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

Oct 20, 2015

$\frac{1 , 2 , 4 , 8}{1 , 3 , 5 , 9 , 15 , 45} = 1 , \frac{1}{3} , \frac{1}{5} , \frac{1}{9} , \frac{1}{15} , \frac{1}{45} , 2 , \frac{2}{3} , \frac{2}{5} , \frac{2}{9} , \frac{2}{15} , \frac{2}{45} , 4 , \frac{4}{3} , \frac{4}{5} , \frac{4}{9} , \frac{4}{15} , \frac{4}{45} , 8 , \frac{8}{3} , \frac{8}{5} , \frac{8}{9} , \frac{8}{15} , \frac{8}{45}$

#### Explanation:

The general form for a polynomial is as follows:
$p {x}^{n} + {a}_{n} - 1 + \ldots {a}_{1} x + q$

The rational roots theorem states that, to find any potential zeroes of a given polynomial function, the formula is: $\left(\text{factors of p")/("factors of q}\right)$, where $p$ and $q$ are the first and last coefficients of the given function. In this case, the factors of $p$ are 1,2,4,and 8, and the factors of q are 1,3,5,9,15,and 45.

Once you have all the potential zeroes of the function, you just have to test them using synthetic division.

Oct 20, 2015

You can't.

You can determine that $8 {x}^{3} - 3 {x}^{2} + 5 x + 15 = 0$ has one Real irrational root in $\left(- 1 , - \frac{3}{4}\right)$ and no rational roots.

#### Explanation:

By the rational root theorem, any rational roots of $8 {x}^{3} - 3 {x}^{2} + 5 x + 15 = 0$ must be expressible in lowest terms in the form $\frac{p}{q}$ 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 leading term.

The prime factorisation of $8$ is $2 \cdot 2 \cdot 2$.
The prime factorisation of $15$ is $3 \cdot 5$.

That means that the only possible rational roots are:

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

That's quite a lot of possibilities to try, so let's narrow it down.

Let $f \left(x\right) = 8 {x}^{3} - 3 {x}^{2} + 5 x + 15$

Then $f ' \left(x\right) = 24 {x}^{2} - 6 x + 5$, which has a negative discriminant.

So we can deduce that $f \left(x\right)$ is strictly monotonic increasing.

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

$f \left(- \frac{3}{4}\right) = - 8 \cdot {3}^{3} / {4}^{3} - 3 \cdot {3}^{2} / {4}^{2} + 5 \cdot \frac{3}{4} + 15$

$= - \frac{27}{8} - \frac{27}{16} + \frac{15}{4} + 15$

$= \frac{- 54 - 26 + 60 + 240}{16} = \frac{220}{16} = \frac{55}{4}$

So $f \left(x\right) = 0$ has one Real root in $\left(- 1 , - \frac{3}{4}\right)$, but it is not rational.