Solve 6(-7+6i)(-4+2i)

$$6 ( - 7 + 6 i ) ( - 4 + 2 i )$$

Evaluate

$$96-228i$$

Solution Steps

Multiply 6 times -7+6i.

$$\left(6\left(-7\right)+6\times \left(6i\right)\right)\left(-4+2i\right)$$

Do the multiplications.

$$\left(-42+36i\right)\left(-4+2i\right)$$

Multiply complex numbers -42+36i and -4+2i like you multiply binomials.

$$-42\left(-4\right)-42\times \left(2i\right)+36i\left(-4\right)+36\times 2i^{2}$$

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

$$-42\left(-4\right)-42\times \left(2i\right)+36i\left(-4\right)+36\times 2\left(-1\right)$$

Do the multiplications.

$$168-84i-144i-72$$

Combine the real and imaginary parts.

$$168-72+\left(-84-144\right)i$$

Do the additions.

$$96-228i$$

Real Part

$$96$$

Solution Steps

Multiply 6 times -7+6i.

$$Re(\left(6\left(-7\right)+6\times \left(6i\right)\right)\left(-4+2i\right))$$

Do the multiplications in 6\left(-7\right)+6\times \left(6i\right).

$$Re(\left(-42+36i\right)\left(-4+2i\right))$$

Multiply complex numbers -42+36i and -4+2i like you multiply binomials.

$$Re(-42\left(-4\right)-42\times \left(2i\right)+36i\left(-4\right)+36\times 2i^{2})$$

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

$$Re(-42\left(-4\right)-42\times \left(2i\right)+36i\left(-4\right)+36\times 2\left(-1\right))$$

Do the multiplications in -42\left(-4\right)-42\times \left(2i\right)+36i\left(-4\right)+36\times 2\left(-1\right).

$$Re(168-84i-144i-72)$$

Combine the real and imaginary parts in 168-84i-144i-72.

$$Re(168-72+\left(-84-144\right)i)$$

Do the additions in 168-72+\left(-84-144\right)i.

$$Re(96-228i)$$

The real part of 96-228i is 96.

$$96$$