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

Question

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!

Answer 1

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.

Answer 2

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?