Problem 44 - Algebra A Combined Function

Question

Rewrite each rational expression as an equivalent rational expression with the given denominator. $$ \frac{5 x}{x^{3}+2 x^{2}-3 x}=\frac{\underline{\phantom{xx}}}{x(x-1)(x-5)(x+3)} $$

Step-by-Step Solution

Answer

The equivalent expression is \(\frac{5x(x-5)}{x(x-1)(x-5)(x+3)}\).

Step 1: Factor the Original Denominator

First, we need to factor the denominator of the original fraction \(x^3 + 2x^2 - 3x\). Begin by factoring out the common factor \(x\):\[x(x^2 + 2x - 3)\]Next, factor the quadratic \(x^2 + 2x - 3\). We look for two numbers whose product is \(-3\) and sum is \(2\); these numbers are \(3\) and \(-1\):\[x(x + 3)(x - 1)\]So, the original denominator is factored as \(x(x + 3)(x - 1)\).

Step 2: Identify Missing Factors

Compare the factored original denominator \(x(x + 3)(x - 1)\) with the given denominator \(x(x-1)(x-5)(x+3)\). The difference lies in the factor \((x-5)\), which is missing from the original denominator.

Step 3: Express Original Expression with the New Denominator

To make the denominators equivalent, multiply both the numerator and the denominator of the original expression \(\frac{5x}{x(x + 3)(x - 1)}\) by the missing factor \((x-5)\):\[\frac{5x(x-5)}{x(x-1)(x+3)(x-5)}\]Now, the denominator matches the given denominator exactly.

Step 4: Write the Final Equivalent Expression

Combine the steps above to rewrite the original rational expression with the provided denominator. The equivalent expression is:\[\frac{5x(x-5)}{x(x-1)(x-5)(x+3)}\]

Key Concepts

Factoring Quadratics

Factoring quadratics is a crucial skill when working with rational expressions. A quadratic expression takes the general form of \(ax^2 + bx + c\). The objective is to express this quadratic as a product of two binomials.
To factor a quadratic expression like \(x^2 + 2x - 3\), you follow a simple method. Look for two numbers that multiply to give the constant term, \(-3\), while their sum gives the linear coefficient, \(2\).

• These numbers are \(3\) and \(-1\).

Using these numbers, the quadratic is rewritten as the product of binomials: \((x + 3)(x - 1)\).
This process helps simplify polynomial expressions and is essential for working with rational expressions.

Common Denominator

When adding or subtracting rational expressions, having a common denominator is key. A rational expression has the form \(\frac{P}{Q}\), where \(P\) and \(Q\) are polynomials.
To combine expressions with different denominators, you adjust them to share a common one.

• This process can be achieved by identifying and multiplying by any missing factors.

For the exercise at hand, compare factored forms to identify missing elements.
The original denominator \(x(x + 3)(x - 1)\) needs to be adjusted to match the denominator \(x(x-1)(x-5)(x+3)\).
Realizing \((x-5)\) is the missing factor, you multiply both the numerator and denominator by \((x-5)\) to achieve commonality.

Polynomial Factorization

Polynomial factorization allows splitting polynomials into simpler parts or factors that are multiplied together. This is particularly useful for understanding and manipulating rational expressions.
In our exercise, we began by factoring \(x^3 + 2x^2 - 3x\).

• Start with factoring out the greatest common factor, here it is \(x\).
• This gives us \(x(x^2 + 2x - 3)\).

Next, factor the remaining quadratic \(x^2 + 2x - 3\) as described earlier. This results in \(x(x+3)(x-1)\).
Factorization simplifies complex expressions, making them easier to work with and solve.

Equivalence in Algebra

Equivalence in algebra means that two expressions may look different but are actually the same. This is possible because of how expressions can be manipulated. Using operations like factorization and finding common denominators, we can transform expressions into equivalent forms.

• The exercise demonstrated equivalence by rewriting the rational expression with a new, common denominator.

To simplify or solve equations with rational expressions, recognizing equivalences is vital.
It ensures the expressions maintain their value no matter how they are manipulated. Equivalence gives confidence that two forms of an expression behave identically in all scenarios.