# How do you factor completely: 21x^3 + 35x^2 + 9x + 15 ?

Jul 29, 2015

Factor by grouping to find:

$21 {x}^{3} + 35 {x}^{2} + 9 x + 15 = \left(7 {x}^{2} + 3\right) \left(3 x + 5\right)$

#### Explanation:

$21 {x}^{3} + 35 {x}^{2} + 9 x + 15$

$= \left(21 {x}^{3} + 35 {x}^{2}\right) + \left(9 x + 15\right)$

$= 7 {x}^{2} \left(3 x + 5\right) + 3 \left(3 x + 5\right)$

$= \left(7 {x}^{2} + 3\right) \left(3 x + 5\right)$

The term $7 {x}^{2} + 3$ has no linear factors with real coefficients since $7 {x}^{2} + 3 \ge 3 > 0$ for all $x \in \mathbb{R}$.