Problem 60 - Algebra 2

Question

Which of the following is equivalent to \((2+3 \sqrt{5})(3+3 \sqrt{5}) ?\) $$\begin{array}{llll}{\text { A. } 51} & {\text { B. } 6+9 \sqrt{5}} & {\text { C. } 6+24 \sqrt{5}} & {\text { D. } 51+15 \sqrt{5}}\end{array}$$

Step-by-Step Solution

Answer

The equivalent of the given expression is: \(51 + 15 \sqrt{5}\) or option D.

Step 1 Identification

The given expression, (2+3√5)(3+3√5), is a multiplication of two binomials. Binomials are algebraic expressions with two terms.

Step 2 Application of FOIL Method

In order to multiply these binomials, apply the FOIL method (First, Outer, Inner, Last). Multiplying the 'First' terms (2*3) gives 6. Multiplying the 'Outer' terms (2*3√5) gives 6√5. Multiplying the 'Inner' terms (3√5*3) gives 9√5. Multiplying the 'Last' terms (3√5*3√5) gives 45.

Step 3 Simplification

Simplify the expression obtained from the FOIL method, 6 + 6√5 + 9√5 + 45. Combine like terms to finalize the expression 51 + 15√5.

Key Concepts

FOIL Method

The FOIL Method is a fundamental technique used to multiply two binomials, which are algebraic expressions containing two terms each. It's called FOIL because it stands for:

• First - Multiply the first terms of each binomial.
• Outer - Multiply the outer terms of the binomial pair.
• Inner - Multiply the inner terms.
• Last - Multiply the last terms of each binomial.

Let's apply the FOIL method with an example expression: \((2 + 3\sqrt{5})(3 + 3\sqrt{5})\). Here’s how it's done:

• First: Multiply 2 by 3 to get 6.
• Outer: Multiply 2 by \(3\sqrt{5}\) to get \(6\sqrt{5}\).
• Inner: Multiply \(3\sqrt{5}\) by 3 to get \(9\sqrt{5}\).
• Last: Multiply \(3\sqrt{5}\) by \(3\sqrt{5}\) to get 45, as \((\sqrt{5})^2 = 5\).

Ultimately, the FOIL method reveals all the pairwise products within a product of two binomials, providing structure and clarity in processing algebraic multiplications.

Simplification of Radicals

Simplifying expressions often involves dealing with radicals, especially in algebra. Radicals are root expressions, most commonly square roots, that may appear in algebraic operations. When simplifying products with radicals, recognizing like terms is crucial; terms such as \(6\sqrt{5}\) and \(9\sqrt{5}\) are considered like because the radical part, \(\sqrt{5}\), is the same.

To simplify, you:

• Add or subtract coefficients—the numbers in front of the radicals—while leaving the radical part unchanged. For instance, \(6\sqrt{5} + 9\sqrt{5}\) simplifies to \(15\sqrt{5}\).
• Simplify any nub radical expressions if possible. This might involve extracting square roots or rationalizing denominators.

In our specific context, after applying the FOIL method, we have the expression \(6 + 6\sqrt{5} + 9\sqrt{5} + 45\). Here, you combine like terms: \(6\sqrt{5} + 9\sqrt{5} = 15\sqrt{5}\), while adding the constant terms: \(6 + 45 = 51\). As a result, the simplified expression is \(51 + 15\sqrt{5}\). This method streamlines complicated algebraic expressions, showcasing their simplest form.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operators. Variables, often denoted by symbols like \(x\) or \(y\), represent numbers or quantities that can change, while operators include symbols for addition, subtraction, multiplication, and division.

A binomial, a specific type of algebraic expression, contains two terms and is central in this problem. For example, in the expression \((2 + 3\sqrt{5})(3 + 3\sqrt{5})\), each part in parentheses is a binomial.

• The 2 and 3 in the expression are constants.
• The \(3\sqrt{5}\) represents a constant multiplied by a radical.

Algebraic expressions can be manipulated through operations like factorization, expansion, and simplification. These processes depend on standard algebraic rules and properties.

Working with algebraic expressions involves understanding the relationships and interactions between their components—crucial for tasks like multiplying binomials using the FOIL Method or simplifying radical expressions. Practice with these operations strengthens your ability to manipulate such expressions, making complex arithmetic easier to handle.