Question

e="\frac{1}{x^2+3x-4} - \frac{1}{x^2-3x+2}" alt="\frac{1}{x^2+3x-4} - \frac{1}{x^2-3x+2}" align="absmiddle" class="latex-formula">
$\frac{3}{x^2-4} - \frac{x+5}{x+2}$

Answer 1

The first one is $\frac{-6}{(x-2)(x+4)(x-1)}$ and the second one is $\frac{-x^2-3x+13}{(x+2)(x-2)}$;; it's tricky because the denominators aren't the same. You have to factor all of the denominators and multiply only the expressions that the other numbers are missing.

The first expression would be factored into $\frac{1}{(x+4)(x-1)}- \frac{1}{(x-2)(x-1)}$. Multiply $(x-2)$ to the left side and $(x+4)$ to the right, since those are the values the denominator is missing. This results in $\frac{x-2}{(x-2)(x+4)(x-1)} - \frac{x+4}{(x-2)(x+4)(x-1)}$. When simplified, the answer would be $\frac{-6}{(x-2)(x+4)(x-1)}$.

The second one would be factored as $\frac{3}{(x+2)(x-2)} - \frac{x+5}{x+2}$. Like the first one, multiply the missing value to the right side, $(x-2)$, and simplify, resulting in $\frac{-x^2-3x+13}{(x+2)(x-2)}$.