4/1 - x + 2/(1 - x)^2
I have the answer
6-4x/(x-1)(x-1)
but i dont understand how to get there
anyhelp is greatly appreciated
thank you
$\displaystyle \frac{4}{1 - x} + \frac{2}{(1 - x)^2}$.
You need these terms to have the same denominator in order to add the numerators. To do this, multiply the first term by a cleverly disguised 1, in this case, $\displaystyle \frac{1 - x}{1 - x}$.
So $\displaystyle \frac{4}{1 - x} + \frac{2}{(1 - x)^2} = \frac{4}{1 - x}\times \frac{1 - x}{1 - x} + \frac{2}{(1 - x)^2}$
$\displaystyle = \frac{4(1 - x)}{(1 - x)^2} + \frac{2}{(1 - x)^2}$
$\displaystyle = \frac{4(1 - x) + 2}{(1 - x)^2}$
$\displaystyle = \frac{4 - 4x + 2}{(1 - x)^2}$
$\displaystyle = \frac{6 - 4x}{(1 - x)^2}$.
Also note that $\displaystyle (1 - x)^2 = (1 - x)(1 - x) = -(x - 1)(1 - x) = -[-(x - 1)(x - 1)] = (x - 1)(x - 1)$.