Problem 24 - Algebra 2

Question

Simplify. $$ \left(y^{5}+32\right)(y+2)^{-1} $$

Step-by-Step Solution

Answer

The expression cannot be simplified further beyond factored forms.

Step 1: Identify the Expression

The given expression is \((y^{5}+32)(y+2)^{-1}\). Our task is to simplify this expression.

Step 2: Recognize the Binomial Denominator

Notice that the denominator \((y+2)\) is raised to the power of -1. This implies that \((y+2)^{-1} = \frac{1}{y+2}\). This indicates division by \(y+2\).

Step 3: Factor the Numerator if Possible

The numerator is \(y^{5} + 32\). Recognize that this expression can be factored using the sum of cubes formula: \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\). Here, \(a = y^{rac{5}{3}}\) and \(b = 2\). So this becomes \((y^2 + 2)(y^4 - y^2 \cdot 2 + 4)\).

Step 4: Cancel Common Factors

Now rewrite the expression using the factored form:\[(y^2 + 2)(y^4 - y^2 \cdot 2 + 4) \cdot (y+2)^{-1}\]. Since \(y^2 + 2\) and \(y+2\) do not factor directly, we cannot simplify further by cancellation.

Step 5: Simplify the Remaining Expression

Since no direct common factors are found between the factored form of the numerator and the denominator, the expression remains:\[(y^2 + 2)(y^4 - y^2 \cdot 2 + 4) \cdot \frac{1}{y+2}\].

Key Concepts

Sum of Cubes

When simplifying algebraic expressions, recognizing patterns such as the sum of cubes is very helpful. The formula for the sum of cubes is:

• Formula: If you have an equation of the form \(a^3 + b^3\), it can be factored as \((a + b)(a^2 - ab + b^2)\).
• Example: To factor \(y^5 + 32\), recognize it as a sum of cubes where \(a = y^{\frac{5}{3}}\) and \(b = 2\).

Using algebra, this becomes \((y^2 + 2)(y^4 - y^2 \cdot 2 + 4)\). Identifying such patterns can make complex expressions much easier to work with.
Understanding the sum of cubes formula is crucial for breaking down and simplifying polynomials that might seem complicated at first glance.

Factoring Polynomials

Factoring is a key process in simplifying polynomials. By breaking down a complex expression into products of simpler terms, you reveal the structure of the polynomial.

• Identify Potential Patterns: Always look for common patterns like sums or differences of cubes, or even just common factors.
• Decomposition: As in our example, decompose \(y^5 + 32\) using the sum of cubes formula, turning it into a product of two simpler factors.

Think of factoring as unraveling a mystery. By rewriting an expression in its simplest form, it becomes easier to further manipulate or solve it, like we tried on \((y^5 + 32)(y+2)^{-1}\). Even when not leading to simplification through cancellation, it's a step towards understanding the expression better.

Algebraic Fractions

Algebraic fractions involve expressions where the numerator and/or the denominator contain variables. This requires us to engage with fraction operations using algebra.

• Denominator Consideration: An expression like \((y+2)^{-1}\) indicates division by \(y+2\). Any simplification must consider the possibility of eliminating this denominator.
• Simplification Attempts: Simplifying algebraic fractions involves checking for common factors in the numerator and the denominator, like trying to simplify \((y^2 + 2)(y^4 - y^2 \cdot 2 + 4)\) over \(y+2\).

Managing algebraic fractions involves methodical checking and careful rewriting, even when straightforward cancellation isn't possible, maintaining the integrity of original expressions.

Binomial Theorem

The binomial theorem provides a way of expanding expressions of the form \((x + y)^n\) into a sum involving terms of the form \(a \cdot x^b \cdot y^c\). While it is different from factoring or simplification processes directly, it offers a helpful structure.

• Combinatorial Insight: The positive integer \(n\) determines how the expansion unfolds, using coefficients derived from combinations.
• Greater Applicability: While we didn't directly apply the binomial theorem in this specific exercise, recognizing its use is pivotal in expanding and then simplifying expressions.

Grasping the binomial theorem empowers one to broaden understanding when shifting from multiplication to addition in complex polynomial expressions.