Solve 8i(7+2i)(10-3i)

$$8 i ( 7 + 2 i ) ( 10 - 3 i )$$

Evaluate

$$8+608i$$

Solution Steps

Multiply 8i times 7+2i.

$$\left(8i\times 7+8\times 2i^{2}\right)\left(10-3i\right)$$

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

$$\left(8i\times 7+8\times 2\left(-1\right)\right)\left(10-3i\right)$$

Do the multiplications. Reorder the terms.

$$\left(-16+56i\right)\left(10-3i\right)$$

Multiply complex numbers -16+56i and 10-3i like you multiply binomials.

$$-16\times 10-16\times \left(-3i\right)+56i\times 10+56\left(-3\right)i^{2}$$

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

$$-16\times 10-16\times \left(-3i\right)+56i\times 10+56\left(-3\right)\left(-1\right)$$

Do the multiplications.

$$-160+48i+560i+168$$

Combine the real and imaginary parts.

$$-160+168+\left(48+560\right)i$$

Do the additions.

$$8+608i$$

Real Part

$$8$$

Solution Steps

Multiply 8i times 7+2i.

$$Re(\left(8i\times 7+8\times 2i^{2}\right)\left(10-3i\right))$$

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

$$Re(\left(8i\times 7+8\times 2\left(-1\right)\right)\left(10-3i\right))$$

Do the multiplications in 8i\times 7+8\times 2\left(-1\right). Reorder the terms.

$$Re(\left(-16+56i\right)\left(10-3i\right))$$

Multiply complex numbers -16+56i and 10-3i like you multiply binomials.

$$Re(-16\times 10-16\times \left(-3i\right)+56i\times 10+56\left(-3\right)i^{2})$$

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

$$Re(-16\times 10-16\times \left(-3i\right)+56i\times 10+56\left(-3\right)\left(-1\right))$$

Do the multiplications in -16\times 10-16\times \left(-3i\right)+56i\times 10+56\left(-3\right)\left(-1\right).

$$Re(-160+48i+560i+168)$$

Combine the real and imaginary parts in -160+48i+560i+168.

$$Re(-160+168+\left(48+560\right)i)$$

Do the additions in -160+168+\left(48+560\right)i.

$$Re(8+608i)$$

The real part of 8+608i is 8.

$$8$$