# How to simplify $\frac{3x^2+5x-2}{6x^3-17x^2-4x+3}$?

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!

• Hint: $6x^2+x-1=(3x-1)(2x+1)$. – Teddy38 Oct 12 '17 at 13:30
• You didn't need the roots in the first place. Just find the GCD of both polynomials. – Ivan Neretin Oct 12 '17 at 13:31
• Since $6x^2 + x - 1$ is quadratic, you could have also used the quadratic formula to find the roots – Paul Sinclair Oct 12 '17 at 16:13
• Good point, thanks @PaulSinclair – Jose Lopez Garcia Oct 12 '17 at 16:26

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 "$+3x-2x$":
$$6x^2+3x-2x-1$$
\begin{align} 6x^2+3x-2x-1 &=3x(2x+1)-1(2x+1)\\ &=(3x-1)(2x+1) \end{align}
• 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). – user247327 Oct 12 '17 at 13:40 • That's great; why don't you post it as an answer? :) – G Tony Jacobs Oct 12 '17 at 13:41 • I answered the way I did, because it appeared that the OP wanted to know how to factor the quadratic in question. :) – G Tony Jacobs Oct 12 '17 at 13:42 • 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. – G Tony Jacobs Oct 12 '17 at 13:50 • @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. – G Tony Jacobs Oct 12 '17 at 13:58 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?