# How do you factor completely 15y^3+12y^2+5y+4 ?

May 1, 2016

15y^3+12y^2+5y+4)

$= \left(3 {y}^{2} + 1\right) \left(5 y + 4\right)$

$= \left(\sqrt{3} y - i\right) \left(\sqrt{3} y + i\right) \left(5 y + 4\right)$

#### Explanation:

Factor by grouping:

15y^3+12y^2+5y+4)

$= \left(15 {y}^{3} + 12 {y}^{2}\right) + \left(5 y + 4\right)$

$= 3 {y}^{2} \left(5 y + 4\right) + 1 \left(5 y + 4\right)$

$= \left(3 {y}^{2} + 1\right) \left(5 y + 4\right)$

If $y \in \mathbb{R}$ then ${y}^{2} \ge 0$, hence $3 {y}^{2} + 1 > 0$. So this quadratic factor has no linear factors with Real coefficients.

If we allow Complex coefficients we can factor as a difference of squares:

$3 {y}^{2} + 1 = {\left(\sqrt{3} y\right)}^{2} - {i}^{2} = \left(\sqrt{3} y - i\right) \left(\sqrt{3} y + i\right)$