Absolute max or min is where first derivative is zero.
$\displaystyle f(x) =& 2x^3 + 9x^2 -168x + 3$
$\displaystyle \begin{align*} \frac{df}{dx} =& 6x^2 + 18x -168 = 0 \\ &x^2 + 3 x - 28 = 0.......\text{[after dividing both sides by } 6 \text{ ]}\\ &(x + 7)(x - 4) = 0......\text{[after factorization]}\end{align*}$
So absolute max or min can be at $\displaystyle x = -7$ or $\displaystyle x = 4$. But to find out which one which you plug-in the values into the function and find out which one's larger than the other.
Now plug-in $\displaystyle x = -7$ and separately $\displaystyle x = 4$ into $\displaystyle f(x)$ you'll find that $\displaystyle f(-7) > f(4)$
The function has absolute max at $\displaystyle x = -7$
And has absolute min at $\displaystyle x = 4$
You can check it in the following graph to be sure: