Problem 59 - College Algebra and Calculus: An Applied Approach

Question

Completely factor the expression.\(3 x^{3}+x^{2}+15 x+5\)

Step-by-Step Solution

Answer

The factored form of the expression \(3 x^{3}+x^{2}+15 x+5\) is \( (x^2 + 5)(3x + 1) \)

Step 1: Identify the common factor

Analyse the given expression \(3 x^{3}+x^{2}+15 x+5\) and note down the monomials. Then, identify the greatest common factor of those terms, which happens to be none.

Step 2: Factorize by grouping

Next, re-arrange the terms in pairs that have common factors. Grouping allows us to find common binomials. In this case, we can write our expression as \( (3x^3+x^2) + (15x + 5) \) . We then factor each group individually to yield \( x^2(3x + 1) + 5(3x + 1) \)

Step 3: Factorise the resultant expression

The resulting expression has a common binomial \( (3x + 1) \) , so we can factor that binomial out of the entire expression to get \( (x^2 + 5)(3x + 1) \) This is the completely factored form of our original expression.

Key Concepts

Greatest Common Factor

In algebra, the Greatest Common Factor (GCF) of a list of terms is the largest expression that can evenly divide each term. When factoring polynomials, detecting the GCF streamlines the process of simplifying the expression. However, not every polynomial has a common factor greater than one.
For example, if we take a polynomial expression like \(3x^3 + x^2 + 15x + 5\), the first step is to observe each term and search for common factors.
Breaking the given polynomial into individual components:

• First term: \(3x^3\)
• Second term: \(x^2\)
• Third term: \(15x\)
• Fourth term: \(5\)

We check for any common factor among these terms. Despite our efforts, we notice that none of these terms share a factor greater than one.
This indicates that the expression has no GCF, so we move on to explore other methods for factoring.

Factoring by Grouping

Factoring by grouping is a technique particularly handy when dealing with polynomials that have four or more terms. It involves grouping terms into smaller sections that can be factored separately and then applying common factor extraction.
Using our polynomial \(3x^3 + x^2 + 15x + 5\), we split it into two parts: \((3x^3 + x^2)\) and \((15x + 5)\).
This re-arrangement allows us to look at each group individually:

• From \((3x^3 + x^2)\), we pull out \(x^2\) as a common factor, resulting in \(x^2(3x + 1)\).
• From \((15x + 5)\), the common factor is \(5\), giving us \(5(3x + 1)\).

Each group has contributed shares for a factor, leading to a mutual binomial factor \((3x + 1)\) present in both terms. Thus, the expression can be partially rewritten as \(x^2(3x + 1) + 5(3x + 1)\).
This setup primes the expression for the next step, where we'll utilize the common binomial to further simplify the expression.

Binomial Factor

The third stage in polynomial factorization when employing grouping brings us to the concept of the binomial factor. This is where we identify common binomials within our separated groups.
From the expression we achieved earlier through grouping, \(x^2(3x + 1) + 5(3x + 1)\), you can observe that each term retains a common factor of \((3x + 1)\).
Binomial factors are powerful as they allow us to consolidate terms. Here, extracting \((3x + 1)\) from each component results in:

• Factor out \((3x + 1)\)
• Left with \((x^2 + 5)(3x + 1)\)

The polynomial is now completely factored, simplifying it to the equation form \((x^2 + 5)(3x + 1)\).
This method emphasizes the role of identifying common binomial factors for achieving comprehensive factorization. It’s a testament to the effectiveness of thoughtful restructuring and factor extraction in simplifying expressions.