How do you factor completely $bx^2-b = x-b^2x$ ?

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Answer 1 · 2018-05-02T09:17:51

$x=1/b$ or $x=-b$

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

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

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

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

i.e. either $bx-1=0$ i.e. $x=1/b$

or $x+b=0$ i.e. $x=-b$

How do you factor completely bx^2-b = x-b^2xbx^2-b = x-b^2x?