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

$$( 1 + 2 ) ( 2 - i ) ( 3 + 2 i )$$

Evaluate

$$24+3i$$

Solution Steps

Add 1 and 2 to get 3.

$$3\left(2-i\right)\left(3+2i\right)$$

Multiply 3 times 2-i.

$$\left(3\times 2+3\left(-i\right)\right)\left(3+2i\right)$$

Do the multiplications.

$$\left(6-3i\right)\left(3+2i\right)$$

Multiply complex numbers 6-3i and 3+2i like you multiply binomials.

$$6\times 3+6\times \left(2i\right)-3i\times 3-3\times 2i^{2}$$

By definition, i^{2} is -1.

$$6\times 3+6\times \left(2i\right)-3i\times 3-3\times 2\left(-1\right)$$

Do the multiplications.

$$18+12i-9i+6$$

Combine the real and imaginary parts.

$$18+6+\left(12-9\right)i$$

Do the additions.

$$24+3i$$

Real Part

$$24$$

Solution Steps

Add 1 and 2 to get 3.

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

Multiply 3 times 2-i.

$$Re(\left(3\times 2+3\left(-i\right)\right)\left(3+2i\right))$$

Do the multiplications in 3\times 2+3\left(-i\right).

$$Re(\left(6-3i\right)\left(3+2i\right))$$

Multiply complex numbers 6-3i and 3+2i like you multiply binomials.

$$Re(6\times 3+6\times \left(2i\right)-3i\times 3-3\times 2i^{2})$$

By definition, i^{2} is -1.

$$Re(6\times 3+6\times \left(2i\right)-3i\times 3-3\times 2\left(-1\right))$$

Do the multiplications in 6\times 3+6\times \left(2i\right)-3i\times 3-3\times 2\left(-1\right).

$$Re(18+12i-9i+6)$$

Combine the real and imaginary parts in 18+12i-9i+6.

$$Re(18+6+\left(12-9\right)i)$$

Do the additions in 18+6+\left(12-9\right)i.

$$Re(24+3i)$$

The real part of 24+3i is 24.

$$24$$