Solve (2i^2-1)(i^3+2)/2-i

$$\frac { ( 2 i ^ { 2 } - 1 ) ( i ^ { 3 } + 2 ) } { 2 - i }$$

Evaluate

$$-3$$

Solution Steps

Calculate i to the power of 2 and get -1.

$$\frac{\left(2\left(-1\right)-1\right)\left(i^{3}+2\right)}{2-i}$$

Multiply 2 and -1 to get -2.

$$\frac{\left(-2-1\right)\left(i^{3}+2\right)}{2-i}$$

Subtract 1 from -2 to get -3.

$$\frac{-3\left(i^{3}+2\right)}{2-i}$$

Calculate i to the power of 3 and get -i.

$$\frac{-3\left(-i+2\right)}{2-i}$$

Divide -3\left(-i+2\right) by 2-i to get \left(-\frac{6}{5}-\frac{3}{5}i\right)\left(-i+2\right).

$$\left(-\frac{6}{5}-\frac{3}{5}i\right)\left(-i+2\right)$$

Use the distributive property to multiply -\frac{6}{5}-\frac{3}{5}i by -i+2.

$$-\frac{3}{5}+\frac{6}{5}i+\left(-\frac{12}{5}-\frac{6}{5}i\right)$$

Add -\frac{3}{5}+\frac{6}{5}i and -\frac{12}{5}-\frac{6}{5}i to get -3.

$$-3$$

Real Part

$$-3$$

Solution Steps

Calculate i to the power of 2 and get -1.

$$Re(\frac{\left(2\left(-1\right)-1\right)\left(i^{3}+2\right)}{2-i})$$

Multiply 2 and -1 to get -2.

$$Re(\frac{\left(-2-1\right)\left(i^{3}+2\right)}{2-i})$$

Subtract 1 from -2 to get -3.

$$Re(\frac{-3\left(i^{3}+2\right)}{2-i})$$

Calculate i to the power of 3 and get -i.

$$Re(\frac{-3\left(-i+2\right)}{2-i})$$

Divide -3\left(-i+2\right) by 2-i to get \left(-\frac{6}{5}-\frac{3}{5}i\right)\left(-i+2\right).

$$Re(\left(-\frac{6}{5}-\frac{3}{5}i\right)\left(-i+2\right))$$

Use the distributive property to multiply -\frac{6}{5}-\frac{3}{5}i by -i+2.

$$Re(-\frac{3}{5}+\frac{6}{5}i+\left(-\frac{12}{5}-\frac{6}{5}i\right))$$

Add -\frac{3}{5}+\frac{6}{5}i and -\frac{12}{5}-\frac{6}{5}i to get -3.

$$Re(-3)$$

The real part of -3 is -3.

$$-3$$