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

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Answer 1 · 2016-04-09T05:03:04

$x = 3$ $x= 3$ , multiplicity of two and

$x = 1$ $x= 1$ , multiplicity of one

Set the equation equal to zero and solve for $x$ $x$ as follow

$f (x) = {(x + 3)}^{2} (x - 1)$ $f(x) = (x+3)^2(x-1)$

$0 = (x + 3) (x + 3); x - 1 = 0$ $0= (x+3)(x+3) ; " " " " " " x-1= 0$

$0 = x + 3 x = 1$ $0=x+3 " " " " " " " " " x= 1$

$x = - 3$ $x= -3$

The zero are $x = 3$ $x= 3$ multiplicity of two and $x = 1$ $x= 1$ multiplicity of one.

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