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I'm given this algebraic fraction, and I'm asked to simplify it as much as possible: $$\dfrac{3x^2+5x-2}{6x^3-17x^2-4x+3}$$

I applied Ruffini's Rule to find the roots of both the Numerator and Denominator. This helped me simplify the fraction to: $$\dfrac{(x+2)(3x-1)}{(x-3)(6x^2+x-1)}$$

I tried to further simplify the factor $(6x^2+x-1)$, to no avail. I tried these two techniques:

  1. I tried to find any possible roots, but neither $(x-1)$ nor $(x+1)$ are roots.
  2. I tried to see if that is a perfect square in the form: $(x+a)^2$, but it isn't.

I'm not aware of any other way/technique that helps me simplify that expression. I'd just need a bit of your help. Thanks!

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    $\begingroup$ Hint: $6x^2+x-1=(3x-1)(2x+1)$. $\endgroup$ – Teddy38 Oct 12 '17 at 13:30
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    $\begingroup$ You didn't need the roots in the first place. Just find the GCD of both polynomials. $\endgroup$ – Ivan Neretin Oct 12 '17 at 13:31
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    $\begingroup$ Since $6x^2 + x - 1$ is quadratic, you could have also used the quadratic formula to find the roots $\endgroup$ – Paul Sinclair Oct 12 '17 at 16:13
  • $\begingroup$ Good point, thanks @PaulSinclair $\endgroup$ – Jose Lopez Garcia Oct 12 '17 at 16:26
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It is possible to factor $6x^2+x-1$. First, we look for two numbers whose sum is $1$ and whose product is $-6$. (In general, for $ax^2+bx+c$, we would seek two numbers with sum $b$ and product $ac$.) We find these numbers to be $3$ and $-2$. We now "split the middle term", replacing "$+x$" with "$+3x-2x$":

$$6x^2+3x-2x-1$$

The resulting expression we can "factor by grouping":

$$\begin{align} 6x^2+3x-2x-1 &=3x(2x+1)-1(2x+1)\\ &=(3x-1)(2x+1) \end{align}$$

Applying this factorization, you see that your denominator now has a common factor with the numerator, which can be canceled.

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  • $\begingroup$ As far as simplifying this fraction is concerned, the only factors that would help would be either x+ 2 or 3x- 1. "x+ 2" clearly Is not a factor but "3x- 1" looks hopeful. $(3x- 1)(ax+ b)= 3ax^2+ (3b- a)x- b$ and we want that equal to $6x^2+ x- 1$. We must have 3a= 6, 3b- a= 1, and -b= -1. From 3a= 6, a= 2 and from the last, b= 1. is 3b- a= 1? 3(1)- 2= 1. Yes! $6x^2+ x- 1= (3x- 1)(2x+ 1). $\endgroup$ – user247327 Oct 12 '17 at 13:40
  • $\begingroup$ That's great; why don't you post it as an answer? :) $\endgroup$ – G Tony Jacobs Oct 12 '17 at 13:41
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    $\begingroup$ I answered the way I did, because it appeared that the OP wanted to know how to factor the quadratic in question. :) $\endgroup$ – G Tony Jacobs Oct 12 '17 at 13:42
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    $\begingroup$ Ruffini's rule, also called "synthetic division", is applicable to a polynomial of any degree. The method of factoring that I've shown here is only applicable to quadratic polynomials. I usually use Ruffini's rule until I reduce the degree of my factors to $2$, and then I use techniques such as this one. $\endgroup$ – G Tony Jacobs Oct 12 '17 at 13:50
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    $\begingroup$ @JoseLopezGarcia, you should also check out the Rational Roots Theorem. This would tell us that possible zeros of $6x^2+x-1$ include, for example, $\frac13$ and $-\frac12$, which correspond to the factors $(3x-1)$ and $(2x+1)$ respectively. $\endgroup$ – G Tony Jacobs Oct 12 '17 at 13:58
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The factors that will allow you to simplify this fraction comes from the expression $6x^2+x-1$, and in order to factor this, you can do the following:

consider pairs of $x$-coefficients that will multiply to give $-6$ and add to give $1$. The pair that should catch your eye is $3$ and $-2$. Using this, the expression now becomes: $6x^2+3x-2x-1$ which allows you to factor out two brackets independently like so: $$3x(2x+1)-1(2x+1)$$ $$(3x-1)(2x+1)$$ hence your fraction becomes: $$\frac{(x+2)(3x-1)}{(x-3)(3x-1)(2x+1)}$$ can you simplify this?

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