Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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. :/

share|cite|improve this question
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
up vote 0 down vote accepted

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

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

share|cite|improve this answer
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.

share|cite|improve this answer


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

share|cite|improve this answer

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

share|cite|improve this answer

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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.