# How do you find all zeros of the function f(x)=3x^3+x^2-48x-16?

May 1, 2016

Factor $f \left(x\right)$ to find that $f \left(x\right) = 0$ if and only if $x \in \left\{- 4 , - \frac{1}{3} , 4\right\}$

#### Explanation:

$f \left(x\right) = 3 {x}^{3} + {x}^{2} - 48 x - 16$

$= 3 x \left({x}^{2} - 16\right) + {x}^{2} - 16$

$= \left(3 x + 1\right) \left({x}^{2} - 16\right)$

$= \left(3 x + 1\right) \left(x + 4\right) \left(x - 4\right)$

Thus $f \left(x\right) = 0$ if and only if $3 x + 1 = 0$ or $x + 4 = 0$ or $x - 4 = 0$.

Solving, this gives us

$3 x + 1 = 0 \implies x = - \frac{1}{3}$

$x + 4 = 0 \implies x = - 4$

$x - 4 = 0 \implies x = 4$

Thus the solution set to $f \left(x\right) = 0$ is $x \in \left\{- 4 , - \frac{1}{3} , 4\right\}$