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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??

Guest Apr 3, 2018
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2+0 Answers

 #1
avatar+333 
+3

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 smiley

Mathhemathh  Apr 3, 2018
edited by Mathhemathh  Apr 3, 2018
 #2
avatar+6943 
+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.  smiley

 

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

hectictar  Apr 3, 2018

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