Answer
$0,\displaystyle \ \frac{11}{4},\ 6$
Work Step by Step
$f(x)=4x^{3}-11x^{2}-26x+24$
Possible rational roots:
$\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24}{\pm 1,\pm 2,\pm 4}$
Try $ x+2 $ with synthetic division.
$\left.\begin{array}{l}
-2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
4 & -11 & -26 &24 \\\hline
& -8 & 38 & -24 \\\hline
4 & -19 & 12 & |\ \ 0 \end{array}$
$f(x)=(x+2)(4x^{2}-19x+12)$
Try $ x-4 $ with synthetic division.
$\left.\begin{array}{l}
4 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
4 & -19 & 12 \\\hline
& 16 & -12 \\\hline
4 & -3 & |\ \ 0 \end{array}$
$f(x)=(x+2)(x-4)(4x-3)$
The zeros of f are $-2, \displaystyle \frac{3}{4}$ and $4$.
The graph of $f(x-2)$ is obtained from $f(x)$ by shifting it right by 2 units.
So the zeros of f, shifted right by 2 units are
$0,\displaystyle \ \frac{11}{4},\ 6$
