Problem 27 - Algebra 2 and Trigonometry

Question

In \(27-39,\) factor each polynomial completely. $$ a^{3}+3 a^{2}-a-3 $$

Step-by-Step Solution

Answer

The factored form of the polynomial is \( (a - 1)(a + 1)(a + 3) \).

Step 1: Group Terms

To factor the polynomial completely, we can start by grouping terms. The polynomial \( a^3 + 3a^2 - a - 3 \) can be grouped as \( (a^3 + 3a^2) + (-a - 3) \).

Step 2: Factor by Grouping

Observe the common factors in each group. In the first group, \( a^3 + 3a^2 \), we can factor out \( a^2 \), giving us \( a^2(a + 3) \). In the second group, \( -a - 3 \), the common factor is \(-1\), so factor it out to get \((-1)(a + 3)\).

Step 3: Combine Factors

Now rewrite the expression as \( a^2(a + 3) - 1(a + 3) \). Notice that \(a+3\) is a common factor, so we can factor that out: \( (a^2 - 1)(a + 3) \).

Step 4: Factor the Difference of Squares

Observe that \(a^2 - 1\) is a difference of squares, which factors as \( (a - 1)(a + 1) \). Therefore, the polynomial \(a^3 + 3a^2 - a - 3\) is factored completely as \( (a - 1)(a + 1)(a + 3) \).

Key Concepts

Grouping Method

The grouping method is a useful technique for factoring polynomials, especially when dealing with expressions that contain four or more terms. The goal is to rearrange or group the terms in such a way that they reveal common factors. This process simplifies complex polynomials into manageable groups, making it easier to factor further.

Let's understand the process by looking at the polynomial given in the original problem: - Start with the polynomial: \(a^3 + 3a^2 - a - 3\). - Group the terms into two pairs: \((a^3 + 3a^2) + (-a - 3)\).

Once grouped, we need to identify any common factors within those groups.

• In the first group \(a^3 + 3a^2\), the common factor is \(a^2\), allowing us to factor it as \(a^2(a + 3)\).
• In the second group \(-a - 3\), we recognize the common factor is \(-1\), leading to the factorization \(-1(a + 3)\).

Recombining these factored groups gives us \(a^2(a + 3) - 1(a + 3)\). Notice how each group includes the term \(a + 3\), which can be factored out as a common factor in the next step.

Difference of Squares

Differences of squares is a special case in polynomial factorization where a binomial is expressed in the form \(a^2 - b^2\). Recognizing such expressions is helpful because they can be factored into \((a - b)(a + b)\). This method simplifies polynomials with difference of squares into a product of two binomials.

Let's consider our factored polynomial from the grouping method: \(a^2 - 1\).

• Notice that \(a^2 - 1\) is a difference of squares since it can be rewritten as \(a^2 - 1^2\).
• This allows us to apply the difference of squares rule: \(a^2 - 1 = (a - 1)(a + 1)\).

Thus, the expression that was partially factored using the grouping method becomes fully factored using the difference of squares. Together, these steps factor the entire polynomial into \((a - 1)(a + 1)(a + 3)\). Understanding and identifying patterns such as differences of squares can significantly simplify the process of polynomial factorization.

Factoring Polynomials

Factoring polynomials is an essential skill in algebra that involves breaking down a polynomial into a product of simpler polynomials. The process not only simplifies complex expressions but also helps solve polynomial equations because it makes finding roots more manageable.

• To factor a polynomial completely, one must explore possible methods such as grouping, looking for common factors, and recognizing special patterns like the difference of squares.
• In the context of the exercise given, performing operations like grouping and difference of squares were essential techniques needed to factor \(a^3 + 3a^2 - a - 3\).

The complete factorization of a polynomial allows us to write it as a product of its simplest components. For the polynomial in our example, this process resulted in the fully factored form \((a - 1)(a + 1)(a + 3)\). Each factor corresponds to a potential solution or root of the polynomial equation; this is crucial for solving mathematical problems involving polynomial equations.

Mastering the process of factoring not only strengthens algebraic skills but also builds a solid foundation for further studies in mathematics. The ability to recognize various factoring methods paves the way for efficiently tackling a wide range of algebraic problems.