Problem 75 - Intermediate Algebra

Question

Simplify each expression. If an expression cannot be simplified, write "Does not simplify." $$ \frac{m^{3}-m n^{2}}{m n^{2}+m^{2} n-2 m^{3}} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{m-n}{n-2m} \).

Step 1: Factor the Numerator

First, let's take a look at the numerator of the expression: \( m^3 - mn^2 \).Notice that both terms share a common factor, \(m\). Factoring out \(m\), we get:\[ m(m^2 - n^2) \].

Step 2: Recognize a Difference of Squares

The expression \(m^2 - n^2\) in the numerator can be further factored using the difference of squares formula, \(a^2 - b^2 = (a - b)(a + b)\).Applying this formula, we factor \(m^2 - n^2\) as:\[ m(m-n)(m+n) \].

Step 3: Factor the Denominator

Now, let's factor the denominator: \( mn^2 + m^2n - 2m^3 \).Start by factoring out the common factor, \(m\):\[ m(n^2 + mn - 2m^2) \].

Step 4: Factor the Quadratic in the Denominator

Next, focus on factoring the quadratic expression \(n^2 + mn - 2m^2\). We need two numbers that multiply to \(-2m^2\) and add up to \(m\). The numbers \(2m\) and \(-m\) satisfy these requirements:\[ n^2 + mn - 2m^2 = (n-2m)(n+m) \].

Step 5: Substitute Factored Expressions Back

Substitute the factored expressions back into the original expression:\[ \frac{m(m-n)(m+n)}{m(n-2m)(n+m)} \].

Step 6: Cancel Common Factors

With the expression \( \frac{m(m-n)(m+n)}{m(n-2m)(n+m)} \), we can cancel the common \(m\) from both the numerator and the denominator, and also \((m+n)\) which is common in both:\[ \frac{(m-n)}{(n-2m)} \].

Key Concepts

Factoring Expressions

Factoring an expression means breaking it down into simpler components, known as factors, that when multiplied together give you the original expression. Understanding how to factor expressions is crucial in simplifying algebraic expressions and solving equations.
When factoring expressions, keep an eye out for:

• Common factors: Always look for the largest factor that is present in all the terms of the expression.
• Special patterns: Recognize patterns such as the difference of squares, perfect square trinomials, or sum/difference of cubes.
• Factoring quadratics and other polynomials: This often involves finding two numbers that multiply into the constant term and add up to the coefficient of the middle term, especially for quadratic expressions.

By breaking down an expression into its factors, simplification becomes more manageable and also reveals deeper insights into the structure and components of the expression itself.

Difference of Squares

A difference of squares is a specific type of algebraic expression, structured as the subtraction of one squared term from another. It takes the general form:\[ a^2 - b^2 = (a - b)(a + b) \]Recognizing this pattern can significantly simplify expressions.
When looking at an expression:

• Identify pairs of terms that are perfect squares.
• Apply the difference of squares formula to factor the expression into two binomials.
• This formula reduces a seemingly complex expression into a product of simpler expressions.

For example, in the expression \( m^2 - n^2 \), both \( m^2 \) and \( n^2 \) are perfect squares. Therefore, you can express the difference as \( (m-n)(m+n) \). Recognizing and applying this technique is a powerful tool in an algebra student's toolkit.

Common Factor

The first step often taken in simplifying algebraic expressions is identifying and extracting the common factor across all terms. The largest common factor is what you look for, and factoring it out simplifies the remaining terms.
Here's how to do it:

• Scan all the terms to find the largest variable or number that is present in all the terms.
• Rewrite each term as a product of this common factor and another factor, then factor out the common part.
• This step can reduce complex expressions into simpler ones, making further manipulations easier.

For example, in the expression \( m^3 - mn^2 \), the common factor is \( m \). Factoring \( m \) out gives you \( m(m^2 - n^2) \). Finding and factoring out the common factor simplifies expressions and is often the first step towards solving more complicated algebraic equations.