Solve (x^4/4-x^3-14x^2)]_-4^0=

$$( \frac { x ^ { 4 } } { 4 } - x ^ { 3 } - 14 x ^ { 2 } ) ] _ { - 4 } ^ { 0 } =$$

Evaluate

$$\frac{x^{2}\left(x^{2}-4x-56\right)}{4}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply -x^{3}-14x^{2} times \frac{4}{4}.

$$\frac{x^{4}}{4}+\frac{4\left(-x^{3}-14x^{2}\right)}{4}$$

Since \frac{x^{4}}{4} and \frac{4\left(-x^{3}-14x^{2}\right)}{4} have the same denominator, add them by adding their numerators.

$$\frac{x^{4}+4\left(-x^{3}-14x^{2}\right)}{4}$$

Do the multiplications in x^{4}+4\left(-x^{3}-14x^{2}\right).

$$\frac{x^{4}-4x^{3}-56x^{2}}{4}$$

Factor

$$\frac{\left(x-\left(2-2\sqrt{15}\right)\right)\left(x-\left(2\sqrt{15}+2\right)\right)x^{2}}{4}$$