Problem 32 - Elementary Algebra

Question

Factor completely. $$ 4 y 5+2 y 4-4 y 2-2 y $$

Step-by-Step Solution

Answer

The expression factors to \(2y(y - 1)(y^2 + y + 1)(2y + 1)\).

Step 1: Identify Common Factors

First, we need to identify the common factors in each term of the expression. The expression is \(4y^5 + 2y^4 - 4y^2 - 2y\). All terms contain the common factor \(2y\).

Step 2: Factor Out the Common Factor

Factor \(2y\) out of the entire polynomial to simplify it. This gives us:\[2y(2y^4 + y^3 - 2y - 1)\]

Step 3: Group Terms

Observe the remaining polynomial \(2y^4 + y^3 - 2y - 1\). Group the terms to facilitate factoring by grouping. We group it as:\[(2y^4 + y^3) - (2y + 1)\]

Step 4: Factor by Grouping

Factor out the common factor from each group:- From \(2y^4 + y^3\), factor out \(y^3\), giving \(y^3(2y + 1)\).- From \(-(2y + 1)\), the expression remains the same as \(-(2y + 1)\).Notice that \(2y + 1\) is a common factor in both groups.

Step 5: Factor Common Binomial

Factor out the common binomial \(2y + 1\):\[(y^3 - 1)(2y + 1)\]Now factor the remaining component \(y^3 - 1\), which is a difference of cubes.

Step 6: Apply Difference of Cubes

Recognize \(y^3 - 1\) as a difference of cubes. Use the formula \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\):\(y^3 - 1 = (y - 1)(y^2 + y + 1)\).

Step 7: Write Final Factored Form

Combine all the factors:\[2y(y - 1)(y^2 + y + 1)(2y + 1)\]This is the expression factored completely.

Key Concepts

Understanding Algebraic Expressions

An algebraic expression is a combination of numbers, variables, and operators like addition, subtraction, multiplication, and division. Expressions can also include parentheses and exponents. For example, in the expression \(4y^5 + 2y^4 - 4y^2 - 2y\), each term consists of a coefficient (the number), a variable ("\(y\)"), and an exponent that tells us how many times to multiply the variable by itself.
Algebraic expressions are foundational in algebra because they represent problems or situations mathematically. By manipulating these expressions, we can learn more about the relationships between variables. Often, simplifying them, like through factorization, helps us solve equations more easily.

Exploring Polynomials

Polynomials are a special type of algebraic expression. They consist of terms built from variables raised to whole number powers and coefficients. For instance, \(4y^5 + 2y^4 - 4y^2 - 2y\) is a polynomial. Here, the polynomial is arranged from the highest exponent to the lowest, which is a standard form.
When working with polynomials, the goal is often to simplify or rewrite them in a form that makes calculations easier. This can include factoring, which involves breaking a polynomial down into simpler, multiplicative components. Recognizing the structure and terms of polynomials is key to manipulating them effectively, especially when they need to be factored completely.

The Role of Common Factors

A common factor in algebraic expressions refers to a term or number that appears in every element of the expression. Finding common factors is a crucial first step in factorization, which simplifies expressions. For example, in \(4y^5 + 2y^4 - 4y^2 - 2y\), the common factor is \(2y\).
Once identified, each term of the polynomial is divided by the common factor to reduce the expression. This helps simplify further manipulation and makes finding additional factors easier. Recognizing common factors can dramatically streamline complex algebraic operations by reducing expressions to more manageable forms.

Factoring the Difference of Cubes

In algebra, the difference of cubes is a specific expression format: \(a^3 - b^3\). Its factored form is \((a-b)(a^2 + ab + b^2)\). This is a standard formula used to simplify expressions into products of binomials.
For example, in the polynomial factorization problem, \(y^3 - 1\) is a difference of cubes where \(a = y\) and \(b = 1\). Applying the formula, this becomes \((y-1)(y^2 + y + 1)\). Understanding this formula helps in recognizing and simplifying these specific types of expressions, making it easier to achieve a completely factored form. Using the difference of cubes formula is a vital skill in algebra for dealing with higher-degree polynomials.