How To Simplify Radical And Algebraic Expressions Step By Step Guide

In the realm of mathematics, simplification is a crucial skill. Whether you're dealing with radical expressions or algebraic fractions, the ability to simplify not only makes the expressions easier to work with but also provides a deeper understanding of the underlying mathematical concepts. This article aims to provide a comprehensive, step-by-step guide on simplifying various types of expressions, complete with detailed explanations and examples. We will cover simplifying radical expressions involving square roots and algebraic expressions involving fractions, offering clear methodologies and insights.

A) Simplifying Radical Expressions: $3 \sqrt{216}$

The core concept in simplifying radical expressions is to identify and extract perfect square factors from the radicand (the number inside the square root). This process breaks down complex radicals into simpler forms, making them easier to understand and manipulate. Let's delve into simplifying $3 \sqrt{216}$ by focusing on prime factorization and perfect square extraction.

Prime Factorization

To simplify $3 \sqrt{216}$, we begin by finding the prime factorization of 216. This involves breaking down 216 into a product of its prime factors. Here’s how we do it:

• 216 = 2 × 108
• 108 = 2 × 54
• 54 = 2 × 27
• 27 = 3 × 9
• 9 = 3 × 3

Thus, the prime factorization of 216 is $2 × 2 × 2 × 3 × 3 × 3$, which can be written as $2^3 × 3^3$.

Extracting Perfect Squares

Now that we have the prime factorization, we can rewrite the original expression:

$3 \sqrt{216} = 3 \sqrt{2^3 × 3^3}$

We can further break this down to identify perfect squares:

$3 \sqrt{2^3 × 3^3} = 3 \sqrt{2^2 × 2 × 3^2 × 3}$

Perfect squares are numbers that can be expressed as the square of an integer. In this case, $2^2$ and $3^2$ are perfect squares. We can take these out of the square root:

$3 \sqrt{2^2 × 2 × 3^2 × 3} = 3 × 2 × 3 × \sqrt{2 × 3}$

Simplified Form

Finally, we multiply the numbers outside the square root:

$3 × 2 × 3 × \sqrt{2 × 3} = 18 \sqrt{6}$

Therefore, the simplified form of $3 \sqrt{216}$ is $18 \sqrt{6}$. This process highlights the importance of prime factorization in simplifying radical expressions. By breaking down the radicand into its prime factors, we can easily identify and extract perfect squares, leading to a simpler form of the expression. This method not only simplifies the expression but also provides a clearer understanding of its components, making it easier to work with in further mathematical operations.

B) Simplifying Expressions with Multiple Radicals: $3 \sqrt{80} - 2 \sqrt{54} + 3 \sqrt{125}$

Simplifying expressions involving multiple radicals requires a strategic approach to ensure accurate and efficient manipulation. The primary goal is to simplify each radical term individually and then combine like terms. This involves finding the prime factorization of each radicand, extracting perfect square factors, and then combining terms with the same radical. Let's simplify $3 \sqrt{80} - 2 \sqrt{54} + 3 \sqrt{125}$ step by step.

Simplifying Each Radical Term

We start by simplifying each radical term separately. This involves finding the prime factorization of the numbers inside the square roots and extracting perfect squares.

Simplifying $3 \sqrt{80}$

First, we find the prime factorization of 80:

• 80 = 2 × 40
• 40 = 2 × 20
• 20 = 2 × 10
• 10 = 2 × 5

So, 80 = $2^4 × 5$. Now we rewrite the radical:

$3 \sqrt{80} = 3 \sqrt{2^4 × 5} = 3 \sqrt{(2^2)^2 × 5} = 3 × 2^2 \sqrt{5} = 3 × 4 \sqrt{5} = 12 \sqrt{5}$

Simplifying $-2 \sqrt{54}$

Next, we find the prime factorization of 54:

• 54 = 2 × 27
• 27 = 3 × 9
• 9 = 3 × 3

So, 54 = $2 × 3^3$. Now we rewrite the radical:

$-2 \sqrt{54} = -2 \sqrt{2 × 3^3} = -2 \sqrt{2 × 3^2 × 3} = -2 × 3 \sqrt{2 × 3} = -6 \sqrt{6}$

Simplifying $3 \sqrt{125}$

Now, we find the prime factorization of 125:

• 125 = 5 × 25
• 25 = 5 × 5

So, 125 = $5^3$. We rewrite the radical:

$3 \sqrt{125} = 3 \sqrt{5^3} = 3 \sqrt{5^2 × 5} = 3 × 5 \sqrt{5} = 15 \sqrt{5}$

Combining Like Terms

After simplifying each term, we have:

