# How do you use the rational root theorem to find the roots of 3x^4-10x^3-24x^2-6x+5?

Jul 29, 2015

${\left(x + 1\right)}^{2} \left(x - 5\right) \left(3 x - 1\right)$
According to the ration root theorem the roots of the polynomial would be out of the factors of $\frac{5}{3}$ which could be $\pm 1. \pm 5 , \pm \frac{1}{3}$. By trial and error,it can be ascertained that x=-1 and x=5 are the roots of the polynomial. That implies that (x+1)(x-5) would be factors of the polynomial. Now divide the polynomial by the product of (x+1)(x-5) that is ${x}^{2} - 4 x - 5$.
The division would give a quotient $3 {x}^{2} + 2 x - 1$ leaving 0 remainder. Using the remainder theorem again, it can be ascertained that x=-1 is a root of $3 {x}^{2} + 2 x - 1$. Therefore x+1 would be one of its factors. The other factor can be had by dividing $3 {x}^{2} + 2 x - 1$ with x+1. The other factor would be 3x-1.
All the factors of the polynomial would thus be ${\left(x + 1\right)}^{2} \left(x - 5\right) \left(3 x - 1\right)$