# How do you factor y= x^3 + 3x^2 + 9x + 27?

##### 1 Answer
May 5, 2018

$y = \left({x}^{2} + 9\right) \left(x + 3\right)$

$\textcolor{w h i t e}{y} = \left(x - 3 i\right) \left(x + 3 i\right) \left(x + 3\right)$

#### Explanation:

Given:

$y = {x}^{3} + 3 {x}^{2} + 9 x + 27$

Note that the ratio of the first and second terms is the same as that of the third and fourth terms.

So this cubic will factor by grouping:

$y = {x}^{3} + 3 {x}^{2} + 9 x + 27$

$\textcolor{w h i t e}{y} = \left({x}^{3} + 3 {x}^{2}\right) + \left(9 x + 27\right)$

$\textcolor{w h i t e}{y} = {x}^{2} \left(x + 3\right) + 9 \left(x + 3\right)$

$\textcolor{w h i t e}{y} = \left({x}^{2} + 9\right) \left(x + 3\right)$

The remaining quadratic ${x}^{2} + 9$ is positive for all real values of $x$, so has no linear factors with real coefficients.

It can be factored with complex coefficients as a difference of squares...

${x}^{2} + 9 = {x}^{2} + {3}^{2} = {x}^{2} - {\left(3 i\right)}^{2} = \left(x - 3 i\right) \left(x + 3 i\right)$

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