# How do you find all the zeros of f(x)=x^4-x^3-3x^2-x+1?

Aug 11, 2016

$f \left(x\right)$ has zeros:

${x}_{1 , 2} = \frac{1 + \sqrt{21} \pm \sqrt{6 + 2 \sqrt{21}}}{4}$

${x}_{3 , 4} = \frac{1 - \sqrt{21}}{4} \pm \frac{\sqrt{2 \sqrt{21} - 6}}{4} i$

#### Explanation:

$f \left(x\right) = {x}^{4} - {x}^{3} - 3 {x}^{2} - x + 1$

Notice the symmetry of the coefficients: $1 , - 1 , - 3 , - 1 , 1$

So:

$f \frac{x}{x} ^ 2 = {x}^{2} - x - 3 - \frac{1}{x} - \frac{1}{x} ^ 2 = {\left(x + \frac{1}{x}\right)}^{2} - \left(x + \frac{1}{x}\right) - 5$

Let $t = x + \frac{1}{x}$

Then:

$0 = {t}^{2} - t - 5$

$= {\left(t - \frac{1}{2}\right)}^{2} - \frac{1}{4} - 5$

$= {\left(t - \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{21}}{2}\right)}^{2}$

$= \left(t - \frac{1}{2} - \frac{\sqrt{21}}{2}\right) \left(t - \frac{1}{2} + \frac{\sqrt{21}}{2}\right)$

So:

$x + \frac{1}{x} = t = \frac{1}{2} \pm \frac{\sqrt{21}}{2}$

Multiply both ends by $x$ and rearrange slightly to get:

${x}^{2} - \left(\frac{1}{2} \pm \frac{\sqrt{21}}{2}\right) x + 1 = 0$

Writing the solutions of these two possibilities separately, we have solutions given by the quadratic formula:

${x}_{1 , 2} = \frac{\frac{1}{2} + \frac{\sqrt{21}}{2} \pm \sqrt{{\left(\frac{1}{2} + \frac{\sqrt{21}}{2}\right)}^{2} - 4}}{2}$

$= \frac{1 + \sqrt{21} \pm \sqrt{{\left(1 + \sqrt{21}\right)}^{2} - 16}}{4}$

$= \frac{1 + \sqrt{21} \pm \sqrt{6 + 2 \sqrt{21}}}{4}$

${x}_{3 , 4} = \frac{\frac{1}{2} - \frac{\sqrt{21}}{2} \pm \sqrt{{\left(\frac{1}{2} - \frac{\sqrt{21}}{2}\right)}^{2} - 4}}{2}$

$= \frac{1 - \sqrt{21} \pm \sqrt{{\left(1 - \sqrt{21}\right)}^{2} - 16}}{4}$

$= \frac{1 - \sqrt{21} \pm \sqrt{6 - 2 \sqrt{21}}}{4}$

$= \frac{1 - \sqrt{21}}{4} \pm \frac{\sqrt{2 \sqrt{21} - 6}}{4} i$