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This question already has an answer here:

Consider the polynomial $( f(x) = x^7 - 4x^3 + x + 1 )$. All the zeroes of the above polynomial are plotted in the complex plane i.e., the Argand plane. How many are within a unit distance from the origin?

Details and Assumptions:

The repeated roots are counted with multiplicity i.e.,$ ( (2x - 1)^2 = 0 )$ has two solutions within a unit distance from origin and not one.

There is a very neat and simple way of doing without W|A. Please refrain from using any such mathematical computational software.

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marked as duplicate by LutzL, Tim Raczkowski, ncmathsadist, Empty, Strants Oct 18 '15 at 2:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Let $g(x) = -4x^3$, then if $|x| = 1$ we have $|f(x)-g(x)|= |x^7+x+1| \le 3 < 4 = |-4 x^3| = |g(x)|$, hence Rouché's theorem tells us that $f$ and $g$ have the same number of zeros inside the unit circle.

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As a numerically closer alternative to the other answer:

The dominant binomials of the polynomial $f(x)$ are $x^7-4x^3$ for the larger and $-4x^3+1$ for the smaller roots. Indeed the product of the reduced binomials is $$ g(x)=(x^4-4)(x^3-\tfrac14)=x^7-\tfrac14x^4-4x^3+1 = f(x)-\tfrac14x^4-x $$


Numerically, the root sets compare as \begin{array}{rl|l} &\text{numerical roots of }g(x) & \text{numerical roots of }f(x)\\ \hline +\sqrt2&=+1.414214 &+1.313850\\ -\sqrt2&=-1.414214 &-1.401144\\ \pm i\sqrt2&=\pm i·1.414214 &+0.024884\pm i·1.456126\\ \tfrac12\sqrt[3]2&=+0.629961 &+0.792062\\ \tfrac14\sqrt[3]2(-1\pm\sqrt3)&=-0.314980\pm i·0.545562 &-0.377268\pm i·0.425477\\ \end{array}


By Rouché, on $|x|=1$, $|x^4-4|\ge 4-1=3$, $|x^3-\frac14|\ge \frac34$ and thus $$ |f(x)-g(x)|\le \tfrac54< \tfrac94=3·\tfrac34\le |g(x)| $$ which implies again the same number of roots inside and outside the unit circle for $f$ and $g$.

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