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Answer 1

X
=
i
√
5

x
=
i
√
5
,
−
i
√
5
,
−
5
,
5
,
8

Answer 2

Answers:

• The two complex or imaginary roots $x = i\sqrt{5}$ and $x = -i\sqrt{5}$ have multiplicity 2.
• The two real roots x = 5 and x = -5 have multiplicity 3
• The root x = 8 has multiplicity 4

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Explanation:

We'll use the zero product property to solve.

$3(x^2+5)^2(x^2-25)^3(x-8)^4 = 0\\\\(x^2+5)^2=0 \text{ or } (x^2-25)^3=0 \text{ or } (x-8)^4 = 0\\\\x^2+5=0 \text{ or } x^2-25=0 \text{ or } x-8 = 0\\\\x^2=-5 \text{ or } x^2=25 \text{ or } x = 8\\\\x=\pm\sqrt{-5} \text{ or } x=\pm\sqrt{25} \text{ or } x = 8\\\\x=\pm i\sqrt{5} \text{ or } x=\pm 5 \text{ or } x = 8$

where $i = \sqrt{-1}$

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The notation $x=\pm i\sqrt{5}$ breaks up into $x=i\sqrt{5} \text{ or } x=-i\sqrt{5}$. The multiplicity of these two roots is 2 as it's the exponent of the factor $(x^2+5)^2$. Focus on the outermost exponent.

The notation $x = \pm 5$ becomes $x = 5 \text{ or } x = -5$. The multiplicity of these two roots is 3 since it's the outermost exponent of the factor $(x^2-25)^3$

And finally, the multiplicity of the root x = 8 is 4 because it is the outermost exponent of the factor $(x-8)^4$