Problem 152 - College Algebra with Modeling and Visualization

Question

Factor the expression completely. \(2 x^{5}-8 x^{3}-16 x^{2}+64\)

Step-by-Step Solution

Answer

The expression factors to \(2(x - 2)^2(x + 2)(x^2 + 2x + 4)\).

Step 1: Identify the Greatest Common Factor (GCF)

First, identify the greatest common factor (GCF) of all the terms in the expression. The expression is \(2x^5 - 8x^3 - 16x^2 + 64\). The GCF is 2 because each coefficient can be divided by 2. Factoring out 2, we get \(2(x^5 - 4x^3 - 8x^2 + 32)\).

Step 2: Group Terms

Next, group the terms inside the parenthesis in pairs. The expression inside is \(x^5 - 4x^3 - 8x^2 + 32\). Group them as \((x^5 - 4x^3)\) and \((-8x^2 + 32)\).

Step 3: Factor Each Group

Factor out the common factors in each group. From \(x^5 - 4x^3\), factor out \(x^3\) to get \(x^3(x^2 - 4)\). From \(-8x^2 + 32\), factor out \(-8\) to get \(-8(x^2 - 4)\).

Step 4: Factor by Grouping

Notice \(x^2 - 4\) is a common factor in both groups. So, factor out \(x^2 - 4\) to get: \((x^3 - 8)(x^2 - 4)\).

Step 5: Recognize and Apply Special Factoring Formulas

Identify \(x^3 - 8\) as a difference of cubes and \(x^2 - 4\) as a difference of squares. For the difference of cubes, \(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\). For the difference of squares, \(x^2 - 4 = (x - 2)(x + 2)\). Substitute these back into the expression to factor completely.

Step 6: Write the Final Factored Form

Combine all the factored terms along with the GCF. The final factorization of the expression \(2x^5 - 8x^3 - 16x^2 + 64\) is \(2(x - 2)^2(x + 2)(x^2 + 2x + 4)\).

Key Concepts

Greatest Common Factor

The greatest common factor (GCF) is the largest number that can divide each term in an expression without leaving a remainder. In the expression \(2x^5 - 8x^3 - 16x^2 + 64\), the GCF is 2. Here's why it's important:

• It simplifies expressions by reducing their degrees, making them easier to handle.
• Factoring out the GCF is often the first step in breaking down more complex polynomials.

In this case, dividing each term by 2 gives the factored expression \(2(x^5 - 4x^3 - 8x^2 + 32)\). Finding the GCF serves as a useful tool in both simplifying polynomials and preparing them for other factoring processes.

Difference of Cubes

The difference of cubes formula is an essential tool when factoring polynomials. It is used when an expression can be written as \(a^3 - b^3\). The formula for this expression is \((a - b)(a^2 + ab + b^2)\).

In our exercise, we identified \(x^3 - 8\) as a difference of cubes with \(x^3\) and \(8 = 2^3\). Applying the formula gives us \((x - 2)(x^2 + 2x + 4)\). This method helps to fully factor expressions where the direct factoring is not apparent.

When employing the difference of cubes:

• Recognize your terms as cubes, which might require rewriting numbers in exponential form.
• Apply the formula accordingly, ensuring that no terms are overlooked.

Factoring using the difference of cubes allows for simplification of polynomials for further analysis or solution.

Difference of Squares

The difference of squares is a special factoring technique used for expressions of the form \(a^2 - b^2\). The formula is \((a - b)(a + b)\). This method is straightforward yet powerful for reducing quadratic expressions.

In our exercise, the term \(x^2 - 4\) is identified as a difference of squares, where \(4 = 2^2\). Applying the formula leads to \((x - 2)(x + 2)\). It simplifies the expression, making it more manageable.

Some key characteristics of difference of squares include:

• The expression involves two perfect squares subtracted from one another.
• The result of factoring is two binomials, each with opposite signs.

Utilizing the difference of squares in factoring makes solving and simplifying polynomials more efficient, supporting further mathematical exploration.