Problem 25 - Algebra 2

Question

Simplify. Assume that no variable equals 0. $$ \frac{2 c^{3} d\left(3 c^{2} d^{5}\right)}{30 c^{4} d^{2}} $$

Step-by-Step Solution

Answer

\(\frac{1}{5}cd^4\)

Step 1: Distribute the Terms in the Numerator

First, distribute the monomials in the parentheses in the numerator. Multiply \(2c^3d\) by \(3c^2d^5\):\[2c^3d \times 3c^2d^5 = 6c^{3+2}d^{1+5} = 6c^5d^6.\]

Step 2: Write the Simplified Expression

Now, rewrite the problem with the simplified numerator:\[\frac{6c^5d^6}{30c^4d^2}.\]

Step 3: Simplify the Coefficients

Simplify the coefficients \(\frac{6}{30}\):\[\frac{6}{30} = \frac{1}{5}.\]

Step 4: Simplify the Variables

Apply the property of exponents \(\frac{a^m}{a^n} = a^{m-n}\) to each variable:- For \(c\):\[\frac{c^5}{c^4} = c^{5-4} = c^1 = c,\]- For \(d\):\[\frac{d^6}{d^2} = d^{6-2} = d^4.\]

Step 5: Write the Final Simplified Expression

Combine the simplified coefficients and variables to get the final answer:\[\frac{1}{5}cd^4.\]

Key Concepts

Exponents

Exponents are a way to express repeated multiplication of the same number or variable. For example, when you see something like \(c^5\), it means \(c\) is multiplied by itself five times (\(c \times c \times c \times c \times c\)). Exponents have rules called power rules or laws of exponents, which help in simplifying expressions. One of the core rules that we use often is

• The product of powers: \(a^m \times a^n = a^{m+n}\).
• The quotient of powers: \(\frac{a^m}{a^n} = a^{m-n}\).

In the problem, we used the product of powers to combine exponents when multiplying terms in the numerator. This helped us simplify the expression by reducing the complexity of the exponents in the variables \(c\) and \(d\). Thus, knowing how to handle exponents efficiently is crucial when simplifying algebraic expressions.

Coefficients

Coefficients are the numerical parts of a term in an algebraic expression. In the expression \(2c^3d\), the number 2 is the coefficient. Coefficients appear in algebra as the 'multipliers' in a term, dictating how many times a particular variable or product of variables appears.
When simplifying coefficients that are in a fraction, we apply normal division rules. For instance, simplifying \(\frac{6}{30}\) involves dividing both the numerator and the denominator by their greatest common divisor, which is 6 in this case.
After simplification, we get the coefficient \(\frac{1}{5}\). This is the simplified form of our original numerical part in the given expression. Understanding coefficients is important because they help us quantify the expression accurately, enabling us to handle complex algebraic terms more effectively.

Fraction Simplification

Fraction Simplification in algebra involves reducing a fraction to its simplest form. This is often done by canceling out common factors from the numerator and the denominator.
For example, in the problem, the term \(\frac{6}{30}\) was simplified to \(\frac{1}{5}\). To do this, identify the greatest common divisor (GCD) of the two numbers (6 and 30), which is 6, and divide both by this number.

• Numerator: \(\frac{6}{6} = 1\)
• Denominator: \(\frac{30}{6} = 5\)

This simplification makes the fraction much easier to work with in further operations. Simplifying fractions is crucial because it reduces unnecessary complexity and makes calculations more manageable, especially in larger algebraic expressions.

Variables

Variables are symbols that represent unknown numbers or values in an equation or expression, often denoted by letters such as \(c\) or \(d\). They allow us to formulate general solutions applicable to many specific cases.
In this exercise, we deal with variables holding exponents like \(c^5\) and \(d^6\). Simplifying these often involves using exponent rules where we combine or reduce similar bases.
Understanding variables is not just about recognizing that they stand for unknowns, but also being able to treat them as known quantities when applying algebraic rules. Tools like the laws of exponents help us manipulate variables effectively, allowing us to solve more complicated problems by breaking them into simpler components.