# How do you solve bx^2-b = x-b^2x by factoring?

Aug 25, 2015

color(green)( x = -b or x = 1/b

#### Explanation:

We can solve for $x$ by making Groups

$b {x}^{2} - b = x - {b}^{2} x$

Transposing the terms on the right to the left, we get:

$\implies b {x}^{2} + {b}^{2} x - x - b = 0$

Next, we form groups of 2 terms:

$\implies \left(b {x}^{2} + {b}^{2} x\right) - \left(x + b\right) = 0$

Then we take the common factors out from each group

$\implies b x \left(x + b\right) - 1 \left(x + b\right) = 0$

The factor $x + b$ is common to both the terms:

$\implies \left(x + b\right) \left(b x - 1\right) = 0$

This tells us that :

$x + b = 0 \mathmr{and} b x - 1 = 0$

color(green)( x = -b or x = 1/b