Find the roots of $6x^4+19x^3-51x^2+20x$. Enter your answer as a list of numbers separated by commas.
Since the polynomial has no constant term, we can immediately factor out an $x$ from every term
$$x(6x^3+19x^2-51x+20),$$and our first root $x=0$. Let $g(x) = 6x^3+19x^2-51x+20$. Then the remaining roots of our original polynomial are the roots of $g(x)$. By trying out simple values, we can see that $g(0) = 20 > 0$ and $g(1) = 6+19-51+20 = -6<0$. Thus, there must be a root of $g(x)$ between $0$ and $1$. From the Rational Root Theorem, we know that if $g(p/q) = 0$ then $p$ must divide $20$ and $q$ must divide $6$.

Checking rational numbers of the form $p/q$, where $p$ divides $20$ and $q$ divides $6$, and $p/q$ is between $0$ and $1$, we find that
$$\begin{aligned} g\left(\frac{1}{2}\right) &= 6\cdot\frac{1}{8}+19\cdot\frac{1}{4}-51\cdot\frac{1}{2}+20 = 0.
\end{aligned}$$This means that $2x - 1$ is a factor of $g(x).$ Dividing by $2x-1$ gives us $g(x) = (2x-1)(3x^2+11x-20)$.

The quadratic $3x^2+11x-20$ factors as $(3x-4)(x+5),$ so our last two roots are $4/3$ and $-5$.

Thus, the roots of $6x^4+19x^3-51x^2+20x$ are $\boxed{0, \frac{1}{2}, \frac{4}{3}, -5}$.
0,\frac{1}{2},\frac{4}{3},-5
Level 2
Intermediate Algebra
Problem: Find the roots of $6x^4+19x^3-51x^2+20x$. Enter your answer as a list of numbers separated by commas.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the roots of the polynomial $6x^4 + 19x^3 - 51x^2 + 20x$, we first factor out the common term $x$, leading to $x(6x^3 + 19x^2 - 51x + 20) = 0$. This gives us one root, $x = 0$.

Next, we need to find the roots of the cubic polynomial $6x^3 + 19x^2 - 51x + 20$. We can apply the Rational Root Theorem to identify possible rational roots, which are factors of the constant term (20) divided by factors of the leading coefficient (6). The possible rational roots are $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20, \pm \frac{1}{2}, \pm \frac{5}{3}, \pm \frac{10}{3}, \pm \frac{20}{3}, \pm \frac{1}{6}, \pm \frac{5}{6}$.

Testing these values, we find that $x = \frac{1}{2}$ is a root. We can use synthetic division to divide the cubic polynomial by $x - \frac{1}{2}$, resulting in a quadratic polynomial $6x^2 + 22x - 40$. 

Next, we apply the quadratic formula to find the roots of $6x^2 + 22x - 40$. The discriminant is calculated as $b^2 - 4ac = 22^2 - 4 \cdot 6 \cdot (-40) = 484 + 960 = 1444$. The square root of 1444 is 38. 

Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$, we find the roots:
1. $x = \frac{-22 + 38}{12} = \frac{16}{12} = \frac{4}{3}$.
2. $x = \frac{-22 - 38}{12} = \frac{-60}{12} = -5$.

Thus, the roots of the original polynomial are $0$, $\frac{1}{2}$, $\frac{4}{3}$, and $-5$.<|END_OF_LATENT|> The final answer is $0,\frac{1}{2},\frac{4}{3},-5$.