Solve (-6-2i)(5+2i)

$$( - 6 - 2 i ) ( 5 + 2 i )$$

Evaluate

$$-26-22i$$

Solution Steps

Multiply complex numbers -6-2i and 5+2i like you multiply binomials.

$$-6\times 5-6\times \left(2i\right)-2i\times 5-2\times 2i^{2}$$

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

$$-6\times 5-6\times \left(2i\right)-2i\times 5-2\times 2\left(-1\right)$$

Do the multiplications.

$$-30-12i-10i+4$$

Combine the real and imaginary parts.

$$-30+4+\left(-12-10\right)i$$

Do the additions.

$$-26-22i$$

Real Part

$$-26$$

Solution Steps

Multiply complex numbers -6-2i and 5+2i like you multiply binomials.

$$Re(-6\times 5-6\times \left(2i\right)-2i\times 5-2\times 2i^{2})$$

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

$$Re(-6\times 5-6\times \left(2i\right)-2i\times 5-2\times 2\left(-1\right))$$

Do the multiplications in -6\times 5-6\times \left(2i\right)-2i\times 5-2\times 2\left(-1\right).

$$Re(-30-12i-10i+4)$$

Combine the real and imaginary parts in -30-12i-10i+4.

$$Re(-30+4+\left(-12-10\right)i)$$

Do the additions in -30+4+\left(-12-10\right)i.

$$Re(-26-22i)$$

The real part of -26-22i is -26.

$$-26$$