Solve |-3|-sqrt{16}+1/2x^3sqrt{-8}+(-2)^2

$$| - 3 | - \sqrt { 16 } + \frac { 1 } { 2 } x ^ { 3 } \sqrt { - 8 } + ( - 2 ) ^ { 2 }$$

Evaluate

$$\text{Indeterminate}$$

Solution Steps

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -3 is 3.

$$3-\sqrt{16}+\frac{1}{2}x^{3}\sqrt{-8}+\left(-2\right)^{2}$$

Calculate the square root of 16 and get 4.

$$3-4+\frac{1}{2}x^{3}\sqrt{-8}+\left(-2\right)^{2}$$

Subtract 4 from 3 to get -1.

$$-1+\frac{1}{2}x^{3}\sqrt{-8}+\left(-2\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$-1+\frac{1}{2}x^{3}\sqrt{-8}+4$$

Add -1 and 4 to get 3.

$$3+\frac{1}{2}x^{3}\sqrt{-8}$$