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I have to simplify:

$$\frac{(x - 2) }{(3x + 6)} * \frac{(x^2 - x - 6) }{(x + 3)}$$

In working, I just played around with the numbers for a while. I came up with an answer I thought was correct.

It seemed like the left hand side of the equation simplified to $ \frac13x$ - $\frac13$ and that the right hand side simplified to x**2 - x - 2. It seemed like those could be further factorized into 1/3(x - 1) and (x - 2) * (x + 1).

Then I checked my answer by plugging in some random numbers and seeing if the expression returned the same value, which it did not.

Since the quadratic simplifies to $x^2-x-6$, presumably the expression simplifies to

$$\frac{(x-2) (x-3) }{(x + 3)}$$

But this also does not return the same value as the original expression when I plug in random numbers. What am I doing wrong?

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What is your equation? Because there is no "=" in above. – Babak S. Jan 28 '13 at 8:09
You are missing a factor of $\frac{1}{3}$ in your last equation. That is why plugging numbers in doesn't match. – copper.hat Jan 28 '13 at 8:52
up vote 3 down vote accepted


$$\frac{(x - 2) }{\underbrace{(3x + 6)}_{3(x+2)}} \cdot \frac{\overbrace{(x^2 - x - 6)}^{(x + 2)(x - 3)} }{(x + 3)} = \frac{(x - 2)}{3\cancel{(x + 2)}} \cdot \frac{\cancel{(x +2)}(x - 3)}{(x+3)} \overset{x\neq -2}{=} \dfrac{(x - 2)(x-3)}{3(x + 3)}$$

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Hint: The quadratic $x^2-x-6$ on top factors as $(x-3)(x+2)$. Note that we almost have $x+2$ at the bottom, since $3x+6=3(x+2)$. Now you can "cancel" the $x+2$'s, at least when $x\ne -2$.

Remark: The algebraic steps mentioned in the OP are not correct.

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