# How do you factor 2n^3 +6n^2 +10n?

Sep 23, 2016

$2 {n}^{3} + 6 {n}^{2} + 10 n = 2 n \left({n}^{2} + 3 n + 5\right)$

$\textcolor{w h i t e}{2 {n}^{3} + 6 {n}^{2} + 10 n} = 2 n \left(n + \frac{3}{2} - \frac{\sqrt{11}}{2} i\right) \left(n + \frac{3}{2} + \frac{\sqrt{11}}{2} i\right)$

#### Explanation:

Given:

$2 {n}^{3} + 6 {n}^{2} + 10 n$

Note that all of the terms are divisible by $2 n$, so we can separate that out as a factor:

$2 {n}^{3} + 6 {n}^{2} + 10 n = 2 n \left({n}^{2} + 3 n + 5\right)$

Looking at the remaining quadratic in $n$ we find:

${n}^{2} + 3 n + 5 = {n}^{2} + 3 n + \frac{9}{4} + \frac{11}{4}$

$\textcolor{w h i t e}{{n}^{2} + 3 n + 5} = {\left(n + \frac{3}{2}\right)}^{2} + \frac{11}{4}$

For any Real value of $n$ this will be positive, hence ${n}^{2} + 3 n + 5$ has no linear factors with Real coefficients.

We can factor it with Complex coefficients, which can be done using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(n + \frac{3}{2}\right)$ and $b = \frac{\sqrt{11}}{2} i$ as follows:

${n}^{2} + 3 n + 5 = {\left(n + \frac{3}{2}\right)}^{2} + \frac{11}{4}$

$\textcolor{w h i t e}{{n}^{2} + 3 n + 5} = {\left(n + \frac{3}{2}\right)}^{2} - {\left(\frac{\sqrt{11}}{2} i\right)}^{2}$

$\textcolor{w h i t e}{{n}^{2} + 3 n + 5} = \left(\left(n + \frac{3}{2}\right) - \frac{\sqrt{11}}{2} i\right) \left(\left(n + \frac{3}{2}\right) + \frac{\sqrt{11}}{2} i\right)$

$\textcolor{w h i t e}{{n}^{2} + 3 n + 5} = \left(n + \frac{3}{2} - \frac{\sqrt{11}}{2} i\right) \left(n + \frac{3}{2} + \frac{\sqrt{11}}{2} i\right)$

where $i$ is the imaginary unit, which satisfies ${i}^{2} = - 1$

Hence:

$2 {n}^{3} + 6 {n}^{2} + 10 n = 2 n \left(n + \frac{3}{2} - \frac{\sqrt{11}}{2} i\right) \left(n + \frac{3}{2} + \frac{\sqrt{11}}{2} i\right)$
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