Problem 21 - Algebra for College Students

Question

Find the following products and express answers in simplest radical form. All variables represent nonnegative real numbers. \(-4 \sqrt{5}(2 \sqrt{5}+4 \sqrt{12})\)

Step-by-Step Solution

Answer

The simplest form is \(-40 - 32 \sqrt{15}\).

Step 1: Distribute the Multiplication

First, distribute the \(-4 \sqrt{5}\) to both terms inside the parentheses: \(-4 \sqrt{5} \cdot 2 \sqrt{5} + (-4 \sqrt{5} \cdot 4 \sqrt{12})\).

Step 2: Multiply the Radicals

Multiply the coefficients and the radicals separately: \((-4 \times 2) \times (\sqrt{5} \times \sqrt{5}) = -8 \times 5 = -40\)and \((-4 \times 4) \times (\sqrt{5} \times \sqrt{12}) = -16 \times (\sqrt{60}))\).

Step 3: Simplify the Second Product

Simplify \(\sqrt{60}\) with its prime factorization: \(\sqrt{60} = \sqrt{4 \times 15} = 2 \sqrt{15}\). Multiply by the coefficient: \(-16 \times 2 \sqrt{15} = -32 \sqrt{15}\).

Step 4: Combine the Terms

Combine the results from Step 2 and Step 3: \(-40 + (-32 \sqrt{15})\).

Step 5: Express in Simplest Radical Form

The expression \(-40 - 32 \sqrt{15}\) is already in simplest radical form since all radicals and coefficients have been simplified correctly.

Key Concepts

Distributing Multiplication

When dealing with expressions that include parentheses, often the first step is to distribute a term outside the parentheses across all terms inside. In our example, we start with

• Multiplying the term \(-4 \sqrt{5}\) by each term inside the parentheses
• This involves

• distributing, which means applying the multiplication to both \(2 \sqrt{5}\) and \(4 \sqrt{12}\).

This results in two separate products: - \(-4 \sqrt{5} \cdot 2 \sqrt{5}\)- \(-4 \sqrt{5} \cdot 4 \sqrt{12}\) Breaking down complex expressions like this not only helps ensure you multiply correctly but also aids in clear and logical problem-solving. This method of distributing multiplication is crucial for tackling similar problems involving radicals or any algebraic expressions.

Multiplying Radicals

Once we've distributed, we need to address multiplying radicals. This task is simplified if we remember to:

• Multiply the coefficients, which are the numbers outside the radical signs.
• Multiply the radicands, which are the numbers inside the radical signs.

Take \(-4 \sqrt{5} \cdot 2 \sqrt{5}\) as an example. Here, the coefficients are \(-4\) and \(2\) giving \(-8\) when multiplied. The radicands are \(\sqrt{5}\) and \(\sqrt{5}\), which equal \(5\) when multiplied together since \(\sqrt{5} \times \sqrt{5} = 5\). Thus, the total product becomes \(-8 \times 5 = -40\). Similarly, with the second term \(-4 \sqrt{5} \cdot 4 \sqrt{12}\),

• Multiply \(-4\) by \(4\), giving \(-16\),
• Then \(\sqrt{5} \times \sqrt{12}\) results in \(\sqrt{60}\).

The key takeaway is to multiply separately before combining the results for simplicity and efficiency.

Simplifying Radicals

Simplifying radicals involves breaking down the radicand into its prime factors to find a simpler equivalent expression. For example, with \(\sqrt{60}\):

• First, find the prime factorization of \(60\): \(4 \times 15\).
• Further break these down to \(2^2 \times 3 \times 5\).

Since \(4 = 2^2\), we can take \(2\) out of the radical as \(\sqrt{4} = 2\), leaving us with:\(\sqrt{60} = 2 \sqrt{15}\).This simplification helps reduce the complexity of the expression and makes further multiplication straightforward:\(-16 \times 2 \sqrt{15} = -32 \sqrt{15}\).Simplifying is essentially about finding perfect squares within the radical, making the math easier to handle.

Prime Factorization

Prime factorization is an invaluable tool for simplifying radicals as it breaks down a number into its basic building blocks. When simplifying, we look for perfect squares within these prime factors. For instance:

• Prime factorize \(60\) and get \(2 \times 2 \times 3 \times 5\).
• Identify \(2^2\), a perfect square, to pull out of the radical.

This step is critical as it helps us transform \(\sqrt{60}\) into a simpler form: \(2 \sqrt{15}\), which not only looks neater but also offers a simplified structure that’s often easier to work with. Prime factorization underpins much of radical simplification by offering clarity on how to break down and rebuild the expressions we work with. Understanding and practicing this concept will prove helpful in many areas of algebra.