Problem 34 - Algebra for College Students

Question

For Problems 9-50, simplify each rational expression. \(\frac{9 y^{2}-1}{3 y^{2}+11 y-4}\)

Step-by-Step Solution

Answer

The simplified expression is \(\frac{3y + 1}{y + 4}\), except when undefined for \(y = \frac{1}{3}\).

Step 1: Identify Common Patterns

First, examine the numerator and the denominator to identify any common patterns like difference of squares or trinomials that can be factored. The numerator is a difference of squares: \(9y^2 - 1 = (3y - 1)(3y + 1)\).

Step 2: Factor the Denominator

Look at the denominator, \(3y^2 + 11y - 4\), and attempt to factor it. We need factors of \(3 \cdot (-4) = -12\) that add up to \(11\). The correct pair is \(12\) and \(-1\). Thus, we rewrite and factor: \(3y^2 + 12y - y - 4 = (3y^2 + 12y) + (-y - 4)\) which simplifies to \((3y(y + 4) - 1(y + 4))\) and ultimately to \((3y - 1)(y + 4)\).

Step 3: Simplify the Fraction

Now that both the numerator \((3y - 1)(3y + 1)\) and denominator \((3y - 1)(y + 4)\) have been factored, cancel out the common factor \((3y - 1)\). This simplifies the expression to \((3y + 1) / (y + 4)\).

Step 4: Verify Simplification

Ensure that the simplified rational expression does not change the domain. Avoid cancelling terms that cause division by zero. Here, cancelling \(3y - 1\) is valid unless \(3y - 1 = 0\), thus excluded \(y = \frac{1}{3}\), which was already an exclusion in the original expression.

Key Concepts

Factoring Trinomials

Factoring trinomials is a fundamental technique used to simplify expressions, including rational expressions. When you look at a trinomial, a common method to factor it is to find two numbers that multiply to give the product of the leading coefficient and the constant term, while also adding up to the middle term.

Consider the denominator from the exercise: \(3y^2 + 11y - 4\). The product of 3 and -4 is -12. We need two numbers that multiply to -12 and add to 11. The numbers 12 and -1 satisfy these conditions.

• Write the middle term using these two numbers: \(3y^2 + 12y - y - 4\).
• Next, group the terms: \((3y^2 + 12y) + (-y - 4)\).
• Factor out the common factors from each group: \(3y(y + 4) - 1(y + 4)\).
• The expression can now be factored fully as: \((3y - 1)(y + 4)\).

Factoring trinomials helps in rewriting the expressions so that simplifications, like cancellations, are easier.

Difference of Squares

The difference of squares is a special case of factorization where an expression can be rewritten as the product of two binomials. An expression is a difference of squares if it is in the form of \(a^2 - b^2\), which can be factored to \((a - b)(a + b)\).

In the given exercise, the numerator \(9y^2 - 1\) can be classified as a difference of squares. Here, \(9y^2\) is the square of \(3y\), and \(1\) is the square of \(1\). This fits the pattern:

• \(9y^2 - 1 = (3y)^2 - (1)^2\)
• Thus, it factors to \((3y - 1)(3y + 1)\).

Recognizing these patterns allows you to factor expressions quickly, facilitating easier simplification of rational expressions.

Common Factor Cancellation

Once you have factored both the numerator and the denominator of a rational expression, the next step is to simplify it by cancelling out common factors.

In rational expressions, common factor cancellation helps in reducing the expression to its simplest form. However, one should always ensure that cancelling these factors does not introduce any excluded values or change the domain of the original expression.

• For example, consider the factored expression from our problem: \((3y - 1)(3y + 1)/(3y - 1)(y + 4)\).
• Here, \((3y - 1)\) is a common factor in both the numerator and the denominator.
• By canceling \((3y - 1)\), the expression simplifies to \((3y + 1)/(y + 4)\).

This reduction is valid as long as you exclude the values that make the original denominator zero, like \(y = \frac{1}{3}\) for the term \((3y - 1)\). Ensuring that cancellation does not affect the domain of the expression is crucial in maintaining its validity.