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

Nov 2, 2016

Here's a sketch of how you can solve this algebraically...

#### Explanation:

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

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 2$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational zeros are:

$\pm 1$, $\pm 2$

None of these work, so $f \left(x\right)$ has no rational zeros.

Tschirnhaus transformation

Let $t = 4 x - 1$

$256 f \left(x\right) = 256 {x}^{4} - 256 {x}^{3} - 1024 {x}^{2} - 768 x - 512$

$\textcolor{w h i t e}{256 f \left(x\right)} = {\left(4 x - 1\right)}^{4} - 70 {\left(4 x - 1\right)}^{2} - 328 \left(4 x - 1\right) - 771$

$\textcolor{w h i t e}{256 f \left(x\right)} = {t}^{4} - 70 {t}^{2} - 328 t - 771$

$\textcolor{w h i t e}{256 f \left(x\right)} = \left({t}^{2} - a t + b\right) \left({t}^{2} + a t + c\right)$

$\textcolor{w h i t e}{256 f \left(x\right)} = {t}^{4} + \left(b + c - {a}^{2}\right) {t}^{2} + a \left(b - c\right) t + b c$

Hence:

$\left\{\begin{matrix}b + c = {a}^{2} - 70 \\ b - c = - \frac{328}{a} \\ b c = - 771\end{matrix}\right.$

So:

${\left({a}^{2} - 70\right)}^{2} = {\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c = {\left(- \frac{328}{a}\right)}^{2} - 4 \left(771\right)$

Multiplied out:

${a}^{4} - 140 {a}^{2} + 4900 = \frac{107584}{a} ^ 2 - 3084$

Multiplied through by ${a}^{2}$ and rearranged a little:

${\left({a}^{2}\right)}^{3} - 140 {\left({a}^{2}\right)}^{2} + 7984 \left({a}^{2}\right) - 107584 = 0$

Solve this cubic using Cardano's method (see https://socratic.org/s/azh362cn) to find:

${a}^{2} = \frac{1}{3} \left(140 + 8 \sqrt{1628 + 12 \sqrt{20589}} + 8 \sqrt{1628 - 12 \sqrt{20589}}\right) \approx 18.885$

Without loss of generality we can choose the positive square root as our value of $a$, then:

$\left\{\begin{matrix}b = \frac{1}{2} \left({a}^{2} - 70 - \frac{328}{a}\right) \\ c = \frac{1}{2} \left({a}^{2} - 70 + \frac{328}{a}\right)\end{matrix}\right.$

${t}^{2} - a t + b = 0$
${t}^{2} + a t + c = 0$
$x = \frac{1}{4} \left(t + 1\right)$
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