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

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

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Answer 1 · 2016-03-11T16:16:22

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

Explanation:

$x^{2} + 9 = 0, x + 3 = 0$ $x^2+9=0, x+3=0$
$x^{2} = - 9, x = - 3$ $x^2=-9, x=-3$
$x = \pm \sqrt{-} 9 \to n o s o l u t i o n, x = - 3$ $x=+-sqrt-9-> no solution, x=-3$
$x = - 3$ $x=-3$
If it is in the set of complex numbers then answers will be
$x = \pm 3 i or x = - 3$ $x=+-3i or x=-3$

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

1 Answer

Explanation:

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