Solve 27x^4+54x^3-64x-128

$$27 x ^ { 4 } + 54 x ^ { 3 } - 64 x - 128$$

Factor

$$\left(3x-4\right)\left(x+2\right)\left(9x^{2}+12x+16\right)$$

Solution Steps

Do the grouping 27x^{4}+54x^{3}-64x-128=\left(27x^{4}+54x^{3}\right)+\left(-64x-128\right), and factor out 27x^{3} in the first and -64 in the second group.

$$27x^{3}\left(x+2\right)-64\left(x+2\right)$$

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

$$\left(x+2\right)\left(27x^{3}-64\right)$$

Consider 27x^{3}-64. Rewrite 27x^{3}-64 as \left(3x\right)^{3}-4^{3}. The difference of cubes can be factored using the rule: a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right).

$$\left(3x-4\right)\left(9x^{2}+12x+16\right)$$

Rewrite the complete factored expression. Polynomial 9x^{2}+12x+16 is not factored since it does not have any rational roots.

$$\left(3x-4\right)\left(x+2\right)\left(9x^{2}+12x+16\right)$$

Evaluate

$$27x^{4}+54x^{3}-64x-128$$