# How do you use the rational roots theorem to find all possible zeros of P(x) = 12x^4 + x^3 + 4x^2 + 7x + 8 ?

Jun 7, 2016

Use the rational roots theorem to find possible candidate rational zeros, any thereby find that it has none.

#### Explanation:

$P \left(x\right) = 12 {x}^{4} + {x}^{3} + 4 {x}^{2} + 7 x + 8$

By the rational roots theorem, any rational zeros of $P \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $8$ and $q$ a divisor of the coefficient $12$ of the leading term.

So in our example, it means that the only possible rational zeros are:

$\pm \frac{1}{12}$, $\pm \frac{1}{6}$, $\pm \frac{1}{4}$, $\pm \frac{1}{3}$, $\pm \frac{1}{2}$, $\pm \frac{2}{3}$, $\pm 1$, $\pm \frac{4}{3}$, $\pm 2$, $\pm \frac{8}{3}$, $\pm 4$, $\pm 8$

That's rather a lot of possible zeros to try, but you can narrow it down a little by noting that the coefficients of $P \left(x\right)$ are all positive. Hence it has no positive zeros, leaving the possible rational zeros:

$- \frac{1}{12}$, $- \frac{1}{6}$, $- \frac{1}{4}$, $- \frac{1}{3}$, $- \frac{1}{2}$, $- \frac{2}{3}$, $- 1$, $- \frac{4}{3}$, $- 2$, $- \frac{8}{3}$, $- 4$, $- 8$

Substituting each of these for $x$ in $P \left(x\right)$, we find that none is a zero. So $P \left(x\right)$ has no rational zeros.

That's all the rational roots theorem tells us.

In fact, this particular quartic only has non-Real Complex zeros.