Solve 5x^3+10x^2-20x-40

$$5 x ^ { 3 } + 10 x ^ { 2 } - 20 x - 40$$

Factor

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

Solution Steps

Factor out 5.

$$5\left(x^{3}+2x^{2}-4x-8\right)$$

Consider x^{3}+2x^{2}-4x-8. Do the grouping x^{3}+2x^{2}-4x-8=\left(x^{3}+2x^{2}\right)+\left(-4x-8\right), and factor out x^{2} in the first and -4 in the second group.

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

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

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

Consider x^{2}-4. Rewrite x^{2}-4 as x^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-2\right)\left(x+2\right)$$

Rewrite the complete factored expression.

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

Evaluate

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