$$\frac{x^2 + 4x + 3}{x^2 - 2x - 3}$$
I'm coming up with $\frac{x + 3}{ x - 3}$ however it seems wrong.
$$x^2 + 4x + 3 = (x + 3) ( x + 1)$$
$$x^2 - 2x - 3 = (x - 3 )(x + 1) $$
cancel out $x + 1$ and left with $\frac{x + 3}{x - 3}$
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2And why does that seem wrong? – Harry Alli Sep 25 '17 at 07:13
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$\frac{x+3}{x-3}$ is the correct answer. – Sep 25 '17 at 07:14
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2Your answer is correct, but you should also add the condition $x\neq -1$ for your final expression. – MrYouMath Sep 25 '17 at 07:21
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It's correct,
$$\require{cancel} \frac{x^2 + 4x + 3} {x^2 - 2x - 3} = \frac{(x+3)\cancel{(x+1)}} {(x-3) \cancel{(x+1)}}$$
Provided $x \neq 3$, $x \neq -1$.
George C
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