Problem 34 - College Algebra and Calculus: An Applied Approach

Question

Write the rational expression in simplest form.\(\frac{y^{3}-2 y^{2}-8 y}{y^{3}+8}\)

Step-by-Step Solution

Answer

The simplified form of the given rational expression is \(\frac{y*(y-4)}{y^{2}-2y+4}\)

Step 1: Factorize the Numerator

Firstly, factorize the numerator expression \(y^{3}-2 y^{2}-8 y\). The y is common in each term, so take y as common factor. It will simplify to: \(y*(y^{2}-2y-8)\). Now we can further factorize \(y^{2}-2y-8\) by using factorization method. It will factorize to: \( (y-4)*(y+2)\). So numerator will become: \(y*(y-4)*(y+2)\).

Step 2: Factorize the Denominator

Secondly, the expression of denominator \(y^{3}+8\) is of type \(y^{3}+a^{3}\), so we can use the standard identity for cubes sum. Hence, it will factorize to: \((y+2)*(y^{2}-2y+4)\)

Step 3: Cancel out the Common Factor

Thirdly, \(y+2\) is a common factor in both the numerator and denominator, therefore cancel out this common factor. The simplified form will become \(\frac{y*(y-4)}{y^{2}-2y+4}\)

Step 4: Final Simplified form

Finally, Simplify the remaining expression for final form that will be: \(\frac{y*(y-4)}{y^{2}-2y+4}\)

Key Concepts

Factoring Polynomials

Factoring polynomials involves breaking down a complex polynomial into simpler terms that can multiply to create the original polynomial. In this exercise, the first step was to factor the numerator, which is the expression \(y^{3} - 2y^{2} - 8y\). Here’s how it's done:

• Identify common factors in the terms. In this polynomial, each term features the variable \(y\), making it the common factor. By factoring \(y\) out, the remaining expression becomes \(y(y^{2} - 2y - 8)\).
• Next, factor \(y^{2} - 2y - 8\). The simplest way is to find two numbers that multiply to \(-8\) and add up to \(-2\). Here, the numbers \(-4\) and \(2\) work perfectly. Therefore, further factorizing gives \((y - 4)(y + 2)\).

Combining it all, the numerator \(y^{3} - 2y^{2} - 8y\) is factored into \(y(y - 4)(y + 2)\). Factoring is an important skill as it often simplifies complex algebraic expressions.

Simplifying Expressions

Simplifying expressions is all about making a mathematical statement simpler and easier to work with. In our case, after factoring both the numerator and the denominator, the goal was to cancel out any common factors.

• In the equation \(\frac{y(y-4)(y+2)}{(y+2)(y^{2}-2y+4)}\), notice that \(y + 2\) is a common factor in both the numerator and denominator.
• Canceling out \(y + 2\) simplifies the expression to \(\frac{y(y-4)}{y^{2}-2y+4}\).

The process of simplifying by canceling common factors helps reduce the complexity of a rational expression, making further calculations or manipulations easier and more direct.

Algebraic Identities

Algebraic identities are expressions that hold true for all values of the variables involved. They are handy in quickly simplifying complex algebraic expressions without a lot of tedious calculations.

• In the original exercise, factoring the denominator \(y^{3} + 8\), we recognize it as a sum of cubes formula, \(a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})\).
• Here, \(a\) is \(y\) and \(b\) is \(2\). Applying this identity, \(y^{3} + 8\) rewrites to \((y + 2)(y^{2} - 2y + 4)\).

Using algebraic identities makes the factorization process much more efficient and saves the effort of trial-and-error when dealing with complex expressions.

Polynomial Long Division

Polynomial long division is a technique used to divide polynomials, similar to long division with numbers. Although not used directly in the initial solution, it's a useful approach when dealing with non-factorable expressions.

• Imagine dividing \(y^{3} - 2y^{2} - 8y\) by \(y+2\) if factoring didn't work out as neatly. This process involves dividing the leading terms of the dividend (numerator) and the divisor (denominator) first, multiplying, subtracting, and repeating.
• Step by step, you adjust the dividend by subtracting to line up new terms until the degree of the remainder is less than the degree of the divisor.
• The result would be a quotient with potentially a remainder that expresses any non-divisible portion beyond common factors.

Using polynomial long division offers a systematic approach to explore and further simplify expressions that resist simple factorization techniques.