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# Troubles with adding this rational expression!

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$${x - 5 \over (x - 2)(x + 10)}$$+$${2x + 1 \over (x - 2)(x + 2)}$$

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??

Apr 3, 2018

#1
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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 Apr 3, 2018
edited by Mathhemathh  Apr 3, 2018
#2
+3

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. $$\qquad\ \frac{(x-5)(x+2)+(x+10)(2x-1)}{(x-2)(x+10)(x+2)} \\ =\\ \qquad\frac{(x^2-3x-10)+(2x^2+21x+10)}{(x-2)(x+10)(x+2)} \\ =\\ \qquad\frac{3x^2+18x}{(x-2)(x+10)(x+2)} \\ =\\ \qquad\frac{3x(x+6)}{(x-2)(x+10)(x+2)}$$

.
Apr 3, 2018