How do you find all zeros of the function $f (x) = {(x + 5)}^{3} (x - 4) {(x + 1)}^{2}$ $f(x)=(x+5)^3(x-4)(x+1)^2$ ?

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Answer 1 · 2018-07-06T14:34:55

$x = - 5, x = - 5, x = - 5, x = 4, x = - 1$ $color(blue)(x=-5,x=-5,x=-5 , x=4, x=-1$

We first equate ${(x + 5)}^{3} (x - 4) (x + 1)$ $(x+5)^3(x-4)(x+1)$ to zero:

$(x + 5) (x + 5) (x + 5) (x - 4) (x + 1) = 0$ $(x+5)(x+5)(x+5)(x-4)(x+1)=0$

We just find values of x that make each bracket in turn zero.

Solutions here are:

$x = - 5, x = - 5, x = - 5, x = 4, x = - 1$ $x=-5,x=-5,x=-5 , x=4, x=-1$

The graph confirms this:

enter image source here

How do you find all zeros of the function f(x)=(x+5)3(x−4)(x+1)2f(x)=(x+5)^3(x-4)(x+1)^2?