To solve the expression \((-8 + 6i)(3 + i)\), we apply the distributive property (also known as FOIL for binomials):

\[
(-8 + 6i)(3 + i) = (-8)(3) + (-8)(i) + (6i)(3) + (6i)(i)
\]

Let's calculate each term:

1. \((-8)(3) = -24\)
2. \((-8)(i) = -8i\)
3. \((6i)(3) = 18i\)
4. \((6i)(i) = 6i^2\)

Since \(i^2 = -1\), we substitute \(6i^2\) with \(6(-1) = -6\).

Now, combining all the terms:
\[
-24 - 8i + 18i - 6 = (-24 - 6) + (-8i + 18i) = -30 + 10i
\]

Thus, the result of \((-8 + 6i)(3 + i)\) is:

\[
\boxed{-30 + 10i}
\]

Would you like more details or have any questions? Here are some related questions:

1. How do you multiply complex numbers?
2. What is the significance of \(i^2 = -1\) in complex numbers?
3. How do you add or subtract complex numbers?
4. Can you simplify powers of \(i\)?
5. How do you find the magnitude of a complex number?

**Tip:** When multiplying complex numbers, always remember to simplify \(i^2\) to \(-1\).

Learn how to multiply complex numbers (-8 + 6i)(3 + i) with a step-by-step solution. This problem involves the distributive property and multiplication of complex numbers. Suitable for high school students studying advanced algebra.

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