\({x - 5 \over (x - 2)(x + 10)}\)+\({2x + 1 \over (x - 2)(x + 2)}\)
I added it using LCD: \({(x - 2)(x + 10)(x + 2)}\)
Here's what I did:
\({(x - 5)(x + 2) + (x + 10)(2x + 1) \over (x - 2)(x + 10)(x + 2)}\)
\({(x^2 - 3x + 10) + (x^2 + 21x + 10) \over (x - 2)(x + 10)(x + 2)}\)
\({2x^2 + 18x + 20 \over (x - 2)(x + 10)(x + 2)}\) (I tried to factor the numerator, and I don't think it's factorable...?)
The textbook got \({3x(x + 6) \over (x - 2)(x + 10)(x + 2)}\), but how??
+349 This is what I did:
1. Factor out \({1\over x-2}\): \({1\over x-2}({x-5\over x+10}+{2x+1\over x+2})\)
2. Apply the cross-horizontal method: \({1\over x-2}({(x+2)(x+5)+(x+10)(2x+1)\over(x+10)(x+2)})\)
3. Simplify the numerator: \({1\over x-2}({3x^2+18x\over(x+10)(x+2)})\)
4. Factor out 3x from the numerator: \({1\over x-2}({3x(x+6)\over(x+10)(x+2)})\)
5. Multiply: \({3x(x+6)\over(x-2)(x+10)(x+2)}\)
Your error is in the second line wherein you applied FOIL. \(-5*2=-10\). You forgot to consider the signs 
+9479 You made an error in multiplying (x - 5)(x + 2) and (x + 10)(2x + 1) .
(x - 5)(x + 2) = (x)(x) + (x)(2) + (-5)(x) + (-5)(2) = x2 - 3x - 10
and
(x + 10)(2x + 1) = (x)(2x) + (x)(1) + (10)(2x) + (10)(1) = 2x2 + 21x + 10
Otherwise your method is correct.
\(\qquad\ \frac{(x-5)(x+2)+(x+10)(2x-1)}{(x-2)(x+10)(x+2)} \\ =\\ \qquad\frac{(x^2-3x-10)+(2x^2+21x+10)}{(x-2)(x+10)(x+2)} \\ =\\ \qquad\frac{3x^2+18x}{(x-2)(x+10)(x+2)} \\ =\\ \qquad\frac{3x(x+6)}{(x-2)(x+10)(x+2)}\)
