# Evaluating the modulus of roots of a 6 degree polynomial

Let $x_1,x_2,x_3,x_4,x_5,x_6$ be the roots of the polynomial equation

$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0.$$

Then,

(A)$|x_i|=2$ for exactly one value of $i$

(B)$|x_i|=2$ for exactly two values of $i$

(C)$|x_i|=2$ for all values of $i$.

(D)$|x_i|=2$ for no value of $i$.

My attempt: I noticed that the polynomial forms a geometric

progression with common ratio $2/x$.

Then I summed the terms and found this relation $x^7=128$ which gives us

$x=2$ but $x=2$ doesn't satisfy the original polynomial.

I looked at the graph too and found that the polynomial has no real roots .

So now , how to find the modulus of the roots ?

• You found $x^ 7 = 128$. Note that $x \neq 2$, of course. It turns out that, in fact, the above polynomial is a factor of $x^7 - 128$. Consequently, every root must have modulus equal to $2$, and we can write down all the roots explicitly using the seventh roots of unity. – астон вілла олоф мэллбэрг Aug 21 '18 at 12:21

## 2 Answers

$$f(x)=\sum_{r=0}^6(x^r2^{6-r})=2^6\dfrac{\left(\dfrac x2\right)^7-1}{\dfrac x2-1}=\dfrac{x^7-128}{x-2}$$

Clearly, $f(2)\ne0$

$\implies x^7=128$ with $x\ne2$

$\iff\left(\dfrac x2\right)^7=1$

Now take modulus in both sides using $|z^m|=|z|^m$

As $x=2$ is not a root, the equation is

$$\frac{x^7-128}{x-2}=0.$$

From this we draw

$$|x^7|=|x|^7=128$$ and the only real solution $$|x|=2.$$