Problem 44 - Elementary and Intermediate Algebra

Question

Simplify each rational expression. See Example 4 $$\frac{m^{3}-m n^{2}}{m n^{2}+m^{2} n-2 m^{3}}$$

Step-by-Step Solution

Answer

The rational expression simplifies to \(\frac{m-n}{n-2m}\).

Step 1: Identify Common Terms

Identify and factor out the common term in the numerator of the rational expression. The numerator is \(m^3 - mn^2\). Both terms have a common factor of \(m\). So we factor \(m\) out, resulting in \(m(m^2 - n^2)\).

Step 2: Factor the Denominator

Next, we'll tackle the denominator \(mn^2 + m^2 n - 2m^3\). We notice that all the terms are divisible by \(m\), so we can factor \(m\) out, which gives us \(m(n^2 + mn - 2m^2)\).

Step 3: Simplify the Expression

Now substitute the factored forms from Steps 1 and 2 back into the original rational expression. We have \(\frac{m(m^2 - n^2)}{m(n^2 + mn - 2m^2)}\). Simplifying this expression by canceling out the common factor \(m\) in the numerator and denominator results in \(\frac{m^2 - n^2}{n^2 + mn - 2m^2}\).

Step 4: Further Factor and Simplify

Now, we try to factor \(m^2 - n^2\) in the numerator. This is a difference of squares which factors to \((m-n)(m+n)\). Look at the denominator \(n^2 + mn - 2m^2\). This can be factored by finding two numbers that add to \(mn\) and multiply to \(-2m^2\). After factoring, we get \((n-2m)(n+m)\).

Step 5: Final Cancellation

The expression now is \(\frac{(m-n)(m+n)}{(n-2m)(n+m)}\). Spot the common factor in the numerator and denominator. The term \((n+m)\) appears in both, so cancel them out, leaving \(\frac{m-n}{n-2m}\) as the final simplified expression.

Key Concepts

Factoring

Factoring is a critical strategy in simplifying rational expressions. It involves breaking down expressions into products of simpler factors. Think of it like converting a sum or difference of numbers into a product form. For instance, if you have an expression like \(m^3 - mn^2\), look for common factors among the terms. Here, you can factor out \(m\), resulting in \(m(m^2 - n^2)\). This step simplifies the expression and makes it easier to evaluate or further simplify.

When factoring, always:

• Identify common factors among the terms.
• Look for patterns like the difference of squares or the trinomial structure.

Factorization requires practice, so don't hesitate to practice different types of expressions to get the hang of it.

Simplification

Simplification is the process of making an expression easier to work with. When dealing with rational expressions, simplifying often involves reducing the expression to its simplest form by canceling out common terms or factors in the numerator and denominator. After factoring the parts of the expression, substitute them back into the original and look for identical terms that can be canceled out.

As seen in the problem solution, after factoring, the expression becomes \(\frac{m(m^2 - n^2)}{m(n^2 + mn - 2m^2)}\). The factor \(m\) appears in both the numerator and denominator, making it possible to simplify further to \(\frac{m^2 - n^2}{n^2 + mn - 2m^2}\).

To simplify correctly:

• Always factor first – that’s how you find what can be canceled.
• Cancel out common factors present in both parts of the fraction.
• Check your steps to ensure nothing essential is lost or incorrect.

Difference of Squares

The difference of squares is a special factoring technique applied to expressions of the form \(a^2 - b^2\). Such expressions can be factored into \((a-b)(a+b)\). This strategy is used in the solution to handle the term \(m^2 - n^2\), converting it to \((m-n)(m+n)\).

Understanding the difference of squares is important because it simplifies calculations and reveals hidden patterns in algebraic expressions. This method is broadly applicable not only in algebra but also in solving equations, and it's valuable for simplifying rational expressions.
Here's how you can handle difference of squares:

• Recognize the pattern \(a^2 - b^2\).
• Apply the formula to factor it into \((a-b)(a+b)\).
• Use this technique in tandem with other factoring methods to simplify complex expressions.

By mastering the difference of squares, you make algebra easier and build confidence to tackle more advanced problems.