How do you find all the zeros of $p(x)=(x^2+9)(x+3)$ ?

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Answer 1 · 2016-03-11T16:16:22

$x=-3$

$x^2+9=0, x+3=0$
$x^2=-9, x=-3$
$x=+-sqrt-9-> no solution, x=-3$
$x=-3$
If it is in the set of complex numbers then answers will be
$x=+-3i or x=-3$

How do you find all the zeros of p(x)=(x^2+9)(x+3)p(x)=(x^2+9)(x+3)?