Question

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

Answer 1

`2n^3+6n^2+10n = 2n(n^2+3n+5)`

`color(white)(2n^3+6n^2+10n) = 2n(n+3/2-sqrt(11)/2i)(n+3/2+sqrt(11)/2i)`

Explanation:

Given:

`2n^3+6n^2+10n`

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

`2n^3+6n^2+10n = 2n(n^2+3n+5)`

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

`n^2+3n+5 = n^2+3n+9/4+11/4`

`color(white)(n^2+3n+5) = (n+3/2)^2+11/4`

For any Real value of `n` this will be positive, hence `n^2+3n+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 = (a-b)(a+b)`

with `a=(n+3/2)` and `b=sqrt(11)/2i` as follows:

`n^2+3n+5 = (n+3/2)^2+11/4`

`color(white)(n^2+3n+5) = (n+3/2)^2-(sqrt(11)/2i)^2`

`color(white)(n^2+3n+5) = ((n+3/2)-sqrt(11)/2i)((n+3/2)+sqrt(11)/2i)`

`color(white)(n^2+3n+5) = (n+3/2-sqrt(11)/2i)(n+3/2+sqrt(11)/2i)`

where `i` is the imaginary unit, which satisfies `i^2=-1`

Hence:

`2n^3+6n^2+10n = 2n(n+3/2-sqrt(11)/2i)(n+3/2+sqrt(11)/2i)`