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

Mar 11, 2016

$x = - 3$
${x}^{2} + 9 = 0 , x + 3 = 0$
${x}^{2} = - 9 , x = - 3$
$x = \pm \sqrt{-} 9 \to n o s o l u t i o n , x = - 3$
$x = - 3$
$x = \pm 3 i \mathmr{and} x = - 3$