intermediate algebra

If we factor the quadratic, we find nothing cancels.

\(\displaystyle \frac{x^{2}-2x-15}{x+5}=\frac{(x-5)(x+3)}{x+5}\)

So, we have to resort to long division.

I think you are going abut this a little wrong. Note, we divide the same as if they were numbers instead of x's.

..........\(\displaystyle x\;\ \;\ \;\ \;\ -7\)
\(\displaystyle x+5)\overline{x^{2}-2x-15}\)
.........\(\displaystyle \underline{x^{2}+5x+0}\)
..............\(\displaystyle -7x-15\)
..............\(\displaystyle \underline{-7x-35}\)
------------------.....-\(\displaystyle 20\)


\(\displaystyle x-7+\frac{20}{x+5}\)
 
hopen2impov said:
divide
(x^2 -2x -15) / (x+5)
answer x-3
IF answer is x-3 as you say, then you have a typo in original; should be: (x^2 + 2x -15) / (x+5)
This simplifies to : (x+5)(x-3) / (x+5) = x-3 ; kapish?
 
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