Let $p(x) = x^{6} + ax^{5} + bx^{4} + x^{3} + bx^{2} + ax + 1$. Given that 1 is a root of $p(x) = 0$

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Let $p(x) = x^{6} + ax^{5} + bx^{4} + x^{3} + bx^{2} + ax + 1$. Given that 1 is a root of $p(x) = 0$ and -1 is not. What is the maximum number of distinct real roots that p could have

Accepted Answer

If x is a root, then $\frac{1}{x}$ is also a root$	herefore\ $ maximum number of distinct real roots $= 5$.

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