spacewater said:
I cross multiplied x² - x - 2 with 10 , x² + 2x - 8 with 2 to get the numerator and factored out (x² + 2x - 8)(x² - x - 2) for the denominator.
That does not make sense - you do not cross-multiply while you are adding.
The numerator came out to be 12x² - 6x - 36 and the denominator (x-2)(x+1)(x+4)(x-2).
After factoring out the numerator it came out to be (2x+4)(6x-9) .. and nothing cancels out... so there is something wrong along the steps I've taken to solve this problem.. can anybody out there correct my issue? Thanks
(2 / (x² - x - 2)) + (10 / (x² + 2x - 8))
\(\displaystyle \frac{2}{x^2-x-2} + \frac{10}{x^2+2x-8}\)
When you have to simplify with polynomials - you need to look if "things" factorizes. That will make rest easier.
\(\displaystyle =\frac{2}{(x-2)(x+1)} + \frac{10}{(x+4)(x-2)}\)
\(\displaystyle =\frac{2}{(x-2)(x+1)} + \frac{10}{(x+4)(x-2)}\)
Now get LCD of the denominators. In this case it is:
\(\displaystyle (x-2)\cdot (x+1)\cdot (x+4)\)
then
\(\displaystyle \frac{2}{x^2-x-2} + \frac{10}{x^2+2x-8}\)
\(\displaystyle =\frac{2\cdot (x+4)}{(x-2)\cdot (x+1)\cdot (x+4)} + \frac{10\cdot (x+1)}{(x-2)\cdot (x+1)\cdot (x+4)}\)
\(\displaystyle =\frac{2\cdot (x+4) \, + \, 10\cdot (x+1)}{(x-2)\cdot (x+1)\cdot (x+4)}\)
\(\displaystyle =\frac{12x \, + \, 18}{(x-2)\cdot (x+1)\cdot (x+4)}\)
\(\displaystyle =\frac{6\cdot(2x+3)}{(x-2)\cdot (x+1)\cdot (x+4)}\)
done...
