Problem 105 - Algebra A Combined Function

Question

Simplify each expression. $$ \frac{\left(2 a^{5} b^{3}\right)^{4}}{-16 a^{20} b^{7}} $$

Step-by-Step Solution

Answer

The simplified expression is \(-b^5\).

Step 1: Apply the Power of a Power Rule

First, let's simplify the numerator by applying the rule \((x^m)^n = x^{m\cdot n}\). For \((2a^5b^3)^4\), each component inside the parentheses is raised to the fourth power, so we have: \(2^4 \cdot (a^5)^4 \cdot (b^3)^4\). This simplifies to \(16a^{20}b^{12}\).

Step 2: Simplify the Fraction

Now, substitute the simplified numerator back into the expression. We get \(\frac{16a^{20}b^{12}}{-16a^{20}b^7}\).

Step 3: Simplify the Coefficients

The coefficients in the fraction are \(16\) and \(-16\). Divide these coefficients \(\frac{16}{-16} = -1\). Include this in the simplified expression: \(-1\cdot \frac{a^{20}b^{12}}{a^{20}b^7}\).

Step 4: Simplify Using the Quotient of Powers Rule for a

For \(a^{20} \), use the quotient rule \((\frac{x^m}{x^n} = x^{m-n})\). Since the powers of \(a\) in the numerator and denominator are the same, \(a^{20-20} = a^0 = 1\).

Step 5: Simplify Using the Quotient of Powers Rule for b

Now, apply the quotient rule to \(b\): \(b^{12-7} = b^5\).

Step 6: Write the Final Simplified Expression

Combine the results from the previous steps. The simplified expression is: \(-1 \cdot 1 \cdot b^5 = -b^5\).

Key Concepts

Power of a Power Rule

The "Power of a Power" rule is a fundamental principle in algebra that assists in simplifying expressions where the same base is raised to multiple exponents. In simpler terms, when you have \[ (x^m)^n = x^{m\cdot n} \]this means that you multiply the exponents together. This rule is particularly useful when dealing with expressions like \[ (2a^5b^3)^4 \]. Here's how it works:

• Firstly, each term inside the parentheses is raised to the power of 4.
• For numbers, like 2, this is straightforward: \[ 2^4 = 16 \].
• For variables, multiply their exponents: \[ (a^5)^4 = a^{5\cdot4} = a^{20} \] and \[ (b^3)^4 = b^{3\cdot4} = b^{12} \].

Thus, the expression \[ (2a^5b^3)^4 \] simplifies to \[ 16a^{20}b^{12} \]. This step makes complex expressions manageable by reducing them to a more straightforward form.

Quotient of Powers Rule

The "Quotient of Powers" rule allows you to simplify expressions that involve dividing exponential terms with the same base. According to this rule, \[ \frac{x^m}{x^n} = x^{m-n} \]. Essentially, you subtract the exponent in the denominator from the exponent in the numerator. This rule is very helpful when simplifying expressions such as \[ \frac{16a^{20}b^{12}}{-16a^{20}b^7} \].

• Firstly, evaluate the coefficients \[ \frac{16}{-16} = -1 \]. Coefficients are ordinary numbers and are treated separately from variables.
• For the variable \[ a \], since \[ a^{20} \] is the same in both the numerator and denominator, it simplifies to \[ a^{20-20} = a^0 \]. Since any number to the power of zero is 1, the \[ a \] terms cancel out.
• For the variable \[ b \], apply the rule: \[ b^{12-7} = b^5 \].

This results in the simplified expression \[ -1 \cdot b^5 = -b^5 \]. The "Quotient of Powers" rule helps in systematically reducing complex fractions.

Simplifying Expressions

Simplifying expressions involves breaking down complex algebraic expressions into their simplest form. This usually entails deploying various algebraic rules, such as the power of a power and the quotient of powers. Let's explore how to effectively simplify expressions like \[ \frac{(2a^5b^3)^4}{-16a^{20}b^7} \].

• Start by simplifying each part of the expression individually. Use the power of a power rule to handle terms inside parentheses.
• Simplify the numerator first to make calculations easier later on.
• Substitute back the simplified form into the original expression.
• Simplify the fraction by dealing with coefficients separately. For instance, \[ \frac{16}{-16} = -1 \].
• Next, handle the variable terms using the quotient of powers rule.

Through these steps, arrive at the final result: \[ -b^5 \]. Simplifying expressions is crucial to solve algebraic equations quickly and efficiently. It reduces errors and makes mathematics more approachable.