Answer
$\dfrac{3x^2-4x+4}{(x-1)^2}$
Work Step by Step
Factor $x^2-2x+1$ using the special formula $a^2-2ab+b^2=(a-b)^2$ to obtain:
$x^2-2x+1=(x-1)^2$
Substitute back the factors into the given expression.
$=\dfrac{3x}{x-1}-\dfrac{x-4}{(x-1)^2}$
The LCD is $(x-1)^2$.
Make the expressions similar by multiplying $x-1$ to both the numerator and denominator of the first expresison to obtain:
$=\dfrac{3x}{x-1}\cdot \dfrac{x-1}{x-1}-\dfrac{x-4}{(x-1)^2}$
$=\dfrac{3x(x-1)}{(x-1)^2}-\dfrac{x-4}{(x-1)^2}$
Subtract the numerators and retain the denominator to obtain:
$=\dfrac{3x(x-1)-(x-4)}{(x-1)^2}$
Use the distributive property.
$=\dfrac{3x^2-3x-x+4}{(x-1)^2}$
$=\dfrac{3x^2-4x+4}{(x-1)^2}$
The numerator can no onger be factored.
Hence, the lowest term is:
$\dfrac{3x^2-4x+4}{(x-1)^2}$
