# How do you factor 81x^5y+24x^2y^4?

Jul 2, 2016

$3 {x}^{2} y \left(3 x + 2 y\right) \left(9 {x}^{2} - 6 x y + 4 {y}^{2}\right)$

#### Explanation:

Always look for a common factor first, in this case, $3 {x}^{2} y$

$3 {x}^{2} y \left(27 {x}^{3} + 8 {y}^{3}\right) \text{ now this is the sum of cubes}$

=$3 {x}^{2} y \left(3 x + 2 y\right) \left(9 {x}^{2} - 6 x y + 4 {y}^{2}\right)$

Sum of cubes is given as:

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

Difference of cubes is given as:

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