# How do you find the zeros of the polynomial function with equation  f(x) = -3(x+1/2)(x-4)^3?

Aug 11, 2016

The zeors of $f \left(x\right)$ in this case are $x = - \frac{1}{2} \mathmr{and} + 4$

#### Explanation:

$f \left(x\right) = - 3 \left(x + \frac{1}{2}\right) {\left(x - 4\right)}^{3}$

The zeros of $f \left(x\right)$ occur where $x$ satisfies $f \left(x\right) = 0$

Since $f \left(x\right)$ is already factorised in this example, $f \left(x\right) = 0$ when

either:

$\left(x + \frac{1}{2}\right) = 0 \to x = - \frac{1}{2}$

or:

$\left(x - 4\right) = 0 \to x = + 4$

This may be easier to visualise from the graph of $f \left(x\right)$ below:

For interest: $f \left(x\right)$ reaches a maximum value at $x = \frac{5}{8}$