Question

By Descartes rule of sign, the maximum number of negative real roots of equation $x^5-6x^4+7x^3+8x^2+9x+10=0$

• A. 03
• B. 3.0
• C. 003
• D. 3
• E. 3.00

Answer 1

Answer: (A), (B), (C), (D), (E)

Given equation is $x^5-6x^4+7x^3+8x^2+9x+10=0$
$\text{let}f\left(x\right)=x^5-6x^4+7x^3+8x^2+9x+10$
$\Rightarrow f(-x)=-x^5-6x^4-7x^3+8x^2-9x+10$
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 $f\left(x\right)$ is 3.