# How do you use the rational roots theorem to find all possible zeros of f(x)=x^4-8x^3+7x-14?

Aug 9, 2016

This $f \left(x\right)$ has no rational zeros, but we can find the zeros algebraically...

#### Explanation:

$f \left(x\right) = {x}^{4} - 8 {x}^{3} + 7 x - 14$

By the rational root 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 $- 14$ and $q$ a divisor of the coefficient $1$ of the leading term. So the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 7 , \pm 14$

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

Can we solve it using other methods?

Substitute $t = \left(x - 2\right)$ to get a quartic with no cubed term...

${x}^{4} - 8 {x}^{3} + 7 x - 14$

$= {\left(x - 2\right)}^{4} - 24 {\left(x - 2\right)}^{2} - 57 \left(x - 2\right) - 48$

$= {t}^{4} - 24 {t}^{2} - 57 t - 48$

$= \left({t}^{2} - a t + b\right) \left({t}^{2} + a t + c\right)$

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

Equating coefficients and rearranging slightly, we get:

$\left\{\begin{matrix}b + c = {a}^{2} - 24 \\ b - c = - \frac{57}{a} \\ b c = - 48\end{matrix}\right.$

Then:

${\left({a}^{2} - 24\right)}^{2} = {\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c = {\left(- \frac{57}{a}\right)}^{2} - 192$

Expanding both ends, we get:

${\left({a}^{2}\right)}^{2} - 48 \left({a}^{2}\right) + 576 = \frac{3249}{\left({a}^{2}\right)} - 192$

Add $192$ to both sides, multiply through by $\left({a}^{2}\right)$ and rearrange slightly to get:

$0 = {\left({a}^{2}\right)}^{3} - 48 {\left({a}^{2}\right)}^{2} + 768 \left({a}^{2}\right) - 3249$

$= {\left(\left({a}^{2}\right) - 16\right)}^{3} + 847$

Hence one Real root gives us:

${a}^{2} = 16 - \sqrt[3]{847}$

which is positive.

So we can let $a = \sqrt{16 - \sqrt[3]{847}}$

Substituting this into the set of simultaneous equations, we get:

$\left\{\begin{matrix}b + c = {a}^{2} - 24 = \left(16 - \sqrt[3]{847}\right) - 24 = - 8 - \sqrt[3]{847} \\ b - c = - \frac{57}{a} = - \frac{57}{\sqrt{16 - \sqrt[3]{847}}} \\ b c = - 48\end{matrix}\right.$

Add the first and second of these and divide by $2$ to find:

$b = - 4 - \frac{1}{2} \sqrt[3]{847} - \frac{57}{2 \sqrt{16 - \sqrt[3]{847}}}$

Subtract the second from the first and divide by $2$ to find:

$c = - 4 - \frac{1}{2} \sqrt[3]{847} + \frac{57}{2 \sqrt{16 - \sqrt[3]{847}}}$

So we now have two quadratics to solve:

${t}^{2} - \left(\sqrt{16 - \sqrt[3]{847}}\right) t - \left(4 + \frac{1}{2} \sqrt[3]{847} + \frac{57}{2 \sqrt{16 - \sqrt[3]{847}}}\right) = 0$

${t}^{2} + \left(\sqrt{16 - \sqrt[3]{847}}\right) t - \left(4 + \frac{1}{2} \sqrt[3]{847} - \frac{57}{2 \sqrt{16 - \sqrt[3]{847}}}\right) = 0$

We can solve these using the quadratic formula, then add $2$ to get zeros of our original quartic in $x$.

I will leave that as an exercise.