# Problem: Sum of absolute values of polynomial roots

Can you please offer me some tips regarding just how I might approach this trouble? Many thanks!

Offered the polynomial $f = 2X^3 - aX^2 - aX + 2, a \mathbb \in R$ and also origins $x_1, x_2$ and also $x_3,$ locate $a$ such that $|x_1| + |x_2| + |x_3| = 3.$

**Edit: ** We recognize $-1$ is just one of the origins of that polynomial, no matter the value of $a$. So, fundamentally, what we need to show is that $|x_2| + |x_3| = 2.$

The means to address this is in fact fairly straightforward, evidently:

$$|x_2| + |x_3| =2 \implies -2 \le x_1 + x_1 \le 2$$

**HINT **

$|x_1| + |x_2| + |x_3| \geq 3 \sqrt[3]{|x_1||x_2||x_3|} = 3$

Hence, we have $|x_1| + |x_2| + |x_3| \geq 3$. Equal rights holds indicates $|x_1| = |x_2| = |x_3| = 1$

We have $x_1 + x_2 + x_3$, $x_1x_2 + x_2x_3 + x_3x_2$ and also $x_1 x_2 x_3$ to be actual, and also more $x=-1$ pleases the formula.

Therefore $f(x) = 2 \left( x+1 \right) \left( x^2-\left(1 + \frac{a}{2} \right)x + 1 \right)$. We have $x_3 = -1$. And also $|x_1| = 1$ and also $|x_2| = 1$. Therefore, $x_1 = e^{i \theta}$ and also $x_2 = e^{i \phi}$.

$(x-x_1)(x-x_2) = \left( x^2-\left(1 + \frac{a}{2} \right)x + 1 \right)$

$x_1 x_2 = 1 \implies \phi = -\theta$

$x_1 + x_2 = 2 \cos(\theta) = 1 + \frac{a}{2}$

Hence, $\frac{a}{2} = 2 \cos(\theta) - 1 \implies a = 4 \cos(\theta) - 2$.

Therefore, $a = 4 \cos(\theta) - 2$ and also the origins are $-1,e^{i \theta},e^{-i \theta}$

We first have the adhering to factorization,

$ 2x^3 - ax^2 - ax + 2 = (x+1)(2x^2 - (2+a)x + 2).$

Suppose $x_1 = -1$, after that $|x_2| + |x_3| = 2$ and also they are remedies to the square formula $2x^2 - (2+a)x + 2 = 0$. Therefore,

$ x_2 + x_3 = 1 + a/2 $ and also

$ x_2 x_3 = 1$.

By observing the 2nd formula, $x_2$ and also $x_3$ are either both favorable or both adverse, so we have

$1 + a/2 = x_2 + x_3 = 2$ or $-2$.

Consequently, $a = 2$ or $-6$.

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