Solve 16r^4s+2rs^4

$$16 r ^ { 4 } s + 2 r s ^ { 4 }$$

Factor

$$2rs\left(2r+s\right)\left(4r^{2}-2rs+s^{2}\right)$$

Solution Steps

Factor out 2.

$$2\left(8r^{4}s+rs^{4}\right)$$

Consider 8r^{4}s+rs^{4}. Factor out rs.

$$rs\left(8r^{3}+s^{3}\right)$$

Consider 8r^{3}+s^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

$$\left(2r+s\right)\left(4r^{2}-2rs+s^{2}\right)$$

Rewrite the complete factored expression.

$$2rs\left(2r+s\right)\left(4r^{2}-2rs+s^{2}\right)$$

Evaluate

$$2rs\left(8r^{3}+s^{3}\right)$$