# How do you use the rational root theorem to find the roots of F(x) = 8x^3 - 3x^2 + 5x + 15 ?

Aug 17, 2015

Use the rational root theorem to identify possible rational roots, and find that none works.

So $F \left(x\right) = 0$ has no rational roots.

#### Explanation:

By the rational root theorem, any rational roots of $F \left(x\right) = 0$ must be of the form $\frac{p}{q}$ in lowest terms, where $p , q \in \mathbb{Z}$, $q \ne 0$, $p$ a divisor of the constant term $15$ and $q$ a divisor of the coefficient $8$ of the term $8 {x}^{3}$ of highest degree.

That means that the only possible rational roots are:

$\pm 15$, $\pm \frac{15}{2}$, $\pm 5$, $\pm \frac{15}{4}$, $\pm 3$, $\pm \frac{5}{2}$, $\pm \frac{15}{8}$, $\pm \frac{3}{2}$, $\pm \frac{5}{4}$, $\pm 1$, $\pm \frac{3}{4}$, $\pm \frac{5}{8}$, $\pm \frac{1}{2}$, $\pm \frac{3}{8}$, $\pm \frac{1}{4}$, $\pm \frac{1}{8}$

$F ' \left(x\right) = 24 {x}^{2} - 6 x + 5 > 0$ for all $x \in \mathbb{R}$

So $F \left(x\right) = 0$ has exactly one root.

$F \left(- 1\right) = - 8 - 3 - 5 + 15 = - 1$

$F \left(- \frac{3}{4}\right) = - \frac{27}{8} - \frac{27}{16} - \frac{15}{4} + 15$

$= \frac{- 54 - 27 - 60 + 240}{16} = \frac{99}{16}$

So the root lies strictly between $- 1$ and $- \frac{3}{4}$

None of the possible rational roots lies in this range, so $F \left(x\right) = 0$ has no rational roots.

graph{8x^3-3x^2+5x+15 [-1.688, 0.812, -0.62, 0.63]}

If you still want to solve the original problem, first let $t = x - \frac{1}{8}$. This is called a Tschirnhaus transformation, giving us:

$F \left(x\right) = 8 \left({t}^{3} + \frac{37}{64} t + \frac{499}{256}\right)$

Next solve ${t}^{3} + \frac{37}{64} t + \frac{499}{256} = 0$ using Cardano's method

Let $t = u + v$

Then $0 = {t}^{3} + \frac{37}{64} t + \frac{499}{256}$

$= {\left(u + v\right)}^{3} + \frac{37}{64} \left(u + v\right) + \frac{499}{256}$

$= {u}^{3} + {v}^{3} + \left(3 u v + \frac{37}{64}\right) \left(u + v\right) + \frac{499}{256}$

Now let $v = - \frac{37}{192 u}$ so that $3 u v + \frac{37}{64} = 0$

Then multiplying our equation through by ${u}^{3}$ we get:

${\left({u}^{3}\right)}^{2} + \frac{499}{256} \left({u}^{3}\right) - {37}^{3} / {192}^{3} = 0$

Since this derivation is symmetric in $u$ and $v$ we find:

$t = \sqrt[3]{\frac{- \frac{499}{256} + \sqrt{{\left(\frac{499}{256}\right)}^{2} + 4 \cdot {37}^{3} / {192}^{3}}}{2}} + \sqrt[3]{\frac{- \frac{499}{256} - \sqrt{{\left(\frac{499}{256}\right)}^{2} + 4 \cdot {37}^{3} / {192}^{3}}}{2}}$

and

$x = \frac{1}{8} + \sqrt[3]{\frac{- \frac{499}{256} + \sqrt{{\left(\frac{499}{256}\right)}^{2} + 4 \cdot {37}^{3} / {192}^{3}}}{2}} + \sqrt[3]{\frac{- \frac{499}{256} - \sqrt{{\left(\frac{499}{256}\right)}^{2} + 4 \cdot {37}^{3} / {192}^{3}}}{2}}$

So it can be done, but it's a little complex.