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

Apr 9, 2016

$x = 3$, multiplicity of two and

$x = 1$ , multiplicity of one

Explanation:

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

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

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

$0 = x + 3 \text{ " " " " " " " } x = 1$

$x = - 3$

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