Problem 68 - Algebra and Trigonometry

Question

31–76 ? Factor the expression completely. $$ y^{3}-y^{2}+y-1 $$

Step-by-Step Solution

Answer

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

Step 1: Group the Terms

The original expression is \( y^3 - y^2 + y - 1 \). We start by grouping the terms: \( (y^3 - y^2) + (y - 1) \). This allows us to address each section separately for factoring.

Step 2: Factor Each Group

Within the grouped terms \( (y^3 - y^2) \), we can factor out a common factor \( y^2 \), giving us \( y^2(y - 1) \). In the second group, \( (y - 1) \), we note that this is already simplified.

Step 3: Factor by Grouping

Now, observe that both groups contain a common term \( (y - 1) \). We can factor this out as follows: \( y^2(y - 1) + 1(y - 1) = (y^2 + 1)(y - 1) \).

Step 4: Check for Further Factorization

The expression \( y^2 + 1 \) cannot be factored further using real numbers since it has no real roots, and \( (y - 1) \) is already a linear factor. Thus, the factorization is complete.

Key Concepts

Grouping Method

The "grouping method" is a strategic way to tackle polynomial expressions that hints at possible factors through intelligent arrangement of their terms. This technique is particularly useful when dealing with polynomials that might not immediately present evident factor pairs.
To apply the grouping method, you start by organizing the terms of the polynomial into groups. Usually, each group will have a certain structure or a common factor that can be factored out.
In our original exercise, the polynomial is grouped as \((y^3 - y^2) + (y - 1)\). By doing this, we set ourselves up to not only unravel common factors but also simplify the polynomial into segments that are easier to handle. This method is a crucial first step when approaching complex polynomials that don't fit simple patterns.

Factoring by Grouping

"Factoring by grouping" involves taking our grouped terms and factoring out common elements within those groups. This is not only an extension of the grouping method but a necessity when aiming for complete factorization of polynomials that can't be simplified by standard techniques.
In the exercise, from the first group \((y^3 - y^2)\), we factor out \( y^2 \), leaving \( y^2(y - 1) \).
In the second group, \((y - 1)\) is already simplified and can't be further factored.
This method highlights any common factor in each group, in this case, \((y-1)\) across both parts.
This commonality allows us to pull out \( y^2 + 1 \) multiplied by \( (y-1) \), paving the way for the expression \( (y^2 + 1)(y - 1) \).

• Factor common elements within each group.
• Look for a term common across groups to factor completely.
• This method creates a polynomial product that simplifies the original expression.

Essentially, factoring by grouping decomposes complex equations into manageable parts, making them easier to analyze and apply further algebraic operations.

Irreducible Polynomials

"Irreducible polynomials" are expressions that cannot be factored into simpler polynomials using the coefficients available in a given number system. In the context of real numbers, a polynomial like \(y^2 + 1\) is considered irreducible since it doesn't have real roots.
When factoring polynomials, identifying irreducible ones is key to understanding when you’ve reached the simplest form of expression.
In the solution, after factoring by grouping, we were left with \( (y^2 + 1)(y - 1) \). While \( (y - 1) \) is a reducible linear factor, \( y^2 + 1 \) does not factor further without moving to the complex number system.
Recognizing irreducible polynomials ensures precision in polynomial factorization; you know when to stop factoring and understand that any further decomposition isn't possible with real coefficients.
Such understanding assists in identifying perfect squares, recognizing complex roots, and predicting polynomial behavior across different number systems.