$12 \sqrt{5} - 6 \sqrt{6} + 15 \sqrt{5}$

Now, we combine the like terms, which are the terms with the same radical part ($\sqrt{5}$):

$(12 \sqrt{5} + 15 \sqrt{5}) - 6 \sqrt{6} = (12 + 15) \sqrt{5} - 6 \sqrt{6} = 27 \sqrt{5} - 6 \sqrt{6}$

Final Simplified Expression

The simplified expression is $27 \sqrt{5} - 6 \sqrt{6}$. This result demonstrates the importance of breaking down each radical term individually before combining them. By identifying and extracting perfect square factors, we can simplify complex expressions and accurately combine like terms. This method provides a clear and systematic approach to handling multiple radicals in a single expression, ensuring a correct and simplified final result.

C) Simplifying Expressions with Distribution: $3 \sqrt{10}(2 \sqrt{2} - 4 \sqrt{10})$

Simplifying expressions that involve distribution with radicals requires a clear understanding of the distributive property and how to manipulate radical terms. The distributive property states that $a(b + c) = ab + ac$, which allows us to multiply a term by a sum or difference. When dealing with radicals, we apply this property while also simplifying the resulting radical terms. Let’s simplify $3 \sqrt{10}(2 \sqrt{2} - 4 \sqrt{10})$ by carefully applying the distributive property and simplifying each resulting term.

Applying the Distributive Property

We begin by distributing $3 \sqrt{10}$ across the terms inside the parentheses:

$3 \sqrt{10}(2 \sqrt{2} - 4 \sqrt{10}) = (3 \sqrt{10} × 2 \sqrt{2}) - (3 \sqrt{10} × 4 \sqrt{10})$

Multiplying the Radical Terms

Next, we multiply the coefficients and the radicals separately:

$(3 \sqrt{10} × 2 \sqrt{2}) - (3 \sqrt{10} × 4 \sqrt{10}) = (3 × 2 × \sqrt{10} × \sqrt{2}) - (3 × 4 × \sqrt{10} × \sqrt{10})$

This simplifies to:

$6 \sqrt{10 × 2} - 12 \sqrt{10 × 10} = 6 \sqrt{20} - 12 \sqrt{100}$

Simplifying the Radicals

Now, we simplify each radical term. We start by finding the prime factorization of the numbers inside the square roots.

Simplifying $6 \sqrt{20}$

The prime factorization of 20 is $2^2 × 5$. Thus,

$6 \sqrt{20} = 6 \sqrt{2^2 × 5} = 6 × 2 \sqrt{5} = 12 \sqrt{5}$

Simplifying $12 \sqrt{100}$

The square root of 100 is 10, so:

$12 \sqrt{100} = 12 × 10 = 120$

Combining the Simplified Terms

After simplifying each term, we have:

$12 \sqrt{5} - 120$

Final Simplified Expression

The fully simplified expression is $12 \sqrt{5} - 120$. This process demonstrates how to effectively use the distributive property with radical expressions. By multiplying the terms and simplifying each resulting radical, we can reduce complex expressions to a simpler form. This method is crucial for handling more complex algebraic problems involving radicals, ensuring clarity and accuracy in the final result.

D) Simplifying Algebraic Fractions: $\frac{2}{(x+2)} + \frac{3}{(x-5)}$

Simplifying algebraic fractions involves combining fractions by finding a common denominator and then simplifying the resulting expression. This process is fundamental in algebra and is used to solve equations and perform other algebraic manipulations. Let’s simplify the expression $\frac{2}{(x+2)} + \frac{3}{(x-5)}$ by first finding a common denominator and then combining the fractions.

Finding a Common Denominator

To add two fractions, they must have a common denominator. The common denominator for $\frac{2}{(x+2)}$ and $\frac{3}{(x-5)}$ is the least common multiple (LCM) of their denominators, which in this case is $(x+2)(x-5)$.

