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Question

# By Descartes rule of sign, the maximum number of negative real roots of equation x5−6x4+7x3+8x2+9x+10=0

A
03
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B
3.0
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C
003
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D
3
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E
3.00
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Solution

## Given equation is x5−6x4+7x3+8x2+9x+10=0 let f(x)=x5−6x4+7x3+8x2+9x+10 ⇒f(−x)=−x5−6x4−7x3+8x2−9x+10 Since the terms of f(−x) are already in descending order. Here number of sign changes in f(−x) is 3. Hence by Descartes rule of signs, maximum number of negative real roots of f(x) is 3.

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