Answer :
Answers:
- The two complex or imaginary roots [tex]x = i\sqrt{5}[/tex] and [tex]x = -i\sqrt{5}[/tex] 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.
[tex]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\\\\[/tex]
where [tex]i = \sqrt{-1}[/tex]
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The notation [tex]x=\pm i\sqrt{5}[/tex] breaks up into [tex]x=i\sqrt{5} \text{ or } x=-i\sqrt{5}[/tex]. The multiplicity of these two roots is 2 as it's the exponent of the factor [tex](x^2+5)^2[/tex]. Focus on the outermost exponent.
The notation [tex]x = \pm 5[/tex] becomes [tex]x = 5 \text{ or } x = -5[/tex]. The multiplicity of these two roots is 3 since it's the outermost exponent of the factor [tex](x^2-25)^3[/tex]
And finally, the multiplicity of the root x = 8 is 4 because it is the outermost exponent of the factor [tex](x-8)^4[/tex]