# How do you factor n^3 - 3n^2 + 2n - 990 = 0 ?

Jul 24, 2015

Use the rational roots theorem and some approximating to find:

$f \left(n\right) = {n}^{3} - 3 {n}^{2} + 2 n - 990 = \left(n - 11\right) \left({n}^{2} + 8 n + 90\right)$

#### Explanation:

Let $f \left(n\right) = {n}^{3} - 3 {n}^{2} + 2 n - 990$

By the rational roots theorem, any rational roots of $f \left(n\right) = 0$ must be of the form $\frac{p}{q}$ where $p$ and $q$ are integers, $q \ne 0$, $p$ a divisor of the constant term $- 990$ and $q$ a divisor of the coefficient, $1$, of the term ${n}^{3}$ of highest degree.

So the only possible rational roots are the factors of $990$, viz

$\pm 1$, $\pm 2$, $\pm 3$, $\pm 5$, $\pm 6$, $\pm 9$, $\pm 10$, $\pm 11$, $\pm 18$, $\pm 22$, $\pm 30$, $\pm 33$, $\pm 45$, $\pm 55$, $\pm 90$, $\pm 99$, $\pm 110$, $\pm 165$, $\pm 198$, $\pm 330$, $\pm 495$, $\pm 990$

That's rather a lot of factors to check (and I'm not sure I found all of them), so let's try to get closer:

${n}^{3} - 3 {n}^{2} + 2 n = 990$

So try ${n}^{3} \cong 990$, say $n \cong 10$

$f \left(10\right) = 1000 - 300 + 20 - 990 = - 270$
$f \left(11\right) = 1331 - 363 + 22 - 990 = 0$

So $\left(n - 11\right)$ is a factor.

Divide $f \left(n\right)$ by $\left(n - 11\right)$ to find:

$f \left(n\right) = {n}^{3} - 3 {n}^{2} + 2 n - 990 = \left(n - 11\right) \left({n}^{2} + 8 n + 90\right)$

The discriminant of ${n}^{2} + 8 n + 90$ is:

${8}^{2} - \left(4 \cdot 1 \cdot 90\right) = 64 - 360 = - 296$

So there are no more linear factors with real coefficients.