Problem 29 - Precalculus: Functions and Graphs

Question

Simplify. $$\left(5 x^{2} y^{-3}\right)\left(4 x^{-5} y^{4}\right)$$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{20y}{x^3} \).

Step 1: Apply the Distributive Property

To simplify the expression \( (5x^2y^{-3})(4x^{-5}y^4) \), begin by multiplying the numerical coefficients together. This gives us \( 5 \times 4 = 20 \).

Step 2: Multiply the Powers of \(x\)

Next, apply the property of exponents that states \( x^a \times x^b = x^{a+b} \). For the \( x \) terms, we have \( x^2 \times x^{-5} = x^{2 + (-5)} = x^{-3} \).

Step 3: Multiply the Powers of \(y\)

Use the property \( y^a \times y^b = y^{a+b} \) for the \( y \) terms. Thus, \( y^{-3} \times y^4 = y^{-3 + 4} = y^1 = y \).

Step 4: Combine the Results

Combine the results from previous steps: \( 20 \), \( x^{-3} \), and \( y \). The simplified expression is \( 20x^{-3}y \).

Step 5: Finalize the Simplified Expression with Positive Exponents

To express the answer with positive exponents only, move \( x^{-3} \) to the denominator: The final expression is \( \frac{20y}{x^3} \).

Key Concepts

Distributive Property

The Distributive Property is a fundamental algebraic concept that allows you to multiply a sum by multiplying each addend individually and then summing the results. However, in this exercise, we deal with products. We use the distributive nature of multiplication over individual terms such as coefficients and variable bases.
When you see an expression like \( (5x^2y^{-3})(4x^{-5}y^4) \), you multiply the numerical coefficients separately and then the variable parts separately. This approach ensures that every part of the expressions is accounted for in the product, following the rule that multiplication can be distributed over addition or amongst multiple factors.
In this example, the coefficients 5 and 4 from each group were multiplied together first, simplifying to 20. Then, powers of identical bases (like powers of \( x \) and \( y \)) were addressed using the related properties of exponents. This application reflects the essence of the distributive property within the multiplication structure of algebraic expressions.

Exponents

Exponents are used to express repeated multiplication of a number by itself. For instance, \( x^2 \) means \( x \) multiplied by itself, or \( x \times x \). When simplifying expressions with exponents, like \( (5x^2y^{-3})(4x^{-5}y^4) \,\) it is crucial to apply the rules governing how to manage these powers.
The key rule applied here is that when multiplying like bases, you simply add their exponents. This leads to steps such as \( x^2 \times x^{-5} = x^{2 + (-5)} = x^{-3} \). Similarly, multiply the terms with the base \( y \), \( y^{-3} \times y^4 = y^{1} \), following the same exponent rule.
Negative exponents like \( y^{-3} \) negatively impact the position of the term in a fraction format, hinting at a division positioning rather than multiplication, which we address when finalizing expressions to ensure positive exponents.

Multiplying Powers

When multiplying powers of the same base, the rule is straightforward: keep the base and add the exponents together. This principle is crucial in simplifying or multiplying expressions with variables raised to powers.
In an expression such as \( (5x^2y^{-3})(4x^{-5}y^4) \,\) you encounter two powers with bases \( x \): \( x^2 \) and \( x^{-5} \). By the multiplication of powers rule: \( x^2 \times x^{-5} \) results in \( x^{-3} \). The same process applies to terms with the base \( y \).
This method of combining exponents simplifies complex algebraic operations, allowing us to reduce expressions into more manageable forms. Understanding and using this rule makes it easier to handle expressions in equations.

Simplifying with Positive Exponents

Expressions with negative exponents can often confuse students when they appear in simplified results. However, these can be easily managed by converting them into positive exponents. Positive exponents establish a base in its simplest standard form.
In the final step of simplifying our example \( 20x^{-3}y \,\) you'll notice \( x^{-3} \). To convert this to safe-keeping with positive exponents, we utilize the property that \( x^{-a} = \frac{1}{x^a} \). Thus, \( x^{-3} \) becomes \( \frac{1}{x^3} \). This results in the final expression \( \frac{20y}{x^3} \).
The conversion repositions powers with a negative exponent from the numerator to the denominator of a fraction, ensuring all exponents carry a positive sign, making expressions clearer and often required in final answers or solutions.