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#