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

Question

1790 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-11T05:48:06

The zeors of $f (x)$ $f(x)$ in this case are $x = - \frac{1}{2} or + 4$ $x = -1/2 or +4$

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

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

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

either:

$(x + \frac{1}{2}) = 0 \to x = - \frac{1}{2}$ $(x+1/2) = 0 -> x=-1/2$

or:

$(x - 4) = 0 \to x = + 4$ $(x-4)=0 -> x=+4$

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

enter image source here

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

How do you find the zeros of the polynomial function with equation f(x)=−3(x+12)(x−4)3 f(x) = -3(x+1/2)(x-4)^3?