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

Jul 15, 2017

Use the discriminant to find that it has $1$ irrational real zero and $2$ non-real complex zeros.

#### Explanation:

The cubic equation:

${x}^{3} - {x}^{2} - 2 x + 7 = 0$

is in the standard form:

$a {x}^{3} + b {x}^{2} + c x + d = 0$

Its discriminant $\Delta$ 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, $a = 1$, $b = - 1$, $c = - 2$ and $d = 7$, so we find:

$\Delta = 4 + 32 + 28 - 1323 + 252 = - 1007$

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

$\textcolor{w h i t e}{}$
We were also asked if the roots were rational.

By the rational roots theorem, any rational zeros of this cubic must be expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $7$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational roots are:

$\pm 1 , \pm 7$

None of these work, so our cubic has no rational roots.