# What are all the rational zeros of 2x^3-15x^2+9x+22?

Aug 29, 2016

Use the rational roots theorem to find the possible rational zeros.

#### Explanation:

$f \left(x\right) = 2 {x}^{3} - 15 {x}^{2} + 9 x + 22$

By the rational roots theorem, the only possible rational zeros are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $22$ and $q$ a divisor of the coefficient $2$ of the leading term.

So the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm 2 , \pm \frac{11}{2} , \pm 11 , \pm 22$

Evaluating $f \left(x\right)$ for each of these we find that none work, so $f \left(x\right)$ has no rational zeros.

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We can find out a little more without actually solving the cubic...

The discriminant $\Delta$ of a cubic polynomial in the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

In our example, $a = 2$, $b = - 15$, $c = 9$ and $d = 22$, so we find:

$\Delta = 18225 - 5832 + 297000 - 52272 - 106920 = 150201$

Since $\Delta > 0$ this cubic has $3$ Real zeros.

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Using Descartes' rule of signs, we can determine that two of these zeros are positive and one negative.