Answer
$-4, \dfrac{1}{3}, 1$
Work Step by Step
We are given the polynomial:
$p(x)=3x^3+8x^2-15x+4$
The possible rational roots are:
$\pm 1,\pm 2, \pm 4, \pm\dfrac{1}{3}, \pm\dfrac{2}{3}, \pm\dfrac{4}{3}$
We try $x=1$:
$p(1)=3(1^3)+8(1^2)-15(1)+4=0$
This means $x_1=1$ is a root. Divide $p(x)$ by $x-1$:
$p(x)=(x-1)(3x^2+11x-4)$
Determine the other two solutions using the quadratic formula:
$x=\dfrac{-11\pm\sqrt{11^2-4(3)(-4)}}{2(3)}=\dfrac{-11\pm 13}{6}$
$x_2=\dfrac{-11-13}{6}=-4$
$x_3=\dfrac{-11+13}{6}=\dfrac{1}{3}$
The roots are: $-4, \dfrac{1}{3}, 1$
