# How do you find the number of complex, real and rational roots of 4x^3-12x+9=0?

Dec 3, 2016

This cubic has one negative irrational Real root and two Complex ones.

#### Explanation:

$f \left(x\right) = 4 {x}^{3} - 12 x + 9$

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Discriminant

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, $a = 4$, $b = 0$, $c = - 12$ and $d = 9$, so we find:

$\Delta = 0 + 27648 + 0 - 34992 + 0 = - 7344$

Since $\Delta < 0$ this cubic has $1$ Real zero and $2$ non-Real Complex zeros, which are Complex conjugates of one another.

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Descartes' Rule of Signs

The signs of the coefficients of $f \left(x\right)$ are in the pattern $+ - +$. With $2$ changes of sign, $f \left(x\right)$ has $0$ or $2$ positive Real zeros. Since we know that it does not have $2$, it must have $0$ positive Real zeros. The signs of the coefficients of $f \left(- x\right)$ are in the pattern $- + +$. With one change of sign, that means that $f \left(x\right)$ has exactly one negative Real zero.

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Rational root theorem

Since all of the coefficients of $f \left(x\right)$ are integers and the powers of $x$ are in standard descending order, 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 $9$ and $q$ a divisor of the coefficient $4$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{4} , \pm \frac{1}{2} , \pm \frac{3}{4} , \pm 1 , \pm \frac{3}{2} , \pm 3 , \pm \frac{9}{2} , \pm 9$

Note that we already know that the only Real zero is negative, so if there is a rational zero then it is one of:

$- \frac{1}{4} , - \frac{1}{2} , - \frac{3}{4} , - 1 , - \frac{3}{2} , - 3 , - \frac{9}{2} , - 9$

We find:

$f \left(- 3\right) = - 63 < 0$

$f \left(- \frac{3}{2}\right) = \frac{27}{2} > 0$

So the Real zero is somewhere between the possible rational zeros $- 3$ and $- \frac{3}{2}$ and must be irrational.

Conclusion

$f \left(x\right)$ has one Real negative irrational zero and two Complex zeros.