How do you factor the expression 6x^3 + 4x^2 + 3x + 2?

Jan 12, 2016

Factor by grouping to find:

$6 {x}^{3} + 4 {x}^{2} + 3 x + 2$

$= \left(2 {x}^{2} + 1\right) \left(3 x + 2\right)$

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

Explanation:

Factor by grouping:

$6 {x}^{3} + 4 {x}^{2} + 3 x + 2$

$= \left(6 {x}^{3} + 4 {x}^{2}\right) + \left(3 x + 2\right)$

$= 2 {x}^{2} \left(3 x + 2\right) + 1 \left(3 x + 2\right)$

$= \left(2 {x}^{2} + 1\right) \left(3 x + 2\right)$

The remaining quadratic factor has no linear factors with Real coefficients, but can be factored as a difference of squares using Complex coefficients:

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

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