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Please help! I got 14 but it says it's incorrect! Find the maximum number of real zeros of the polynomial. f(x)=2x^(6)-3x^(3)+1-2x^(5)

User Athari
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1 Answer

5 votes

Answer:

There are two or zero positive solutions and zero negative roots (zeros).

Explanation:

Use Descartes' Rule of Signs to determine the number of real zeros of
f(x)=2x^6-3x^3+1-2x^5


f(x)=2x^6-2x^5-3x^3+1\\

Please help! I got 14 but it says it's incorrect! Find the maximum number of real-example-1
User Yujuezhao
by
8.1k points

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