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Mathematics 15 Online
OpenStudy (anonymous):

zeros of x^5-4x^3+4x. I know there are five.

OpenStudy (anonymous):

Yes, first should be easy when you realize that there is an x in all the terms.

OpenStudy (anonymous):

Then the equation will resolve to finding all four zeros in x^4 - 4x^2 + 4

OpenStudy (anonymous):

Hint: substitute n=x^2 in my resulting equation and you will have a quadratic that is easy to solve.

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