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

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Answer 1 · 2015-08-25T13:20:43

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

We can solve for $x$ by making Groups

$bx^2-b = x-b^2x$

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

$=> bx^2 + b^2x - x - b = 0$

Next, we form groups of 2 terms:

$=> (bx^2 + b^2x) - (x + b) = 0$

Then we take the common factors out from each group

$=> bx(x+b) - 1(x+b) = 0$

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

$=> (x + b)(bx - 1) = 0$

This tells us that :

$x + b = 0 or bx - 1 = 0$

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

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