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

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Answer 1 · 2018-05-02T09:17:51

$x = \frac{1}{b}$ $x=1/b$ or $x = - b$ $x=-b$

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

$\Rightarrow b x^{2} + b^{2} x - x - b = 0$ $=>bx^2+b^2x-x-b=0$

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

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

i.e. either $b x - 1 = 0$ $bx-1=0$ i.e. $x = \frac{1}{b}$ $x=1/b$

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

How do you factor completely bx^2-b = x-b^2xbx2−b=x−b2xbx^2-b = x-b^2x?