Solve -x^4-4x^3+16x^2+64x

$$- x ^ { 4 } - 4 x ^ { 3 } + 16 x ^ { 2 } + 64 x$$

Factor

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

Solution Steps

Factor out x.

$$x\left(-x^{3}-4x^{2}+16x+64\right)$$

Consider -x^{3}-4x^{2}+16x+64. Do the grouping -x^{3}-4x^{2}+16x+64=\left(-x^{3}-4x^{2}\right)+\left(16x+64\right), and factor out x^{2} in the first and -16 in the second group.

$$x^{2}\left(-x-4\right)-16\left(-x-4\right)$$

Factor out common term -x-4 by using distributive property.

$$\left(-x-4\right)\left(x^{2}-16\right)$$

Consider x^{2}-16. Rewrite x^{2}-16 as x^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-4\right)\left(x+4\right)$$

Rewrite the complete factored expression.

$$x\left(-x-4\right)\left(x-4\right)\left(x+4\right)$$

Evaluate

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