Find each product. Write answers in standard form. (2+i)(3-2i)

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Recall that to multiply two complex numbers, use the distributive property (FOIL method): multiply each term in the first complex number by each term in the second complex number.

Write the expression explicitly: $(2 + i)(3 - 2i) = 2 \cdot 3 + 2 \cdot (-2i) + i \cdot 3 + i \cdot (-2i)$.

Calculate each product: $2 \cdot 3 = 6$, $2 \cdot (-2i) = -4i$, $i \cdot 3 = 3i$, and $i \cdot (-2i) = -2i^2$.

Remember that $i^2 = -1$, so replace $-2i^2$ with $-2(-1) = 2$.

Combine like terms: add the real parts $6 + 2$ and the imaginary parts $-4i + 3i$ to write the product in standard form $a + bi$.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Complex Numbers

Complex numbers are numbers in the form a + bi, where a and b are real numbers and i is the imaginary unit with the property i² = -1. They extend the real number system and are used to represent quantities involving the square root of negative numbers.

Multiplication of Complex Numbers

To multiply complex numbers, use the distributive property (FOIL method) to expand the product, then combine like terms. Remember to apply i² = -1 to simplify terms involving i², converting them into real numbers.

Standard Form of a Complex Number

The standard form of a complex number is a + bi, where a is the real part and b is the imaginary part. After multiplication, simplify the expression to this form by combining real and imaginary terms separately.

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