# Simplifying polynomial fraction

Working through an old book I got and am at this problem: Simplify:

$$\frac{3x^2 + 3x -6}{2x^2 + 6x + 4}.$$

The answer is supposed to be $\frac{3(x - 1)}{2(x - 1)}$. I thought I had all this polynomial stuff figured out well enough, but I'm having trouble seeing how they got to that answer. :/

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denominator should be $2(x+1)$, I suppose. – puru Jun 28 '14 at 8:10
No, the answer is supposed to be $\frac{3(x-1)}{2(x+1)}$, with a factor of $(x+1)$ on the bottom, not a factor $(x-1)$. – David H Jun 28 '14 at 8:10
@DavidH exactly my point! I can't edit the question,however. – puru Jun 28 '14 at 8:11
Crap, must have messed up writing the question. Sorry. – windy401 Jun 28 '14 at 8:19

$$3x^2+3x-6 = 3(x^2+x-2)=3(x-1)(x+2)$$

$$2x^2+6x+4 = 2(x^2+3x+2)=2(x+1)(x+2)$$

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Ok, wow, I was really over thinking that one. Thanks a lot. – windy401 Jun 28 '14 at 8:11

Simply factorise

$\dfrac{3x^2 + 3x -6}{2x^2 + 6x + 4} = \dfrac{3(x^2 + x -2)}{2(x^2 + 3x + 2)} = \dfrac{3(x-1)(x+2)}{2(x+1)(x+2)}$

Then cancel common terms.

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Hint

First of all, you could start factoring both numerator and denominator $$\frac{3x^2+3x-6}{2x^2+6x+4}=\frac{3(x^2+x-2)}{2(x^2+3x+2)}$$ In order to factor them, you can then compute the roots of the quadratics : for the numerator, the roots are obviously $1$ and $-2$ and for the denominator $-1$ and $-2$ (you can solve for them or just find by inspection).So, $$\frac{3x^2+3x-6}{2x^2+6x+4}=\frac{3(x^2+x-2)}{2(x^2+3x+2)}=\frac{3(x-1)(x+2)}{2(x+1)(x+2)}$$

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$$\frac{3x^2 + 3x -6}{2x^2 + 6x + 4} = \frac{3(x^2 + x -2)}{2(x^2 + 3x + 2)}= \frac{3\require{cancel}\cancel{(x+2)}(x-1)}{2(x+1)\cancel{(x + 2)}}=\frac{3(x-1)}{2(x+1)}$$

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$$\frac{3x^2 + 3x -6}{2x^2 + 6x + 4}=\frac{3(x^2 + x -2)}{2(x^2 + 3x + 2)}=$$ $$=\frac{3(x^2 -1+ x -1)}{2(x^2 + 2x+x + 2)}=\frac{3((x-1)(x+1)+(x-1))}{2(x(x+2)+(x+ 2)}=$$ $$=\frac{3(x-1)(x+1+1)}{2(x+2)(x+ 1)}=\frac{3(x-1)(x+2)}{2(x+2)(x+1)}=\frac{3(x-1)}{2(x+1)}$$

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An alternative to (the admittedly easy) factoring of the quadratics consists in taking the gcd of numerator and denominator, and dividing both by it. This may turn useful in more complicated circumstances.

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