Problem 25 - Introductory Algebra for College Students

Question

Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations using multiplication or a graphing utility. $$x^{3}+2 x^{2}-9 x-18$$

Step-by-Step Solution

Answer

The polynomial \(x^{3}+2 x^{2}-9 x-18\) is not prime and can be completely factored as \((x + 2)(x - 3)(x + 3)\).

Step 1: Search for Common Factors

No common factor, other than 1, is found among all the terms in the polynomial \(x^{3}+2 x^{2}-9 x-18\). So, it can't be factored out to simplify the polynomial.

Step 2: Apply Factorization Techniques

To utilize factorization techniques, we group the terms in such a way that we can apply the rule (a + b)(a − b) = a² - b²: \((x^{3}+2x^{2}) - (9x + 18)\). Then factor out an \(x^2\) from the first group and a 9 from the second group: \(x^2(x + 2) - 9(x + 2)\). Notice now that we have an expression of form a(x + b) - c(x + b). We can factor out (x + 2) from each group, which gets: \((x + 2)(x^2 - 9)\).

Step 3: Continue Factorization Until Prime Polynomials Are Found

The term \(x^2 - 9\) in \((x + 2)(x^2 - 9)\) is the difference of squares, and it can be factored into \((x - 3)(x + 3)\). So, the complete factorization is \((x + 2)(x - 3)(x + 3)\).

Key Concepts

Difference of Squares

The difference of squares technique is a powerful method for factoring specific types of polynomials. It applies when you have a binomial in the form of \(a^2 - b^2\). The special thing about this form is that it can be rewritten as \((a - b)(a + b)\). This factorization is possible because the middle terms cancel out when multiplied, leaving you with the original binomial.Let's look at an example. If you have \(x^2 - 9\), you can recognize this as a difference of squares: \(x^2\) is \(a^2\) and 9 is \(b^2\) where \(b = 3\). Therefore, this can be factored into \((x - 3)(x + 3)\). The ability to recognize and apply the difference of squares formula is crucial for breaking down complex polynomials into simpler, more manageable pieces.

Common Factors

Finding common factors is often the first step in simplifying polynomials. A common factor is a number, variable, or term that divides each part of the polynomial without a remainder.In the polynomial \(x^3 + 2x^2 - 9x - 18\), we start by identifying if there is a common factor in all terms. Here, we check each term:

• \(x^3\)
• \(2x^2\)
• \(-9x\)
• \(-18\)

It turns out there is no common factor other than 1. Despite not finding a factor in this example, it's a critical check that can simplify your work significantly and prevent unnecessary steps in other problems. Sometimes, the simplest techniques lead to the biggest breakthroughs!

Grouping Method

The grouping method is an effective technique for factoring polynomials that don't seem to factor using simpler methods like factoring out a common term or applying the difference of squares directly.In the example \(x^3 + 2x^2 - 9x - 18\), we face a polynomial with four terms. By grouping strategically — \((x^3 + 2x^2)\) and \((-9x - 18)\) — we set ourselves up for successful factoring.Within each group, look for common factors. From \(x^3 + 2x^2\), factor out \(x^2\), and from \(-9x - 18\), factor out \(-9\). This yields \(x^2(x + 2) - 9(x + 2)\). Notice that \((x + 2)\) appears in both parts. You can then factor \((x + 2)\) out, resulting in \((x + 2)(x^2 - 9)\).Grouping simplifies the expression and often reveals hidden factorization opportunities, making it a smart choice when dealing with polynomials that don't have apparent common factors or straightforward factor forms.