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Magazine Chapter 1: Problem 30
Factor the polynomial. $$x^{5}-4 x^{3}+8 x^{2}-32$$
Short Answer
Expert verified
Factorized polynomial: \((x + 2)^2(x - 2)(x^{2} - 2x + 4)\).
Step by step solution
01
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.
02
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.
03
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.
04
Simplifying the Expression
Factor out the common binomial factor \((x^{2} - 4)\): \((x^{3} + 8)(x^{2} - 4)\).
05
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 \((x+2)(x^{2} - 2x + 4)\).
06
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 \(b = 2\).
07
Combining All Factors
Combine all factors obtained: \((x + 2)(x^{2} - 2x + 4)(x - 2)(x + 2)\).
08
Simplifying Repeated Factors
Notice the repeated factor \((x+2)\), resulting in the complete factorization: \((x + 2)^2(x - 2)(x^{2} - 2x + 4)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \(8(x^{2} - 4)\).
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 \((x - 2)(x + 2)\).
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 \((x + 2)(x^{2} - 2x + 4)\).
