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Asked by Asheley

How many total Roots are in this polynomial equation?

(x-4)^5(x^2-11)(x^2+49)=0
14 years ago

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Answered by Bosnian
(x-4)^5(x^2-11)(x^2+49)=0
(x-4)^5*(x^4-11x^2+49x^2-539)
=(x-4)^5*(x^4+38x^2-539)

(x-4)^5* 5 degre

(x^4+38x^2-539) 4 degre

5+4=9
Polynome have 9 degres and have 9 roots.
14 years ago
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How many total Roots are in this polynomial equation?

(x-4)^5(x^2-11)(x^2+49)=0

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