Problem 138 - College Algebra with Modeling and Visualization

Question

Factor the expression completely. \(x^{5}-27 x^{2}\)

Step-by-Step Solution

Answer

The completely factored form is \(x^2(x-3)(x^2 + 3x + 9)\).

Step 1: Identify Common Factors

In the expression \(x^5 - 27x^2\), first look for any common factors in all terms. Both terms have a factor of \(x^2\). Factoring \(x^2\) out gives us \(x^2(x^3 - 27)\).

Step 2: Recognize the Difference of Cubes

Observe the remaining expression \(x^3 - 27\). This can be identified as a difference of cubes, since \(27 = 3^3\). Recall that \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\), where \(a = x\) and \(b = 3\).

Step 3: Apply the Difference of Cubes Formula

Using \(a = x\) and \(b = 3\), substitute into the difference of cubes formula: \(x^3 - 3^3 = (x-3)(x^2 + 3x + 9)\).

Step 4: Combine Factors

Now, substitute back the initial common factor \(x^2\) with our factored form from the cubes, giving us \(x^2(x-3)(x^2 + 3x + 9)\). This is the fully factored form of the original expression.

Key Concepts

Common Factors

When dealing with polynomials, identifying common factors is the first step in the factoring process. Common factors are numbers or variables that can evenly divide every term in the polynomial. Finding these common factors simplifies the expression, making it easier to handle in later steps.
In the expression given, \(x^5 - 27x^2\), both terms, \(x^5\) and \(27x^2\), share a factor of \(x^2\). This common factor is crucial because it can be "factored out," or extracted from each term, streamlining the expression before tackling deeper factorizations.
The factoring process goes as follows:

• Identify the common factor: In our case, \(x^2\).
• Divide each term in the polynomial by this common factor: \(x^5\) becomes \(x^3\) and \(27x^2\) becomes \(27\).
• Rewrite the polynomial with the common factor outside of parentheses: Consequently, \(x^5 - 27x^2\) becomes \(x^2(x^3 - 27)\).

Extracting common factors is a methodical process, creating a cleaner expression to further factorize or solve.

Difference of Cubes

The difference of cubes is a special algebraic form that involves two cubed numbers or expressions subtracted from one another. The formula for factoring a difference of cubes is:
\[a^3 - b^3 = (a-b)(a^2 + ab + b^2)\].
This formula is very handy and is essential for simplifying expressions that exhibit this structure.
In our exercise, the expression left after factoring out the common factor is \(x^3 - 27\). Notice that \(27\) is a perfect cube, as \(27 = 3^3\). This is where the difference of cubes formula comes into play.

• Identify \(a^3\) and \(b^3\): Here, \(a = x\) and \(b = 3\).
• Apply the formula: Substitute \(a\) and \(b\) into the formula:
  \(x^3 - 3^3 = (x-3)(x^2 + 3x + 9)\).

By recognizing and factoring a difference of cubes, you break down the expression into two binomial and trinomial components, making it more manageable.

Factored Form

After identifying and utilizing common factors and special factoring formulas, such as the difference of cubes, the expression is transformed into its factored form. Factored form implies that the polynomial is expressed as a product of its simplest components.
For our example, we began with \(x^5 - 27x^2\). After factoring out \(x^2\) and applying the difference of cubes to \(x^3 - 27\), we reached the full factored state: \(x^2(x-3)(x^2 + 3x + 9)\).
Factored form is highly beneficial for:

• Simplifying expressions, making them easier to solve or analyze.
• Solving polynomial equations, particularly by setting each factor equal to zero.
• Enhancing understanding of the polynomial's structure and roots.

Understanding the factored form allows you to interpret essential characteristics of the polynomial, like its roots and the behavior of its graph.