Solve 3x^6y^5+24y^5

$$3 x ^ { 6 } y ^ { 5 } + 24 y ^ { 5 }$$

Factor

$$3\left(x^{2}+2\right)\left(x^{4}-2x^{2}+4\right)y^{5}$$

Solution Steps

Factor out 3.

$$3\left(x^{6}y^{5}+8y^{5}\right)$$

Consider x^{6}y^{5}+8y^{5}. Factor out y^{5}.

$$y^{5}\left(x^{6}+8\right)$$

Consider x^{6}+8. Rewrite x^{6}+8 as \left(x^{2}\right)^{3}+2^{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).

$$\left(x^{2}+2\right)\left(x^{4}-2x^{2}+4\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: x^{2}+2,x^{4}-2x^{2}+4.

$$3y^{5}\left(x^{2}+2\right)\left(x^{4}-2x^{2}+4\right)$$

Evaluate

$$3y^{5}\left(x^{6}+8\right)$$