Problem 30 - Precalculus: Functions and Graphs

Question

Factor the polynomial. $$x^{5}-4 x^{3}+8 x^{2}-32$$

Step-by-Step Solution

Answer

Factorized polynomial: \((x + 2)^2(x - 2)(x^{2} - 2x + 4)\).

Step 1: Identifying Common Elements

We start by observing the polynomial \(x^{5} - 4x^{3} + 8x^{2} - 32\). Notice that this polynomial doesn't have a common factor in all terms. However, grouping can make factoring easier.

Step 2: Factoring by Grouping

Group the terms to facilitate factoring: \((x^{5} - 4x^{3}) + (8x^{2} - 32)\). Each group has a common factor. In the first group, \(x^{3}\) is common, and in the second, \(8\) is common.

Step 3: Factoring Each Group

Factor \(x^{3}\) out of the first group and \(8\) out of the second: \(x^{3}(x^{2} - 4) + 8(x^{2} - 4)\). Notice that \((x^{2} - 4)\) is a common binomial factor.

Step 4: Simplifying the Expression

Factor out the common binomial factor \((x^{2} - 4)\): \((x^{3} + 8)(x^{2} - 4)\).

Step 5: Recognizing Further Factoring Possibilities

The term \(x^{2} - 4\) is a difference of squares, \( (x - 2)(x + 2) \). Additionally, observe \(x^{3} + 8\) as a sum of cubes which factors as \((x+2)(x^{2} - 2x + 4)\).

Step 6: Applying the Sum of Cubes Formula

Use the sum of cubes factorization formula, \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\), to factor \(x^{3} + 8 = (x + 2)(x^2 - 2x + 4)\). Here \(a = x\) and \(b = 2\).

Step 7: Combining All Factors

Combine all factors obtained: \((x + 2)(x^{2} - 2x + 4)(x - 2)(x + 2)\).

Step 8: Simplifying Repeated Factors

Notice the repeated factor \((x+2)\), resulting in the complete factorization: \((x + 2)^2(x - 2)(x^{2} - 2x + 4)\).

Key Concepts

Factoring by Grouping

Factoring by grouping is a technique used to factor polynomials when no common factor exists for all terms. Here, the goal is to rearrange terms into groups, each with a common factor. For example, consider the polynomial \(x^{5} - 4x^{3} + 8x^{2} - 32\). We group it as \((x^{5} - 4x^{3}) + (8x^{2} - 32)\).

• In the first group, \(x^{3}\) is the common factor, giving us \(x^{3}(x^{2} - 4)\).
• In the second group, the number \(8\) can be factored out, resulting in \(8(x^{2} - 4)\).

After factoring each group, if you notice a common binomial factor, like \((x^{2} - 4)\) in this case, you can factor it out to rewrite the expression as \((x^{3} + 8)(x^{2} - 4)\). This method simplifies complex expressions by breaking them down into manageable parts.

Difference of Squares

The difference of squares is a special factoring technique used when a polynomial has the form \(a^{2} - b^{2}\). Its factorization is given by \((a - b)(a + b)\). This can greatly simplify expressions. For the polynomial \(x^{5} - 4x^{3} + 8x^{2} - 32\), after factoring by grouping, one of the factors becomes \(x^{2} - 4\). Recognizing this as a difference of squares:

• Let \(a = x\) and \(b = 2\) in the formula.
• Thus, \(x^{2} - 4\) can be factored into \((x - 2)(x + 2)\).

This approach is efficient for simplifying polynomials where two squared terms are separated by a subtraction sign. Understanding the difference of squares is crucial for factorization in algebra.

Sum of Cubes

The sum of cubes involves factoring expressions of the form, \(a^3 + b^3\). The factorization formula used here is \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). This formula is essential for breaking down cubic terms. In our example, \(x^{3} + 8\) is a sum of cubes, because:

• We can express \(8\) as \(2^3\), so \(x^{3} + 8 = x^{3} + 2^{3}\).
• Applying the sum of cubes formula with \(a = x\) and \(b = 2\), the expression becomes \((x + 2)(x^{2} - 2x + 4)\).

This technique simplifies the polynomial into factorable terms, yielding results that are easier to manage. Mastery of this concept empowers problem-solving skills in polynomial factorization.