Problem 63 - Algebra A Combined Function

Question

Multiply. Then simplify if possible. Assume that all variables represent positive real numbers. $$ (\sqrt{6}-4 \sqrt{2})(3 \sqrt{6}+1) $$

Step-by-Step Solution

Answer

The expression simplifies to \(18 + \sqrt{6} - 24\sqrt{3} - 4\sqrt{2}\).

Step 1: Recognize the Expression Type

The given expression \((\sqrt{6} - 4\sqrt{2})(3\sqrt{6} + 1)\)is a product of two binomials. The goal is to expand this using the distributive property (also known as FOIL method for binomials).

Step 2: Apply the FOIL Method

Use the FOIL method to multiply the two binomials: First, Outer, Inner, Last.1. **First:**\(\sqrt{6} \times 3\sqrt{6} = 3 \times (\sqrt{6})^2 = 3 \times 6 = 18\)2. **Outer:**\(\sqrt{6} \times 1 = \sqrt{6}\)3. **Inner:**\(-4\sqrt{2} \times 3\sqrt{6} = -12 \times \sqrt{2} \times \sqrt{6} = -12 \times \sqrt{12}\)4. **Last:**\(-4\sqrt{2} \times 1 = -4\sqrt{2}\)

Step 3: Simplify the Product of Radicals

Focus on simplifying the expression from Step 2, especially the product of radicals:- Recall that \(\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}\).- Thus, the Inner term simplifies as \(-12 \times 2 \sqrt{3} = -24 \sqrt{3}\).

Step 4: Combine Like Terms

Write all terms together and combine any like terms:\(18 + \sqrt{6} - 24\sqrt{3} - 4\sqrt{2}\)There are no like terms to combine further since each term has a different radical.

Step 5: Conclude the Simplification

The fully simplified form of the expanded expression is:\(18 + \sqrt{6} - 24\sqrt{3} - 4\sqrt{2}\).There are no further possibilities for simplification as all terms are already in their simplest form.

Key Concepts

FOIL Method

The FOIL method is a technique used to expand the multiplication of two binomials. It is an acronym that stands for First, Outer, Inner, Last, representing the order in which you multiply the terms of the binomials.
In this method:

• 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 in each binomial.

Applying FOIL to the binomials \((\sqrt{6} - 4\sqrt{2})(3\sqrt{6} + 1)\), the product is expanded by multiplying each pair of terms defined by these four groups, one step at a time.
This results in a complete and detailed expansion of the inner product, making sure every combination is covered.

Distributive Property

The distributive property is a fundamental concept in algebra that dictates how multiplication interacts with addition or subtraction.
It states that multiplying a sum by a number is the same as multiplying each addend by that number, and then adding the products. In formal terms, for any numbers a, b, and c, the distributive property is expressed as:

• \(a(b + c) = ab + ac\)

This property also plays a crucial role in simplifying radical expressions. When you distribute radicals, you are essentially applying this property to multiply out each term in an expression by the radicals present.
In our example, the distributive property is employed in the FOIL method, ensuring that each term in one binomial is multiplied by every term in the other. This maximizes accuracy and efficiency in finding the final expanded expression.

Radicals

Radicals, often expressed as square roots, are symbols used to represent the root of a number. They are significant in simplifying expressions and solving equations involving quadratic and higher powers.
Key aspects of working with radicals include understanding how to multiply them and simplifying them when possible. When multiplying radical expressions such as \(\sqrt{a} \times \sqrt{b}\), it gives \(\sqrt{ab}\).
In more complex expressions, like when radicals are squared ( as in the term \((\sqrt{6})^2\) ), they simplify back to their base values (i.e., \(6\) in this case).
When simplifying radicals, such as \(\sqrt{12}\), you can break them into smaller parts (i.e., \(\sqrt{4} \times \sqrt{3}\)) and simplify where possible. \(\sqrt{4}\) becomes \(2\), making the expression \(2\sqrt{3}\).
These simplifications play a crucial role in ensuring the final expression is as simple as possible.