How do you find all the zeros of $f (x) = x (x + 4) {(x - 2)}^{3}$ $f(x)=x(x+4)(x-2)^3$ with its multiplicities?

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Answer 1 · 2016-08-11T19:39:00

$x = 0, &, x = - 4,$ $x=0, &, x=-4,$ each with single multiplicity; $x = 2,$ $x=2,$ with multiplicity $= 3$ $=3$

To find the zeroes of a poly. $f (x)$ $f(x)$ , we need to solve the eqn. $f (x) = 0$ $f(x)=0$

In our case, $f (x) = 0 \Rightarrow x (x + 4) {(x - 2)}^{3} = 0$ $f(x)=0 rArr x(x+4)(x-2)^3=0$

$\Rightarrow x = 0, or, x + 4 = 0, or, x - 2 = 0$ $rArr x=0, or, x+4=0, or, x-2=0$

$\Rightarrow x = 0, &, x = - 4,$ $rArr x=0, &, x=-4,$ each with single multiplicity; $x = 2,$ $x=2,$ with multiplicity $= 3$ $=3$

How do you find all the zeros of f(x)=x(x+4)(x−2)3f(x)=x(x+4)(x-2)^3 with its multiplicities?