How do you simplify $(x^2-3x+2)/(2x^2-2x)$ ?

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Answer 1 · 2018-03-15T16:45:29

See a solution process below:

We can factor the numerator and denominator as;

$((x - 2)(x - 1))/(2x(x - 1))$

We can now cancel common term in the numerator and denominator:

$((x - 2)cancel((x - 1)))/(2xcancel((x - 1))) =>$

$(x - 2)/(2x)$

However, we cannot divide by $0$ so we must exclude:

$2x = 0 => x = 0$ and $x - 1 = 0 => x 1$

$(x^2 - 3x + 2)/(2x^2 - 2x) = (x - 2)/(2x)$ Where: $x != 0$ and $x != 1$

Or

$(x^2 - 3x + 2)/(2x^2 - 2x) = x/(2x) - 2/(2x) = 1/2 - 1/x$ Where: $x != 0$ and $x != 1$

How do you simplify (x^2-3x+2)/(2x^2-2x) (x^2-3x+2)/(2x^2-2x)?