_{2} = - 32$

$- 32 \cdot - 32 = 1024$, and

$\left(- 32\right) + \left(- 32\right) = - 64$

$- 16 {p}^{2} - 64 p - 64 = - 16 {p}^{2} - 32 p - 32 p - 64$

$= - 16 p \left(p + 2\right) - 32 \left(p + 2\right) = 0$

$\left(p + 2\right)$ is a common factor to each of the terms

color(green)((-16p-32)(p+2)=0

we now equate the factors to zero:

-16p-32=0, -16p=32, color(green)(p=-2

p+2=0, color(green)(p=-2