Solve a^2x^3+2ax^3-8a^2-16a

$$a ^ { 2 } x ^ { 3 } + 2 a x ^ { 3 } - 8 a ^ { 2 } - 16 a$$

Factor

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

Solution Steps

Factor out a.

$$a\left(ax^{3}+2x^{3}-8a-16\right)$$

Consider ax^{3}+2x^{3}-8a-16. Do the grouping ax^{3}+2x^{3}-8a-16=\left(ax^{3}+2x^{3}\right)+\left(-8a-16\right), and factor out x^{3} in the first and -8 in the second group.

$$x^{3}\left(a+2\right)-8\left(a+2\right)$$

Factor out common term a+2 by using distributive property.

$$\left(a+2\right)\left(x^{3}-8\right)$$

Consider x^{3}-8. Rewrite x^{3}-8 as x^{3}-2^{3}. The difference of cubes can be factored using the rule: p^{3}-q^{3}=\left(p-q\right)\left(p^{2}+pq+q^{2}\right).

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

Rewrite the complete factored expression. Polynomial x^{2}+2x+4 is not factored since it does not have any rational roots.

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

Evaluate

$$a\left(a+2\right)\left(x^{3}-8\right)$$