Problem 23 - 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. $$2 x^{4}-162$$

Step-by-Step Solution

Answer

The factorized form of the polynomial \(2 x^{4}-162\) is \(2(x - 3)(x + 3)(x^{2} + 9)\).

Step 1: Identifying highest common factor

The first stage in factorizing any polynomial is to identify the Greatest Common Factor (GCF). Here in this case, the GCF of \(2 x^{4}\) and \(-162\) is \(2\). Hence, factor out 2 from the polynomial which would give \(2(x^{4}-81)\).

Step 2: Using the Difference of Squares

The polynomial \(x^{4}-81\) is a difference of squares since \(x^{4}\) can be represented as \((x^{2})^{2}\) and \(81\) as \(9^{2}\). The difference of two squares can be factorized as \((a-b)(a+b)\), where \((a = x^{2}, b = 9)\). So, \(x^{4}-81\) can be factored as \((x^{2} - 9)(x^{2} + 9)\). Thus the polynomial after this step becomes \(2(x^{2} - 9)(x^{2} + 9)\).

Step 3: Further factorization

Further factorizing the polynomial \(x^{2} -9\) using the difference of squares as \(x^{2}\) can be represented as \(x^{2}\) and \(9\) as \(3^{2}\), we find two factors \(x-3\) and \(x+3\). Thus, \(x^{2} - 9\) can be expressed as \((x - 3)(x + 3)\). So the polynomial can now be factorized as \(2(x - 3)(x + 3)(x^{2} + 9)\).

Key Concepts

Greatest Common Factor (GCF)

Understanding the Greatest Common Factor (GCF) is critical when attempting to factorize polynomials. In this context, the GCF represents the largest expression that will divide the terms of the polynomial without leaving a remainder. Finding the GCF is the first step in the factorization process because it simplifies the polynomial and makes the subsequent factorization steps more manageable.

Take, for instance, the polynomial \(2x^{4}-162\). The GCF here is \(2\), as both terms of the polynomial are divisible by this number. By extracting the GCF, we rewrite the polynomial as \(2(x^{4}-81)\), significantly reducing the complexity of the remaining expression that needs to be factorized. Remember, overlooking the GCF can make the task of factorization more complex and could even lead to incorrect solutions. Always look for the GCF first!

Difference of Squares

The Difference of Squares is a useful pattern to recognize in algebra when factorizing polynomials. This classic formula states that if you have a square subtracted from another square, you can factor it as the product of the sum and difference of the two square roots. Mathematically, it is expressed as \(a^2 - b^2 = (a - b)(a + b)\).

In our given problem, after factoring out the GCF, we are left with the polynomial \(x^{4} - 81\), which is a perfect example of the Difference of Squares. This expression can be seen as \(x^{2})^{2} - 9^{2}\), allowing us to apply the formula. Consequently, we obtain two factors: \(x^{2} - 9\) and \(x^{2} + 9\). The ability to identify this pattern and apply the formula will enable students to break down more complex polynomials into manageable factors.

Factorizing Polynomials Algebra

To further understand factorizing polynomials in algebra, one must see it as a systematic process of breaking down a polynomial into the product of its simplest parts, or 'factors'. Once we have utilized the GCF and any identifiable patterns such as the Difference of Squares, as seen in our problem, we then continue to factorize until no further factors can be found.

In our case, the difference of squares pattern occurred again with the term \(x^{2} - 9\), which was factorized to \(x - 3\) and \(x + 3\). However, it's important to note that the term \(x^{2} + 9\) does not factor further since it is not a difference of squares and does not have a GCF other than 1. The complete factorization of the original polynomial \(2x^{4} - 162\) becomes \(2(x - 3)(x + 3)(x^{2} + 9)\).

Remember, consistent practice in identifying patterns and employing factorization techniques is essential for gaining proficiency in factorizing polynomials algebra.