Answer
$-8, \ -4, \ -\displaystyle \frac{7}{3}$
Work Step by Step
$f(x)=3x^{3}+16x^{2}+3x-10$
Possible rational roots: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 2,\pm 5,\pm 10}{\pm 1,\pm 3}$
Try $ x+1 $ with synthetic division.
$\left.\begin{array}{l}
-1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
3 & 16 & 3 &-10 \\\hline
& -3 & -13 & +10 \\\hline
3 & 13 &-10 & |\ \ 0 \end{array}$
$f(x)=(x+1)(3x^{2}+13x-10)$
Try $ x+5 $ with synthetic division.
$\left.\begin{array}{l}
-5 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
3 & 13 &-10 \\\hline
& -15 & +10 \\\hline
3 & -2 & |\ \ 0 \end{array}$
$f(x)=(x+1)(x+5)(3x-2)$
The zeros of f are $-5, -1$ and $\displaystyle \frac{2}{3}.$
The graph of $f(x+3)$ is obtained from $f(x)$ by shifting it left by 3 units.
So the zeros of f, shifted left 3 units are
$-8, \ -4, \ -\displaystyle \frac{7}{3}$
