Problem 44 - Algebra for College Students

Question

Factor each of the following polynomials completely. Indicate any that are not factorable using integers. Don't forget to look first for a common monomial factor. $$2 x^{5}-162 x$$

Step-by-Step Solution

Answer

The polynomial is factored as \(2x(x-3)(x+3)(x^2+9)\).

Step 1: Identify the Common Factor

The first step is to identify any common monomial factor across the terms of the polynomial. Here, both terms, \(2x^5\) and \(-162x\), contain \(2\) and \(x\) as factors. Therefore, the common factor is \(2x\).

Step 2: Factor Out the Common Monomial Factor

Once the common factor \(2x\) is identified, factor it out from both terms of the polynomial. This can be expressed as: \[ 2x(x^4 - 81) \]

Step 3: Recognize the Difference of Squares

The expression inside the parentheses, \(x^4 - 81\), is a difference of squares: \(x^4 = (x^2)^2\) and \(81 = 9^2\). Thus, it can be factored further using the identity \(a^2 - b^2 = (a-b)(a+b)\).

Step 4: Factor the Difference of Squares

Apply the difference of squares to \(x^4 - 81\), which gives:\[ (x^2 - 9)(x^2 + 9) \]Since \(x^2 - 9\) is also a difference of squares (\(x^2 - 9 = (x-3)(x+3)\)), it can be factored further.

Step 5: Further Factor \(x^2 - 9\)

Factor \(x^2 - 9\) as a difference of squares:\[ (x-3)(x+3) \]Thus, the complete expression becomes: \[ 2x(x-3)(x+3)(x^2+9) \]

Step 6: Verify the Factorization

The polynomial is now expressed as \[ 2x(x-3)(x+3)(x^2+9) \]. Check if further factorization is possible. Since \(x^2 + 9\) is not factorable using integers, the factorization process is complete.

Key Concepts

Common Monomial Factor

When starting to factorize a polynomial, the simplest and most effective first step is to find any common monomial factor shared by its terms.
Identifying a common factor can significantly simplify the expression.
For instance, in the polynomial \(2x^5 - 162x\), both terms share a factor of \(2x\).

Finding a common factor involves:

• Looking at each term individually to identify any numbers or variables that appear in all terms.
• Factoring these shared elements out, reducing the degree of each term and simplifying the polynomial's structure.

This is crucial because:

• It reduces complexity by breaking down the original polynomial into simpler parts.
• It often reveals other factorization opportunities, such as spotting differences of squares, which we'll explore next.

By factoring out \(2x\), the polynomial becomes \(2x(x^4 - 81)\), setting the stage for the next steps in factorization.

Difference of Squares

A difference of squares is a specific pattern of polynomial \(a^2 - b^2\), which can be factored using a straightforward identity: \((a-b)(a+b)\). This factorization is based on the difference (subtraction) between two perfect squares.
Recognizing this pattern allows for efficient simplification of certain expressions.

In our example, within \(2x(x^4 - 81)\), notice that \(x^4\) is \((x^2)^2\) and \(81\) is \(9^2\).
The expression fits the difference of squares model.

• Apply the identity: \(x^4 - 81 = (x^2)^2 - 9^2 = (x^2 - 9)(x^2 + 9)\).

Why is this helpful?

• It simplifies the polynomial, making further factorization attainable.
• This pattern commonly appears in algebra, enhancing factorization comprehension.

Thus, the expression becomes \(2x(x^2 - 9)(x^2 + 9)\), leading to even more factorization possibilities.

Complete Factorization

Achieving complete factorization means breaking down a polynomial into a product of simpler polynomial factors until no further factorization is possible using integers.
This final form consists of primes or irreducible polynomials over the integers.
Returning to our example, the expression has now been reduced to \(2x(x^2 - 9)(x^2 + 9)\).

Notice \(x^2 - 9\) is still a difference of squares:

• \(x^2 - 9 = (x-3)(x+3)\), applying the same identity discussed before.

This step refines the polynomial to \(2x(x-3)(x+3)(x^2+9)\).

Why is this form complete?

• \(x^2 + 9\) does not fit any factorization patterns over the integers, remaining unchanged.
• The polynomial is expressed entirely as a product of simpler factors.

Completing this process not only answers the original problem but also reinforces understanding of factorization techniques for more complex algebraic expressions.