Problem 68 - Precalculus Student Solutions Manual 5th

Question

Find each product. Write the answer in standard form. $$(3-i)(3+i)(2-6 i)$$

Step-by-Step Solution

Answer

20 - 60i

Step 1: Multiply the first two complex numbers

To start, multiply (3-i)(3+i) Use the distributive property (also known as the FOIL method) to combine these two binomials:(3)(3) + (3)(i) + (-i)(3) + (-i)(i) = 9 + 3i - 3i - i^2 Simplify the expression by combining like terms: 9 + 3i - 3i - i^2 Since 3i - 3i = 0, you now have 9 - i^2Remember that i^2 = -1, so the expression becomes: 9 - (-1) = 9 +1 = 10

Step 2: Multiply the result by the remaining complex number

Now multiply the previous result (10) by (2-6i) Use the distributive property to do this: 10 * (2-6i) This simplifies to: 10*2 + 10*(-6i) = 20 - 60i

Key Concepts

Understanding the Distributive Property

The distributive property is a fundamental concept in algebra that you'll use often. It states that for any numbers or expressions, say a, b, and c, the formula is: given by: \[ a(b + c) = ab + ac \].
This property is essential when multiplying complex numbers, especially when dealing with binomials. In the example provided, we use this property multiple times.
First, when you multiply \( (3-i)(3+i) \),
you distribute each term in the first binomial by each term in the second binomial: (3)(3) + (3)(i) + (-i)(3) + (-i)(i).
This systematic approach ensures that all combinations of terms are considered.
Remember to combine like terms and simplify as necessary.

Using the FOIL Method

The FOIL method is a specific application of the distributive property, used particularly for multiplying two binomials. FOIL stands for First, Outer, Inner, Last.
Here's a quick breakdown:

• First: Multiply the first terms in each binomial.
• Outer: Multiply the outer terms in the binomials.
• Inner: Multiply the inner terms.
• Last: Multiply the last terms.

When multiplying \( (3-i)(3+i) \),
the steps are as follows: \( (3)(3) + (3)(i) + (-i)(3) + (-i)(i) = 9 + 3i - 3i - i^2 \).
The terms simplify to 9 - (-1), as \( i^2 \) is equivalent to -1, resulting in 9 + 1 which equals 10.
This method is straightforward once you practice a few examples.

Reaching the Standard Form

When working with complex numbers, expressing the final result in standard form is crucial. The standard form of a complex number is \( a + bi \),
where a and b are real numbers, and \( i \) is the imaginary unit.
In the example, the initial multiplication \( (3-i)(3+i) = 10 \) provided a real number.
Then, we multiplied by \( (2-6i) \) next.
Applying the distributive property again, \( 10 \times (2-6i) = 20 - 60i \).
The final expression \( 20 - 60i \) is already in the standard form, where 20 is the real part and -60i is the imaginary part.
Ensuring your final answer is in \( a + bi \) format simplifies the understanding and further use of the result in subsequent calculations.