Problem 41 - Prealgebra and Introductory Algebra

Question

Multiply. $$\frac{x^{2}+5 x+4}{x^{3} y^{2}} \cdot \frac{x^{2} y^{3}}{x^{2}+2 x+1}$$

Step-by-Step Solution

Answer

The final answer to the expression after simplifying is \[\frac{y}{x + \frac{5}{x} + \frac{4}{x^{2}}}\]

Step 1: Multiply the Numerators and Denominators

Multiply the numerators together, and the denominators together. This results in: \[\frac{(x^{2} + 5x + 4) * (x^{2} y^{3})}{(x^{3} y^{2})*(x^{2} + 2x + 1)}\]

Step 2: Distribute the Multiplication in the Numerator

Apply the distribution property in the numerator. With this, we get: \[\frac{x^{4}y^{3} + 5x^{3}y^{3} + 4x^{2}y^{3}}{x^{5}y^{2} + 2x^{4}y^{2} + x^{3}y^{2}}\]

Step 3: Simplifying the Powers

Each term in the numerator and denominator has a common base of \(x\). Therefore, we can simplify by subtracting the exponents. This gives us: \[\frac{x^{4-5}y^{3-2} + 5x^{3-5}y^{3-2} + 4x^{2-5}y^{3-2}}{1 + 2x^{4-5} + x^{3-5}} = \frac{x^{-1}y + 5x^{-2}y + 4x^{-3}y}{1 + 2x^{-1} + x^{-2}}\]

Step 4: Bring negative powers into the denominator

As per the rules of exponents, a negative power in the numerator moves to the denominator and becomes a positive power. With this, we get: \[\frac{y}{x + 5 + 4/x} = \frac{y}{x + \frac{5}{x} + \frac{4}{x^{2}}}\]

Key Concepts

Multiplying Fractions

When working with fractions, multiplying them involves a straightforward process. To multiply fractions, you multiply the numerators together and the denominators together. For example, in the given exercise, we have two fractions: \ \( \frac{x^{2}+5x+4}{x^{3} y^{2}} \) and \( \frac{x^{2} y^{3}}{x^{2}+2x+1} \).
Multiply the numerators, giving us \((x^{2}+5x+4) \,*\, (x^{2} y^{3})\) and the denominators, \((x^{3} y^{2}) \,*\, (x^{2}+2x+1)\).
This results in:\

• Numerator: \( (x^{2} + 5x + 4) \times (x^{2} y^{3}) \)
• Denominator: \( (x^{3} y^{2}) \times (x^{2} + 2x + 1) \)

It's important to keep track of all the variables and powers during multiplication. Make sure to distribute each term properly according to the rules of algebra.

Simplifying Exponents

Simplifying expressions with exponents can make complex algebraic problems more manageable. In this task, each term in the multiplied expression has exponents that need simplification. For example, after distributing the multiplication in the numerator: \

• The expression becomes \( x^{4}y^{3} + 5x^{3}y^{3} + 4x^{2}y^{3} \) in the numerator.
• The denominator becomes \( x^{5}y^{2} + 2x^{4}y^{2} + x^{3}y^{2} \).

To simplify exponents in such terms, use the property of exponents which states that when dividing like bases, subtract the exponents. \For example:\- \( x^{a}/x^{b} = x^{a-b} \) - For bases like \( y \), \( y^{c}/y^{d} = y^{c-d} \)

• Apply this for your variables, and simplify by subtraction: \[ \frac{x^{4-5}y^{3-2} + 5x^{3-5}y^{3-2} + 4x^{2-5}y^{3-2}}{1 + 2x^{4-5} + x^{3-5}} \]

Pay attention to the powers, as they directly influence the simplification process.

Negative Exponents

Negative exponents can often seem intimidating, but they follow simple rules. They indicate the reciprocal or inverse of the base raised to the absolute value of the exponent. Consider this equation from the final stage of the solution:\

• \( x^{-a} = \frac{1}{x^{a}} \)

In the example given, simplifying led to negative exponents like \( x^{-1}, x^{-2}, \) and \( x^{-3} \). According to the rule, \( x^{-1} \to \frac{1}{x} \), \( x^{-2} \to \frac{1}{x^2} \), and so on.

• Moving negative powers from the numerator to the denominator changes their signs, simplifying expressions. Thus,\[ \frac{y}{x + \frac{5}{x} + \frac{4}{x^{2}}} \]

This step not only simplifies the expression but also reinforces your understanding of how to handle negative exponents.