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# How do you find the number of complex, real and rational roots of x^3-6x^2-7x-12=0?

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Feb 1, 2018

This cubic has one positive irrational real root and two non-real complex roots.

#### Explanation:

Given:

$f \left(x\right) = {x}^{3} - 6 {x}^{2} - 7 x - 12 = 0$

Rational root theorem

By the rational root theorem, any rational root of this polynomial is expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 12$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational roots are:

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 12$

Descartes' Rule of Signs

The pattern of signs of the coefficients of the polynomial is $+ - - -$. With one change of signs Descartes' Rule of Signs tells us that this cubic has exactly one positive root.

We find that:

$f \left(6\right) = - 54 < 0$

$f \left(12\right) = 768 > 0$

So the real positive root cannot be rational and must be an irrational number in $\left(6 , 12\right)$.

Reversing the signs on the terms of odd degree, we get the pattern $- - + -$. With two changes of sign, we can deduce that this cubic has $0$ or $2$ negative real roots.

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, $a = 1$, $b = - 6$, $c = - 7$ and $d = - 12$, so we find:

$\Delta = 1764 + 1372 - 10368 - 3888 - 9072 = - 20192$

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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