Rewriting the Fractions with the Common Denominator

We rewrite each fraction with the common denominator:

$\frac{2}{(x+2)} = \frac{2(x-5)}{(x+2)(x-5)}$

$\frac{3}{(x-5)} = \frac{3(x+2)}{(x+2)(x-5)}$

Adding the Fractions

Now that the fractions have the same denominator, we can add them:

$\frac{2(x-5)}{(x+2)(x-5)} + \frac{3(x+2)}{(x+2)(x-5)} = \frac{2(x-5) + 3(x+2)}{(x+2)(x-5)}$

Expanding and Simplifying the Numerator

Next, we expand and simplify the numerator:

$2(x-5) + 3(x+2) = 2x - 10 + 3x + 6$

Combine like terms:

$2x - 10 + 3x + 6 = 5x - 4$

So, the expression becomes:

$\frac{5x - 4}{(x+2)(x-5)}$

Expanding the Denominator (Optional)

We can expand the denominator, although it’s often sufficient to leave it in factored form:

$(x+2)(x-5) = x^2 - 5x + 2x - 10 = x^2 - 3x - 10$

So, the expression can also be written as:

$\frac{5x - 4}{x^2 - 3x - 10}$

Final Simplified Expression

The simplified expression is $\frac{5x - 4}{(x+2)(x-5)}$ or $\frac{5x - 4}{x^2 - 3x - 10}$. This process demonstrates the key steps in adding algebraic fractions: finding a common denominator, rewriting the fractions, adding the numerators, and simplifying the resulting expression. By following these steps, we can effectively simplify complex fractions, making them easier to use in further algebraic manipulations and problem-solving.

E) Simplifying Rational Expressions: $\frac{a^2 - 8a + 7}{(a-2)^2} × \frac{a^2 + 4a - 12}{a^2 - 49}$

Simplifying rational expressions involves factoring polynomials, canceling common factors, and reducing the expression to its simplest form. This process is essential in algebra for solving equations, simplifying complex expressions, and understanding the behavior of rational functions. Let’s simplify the expression $\frac{a^2 - 8a + 7}{(a-2)^2} × \frac{a^2 + 4a - 12}{a^2 - 49}$ by factoring each polynomial and canceling common factors.

Factoring the Polynomials

We begin by factoring each polynomial in the expression.

Factoring $a^2 - 8a + 7$

We look for two numbers that multiply to 7 and add to -8. These numbers are -1 and -7. So,

$a^2 - 8a + 7 = (a - 1)(a - 7)$

Factoring $(a-2)^2$

This is already in factored form:

$(a - 2)^2 = (a - 2)(a - 2)$

Factoring $a^2 + 4a - 12$

We look for two numbers that multiply to -12 and add to 4. These numbers are 6 and -2. So,

$a^2 + 4a - 12 = (a + 6)(a - 2)$

Factoring $a^2 - 49$

This is a difference of squares, which factors as:

$a^2 - 49 = (a + 7)(a - 7)$

Rewriting the Expression with Factored Polynomials

Now we rewrite the original expression with the factored polynomials:

$\frac{a^2 - 8a + 7}{(a-2)^2} × \frac{a^2 + 4a - 12}{a^2 - 49} = \frac{(a - 1)(a - 7)}{(a - 2)(a - 2)} × \frac{(a + 6)(a - 2)}{(a + 7)(a - 7)}$

Canceling Common Factors

Next, we cancel common factors in the numerator and the denominator:

$\frac{(a - 1)(a - 7)}{(a - 2)(a - 2)} × \frac{(a + 6)(a - 2)}{(a + 7)(a - 7)} = \frac{(a - 1) \cancel{(a - 7)}}{(a - 2) \cancel{(a - 2)}} × \frac{(a + 6) \cancel{(a - 2)}}{(a + 7) \cancel{(a - 7)}}$

This simplifies to:

$\frac{(a - 1)(a + 6)}{(a - 2)(a + 7)}$

Final Simplified Expression

The simplified expression is $\frac{(a - 1)(a + 6)}{(a - 2)(a + 7)}$. We can leave the expression in this factored form or expand the numerator and the denominator if needed.

This process illustrates the critical steps in simplifying rational expressions: factoring the polynomials, rewriting the expression with the factored forms, and canceling common factors. By mastering these steps, you can effectively simplify complex rational expressions, which is vital for more advanced algebraic manipulations and problem-solving. This method ensures clarity and accuracy in reducing expressions to their simplest form.

In summary, simplifying mathematical expressions, whether radical or algebraic, involves a systematic approach. For radical expressions, it’s about identifying and extracting perfect square factors. For algebraic fractions, it’s about finding common denominators and combining terms. By mastering these techniques, you can handle complex mathematical problems with greater ease and confidence. The ability to simplify expressions is not just a skill but a foundational tool for success in mathematics and related fields. Each type of expression requires a specific approach, but the underlying principle remains the same: break down the complex into simpler components. By doing so, we not only arrive at the simplified form but also gain a deeper understanding of the mathematical structures involved. This comprehensive guide has provided the methodologies and insights necessary to tackle a wide range of simplification problems, empowering you to excel in your mathematical endeavors.