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Magazine Chapter 6: Problem 88
Factor completely. $$ z^{12}-1 \quad[5.6] $$
Short Answer
Expert verified
\(z^{12} - 1 = (z - 1)(z^2 + z + 1)(z + 1)(z^2 - z + 1)(z^6 + 1)\)
Step by step solution
01
- Identify the form of the expression
Notice that the given expression can be written as a difference of squares: \[ z^{12} - 1 = (z^{6})^2 - 1^2 \]
02
- Factor using the difference of squares
Apply the difference of squares formula:\[ a^2 - b^2 = (a - b)(a + b) \]Here, let \(a = z^6\) and \(b = 1\):\[ (z^6 - 1)(z^6 + 1) \]
03
- Factor \(z^6 - 1\) using the difference of squares
Factor \(z^6 - 1\):\[ z^6 - 1 = (z^3)^2 - 1^2 = (z^3 - 1)(z^3 + 1) \]
04
- Factor \(z^3 - 1\) using the difference of cubes
Factor \(z^3 - 1\) using the difference of cubes formula:\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]Here, let \(a = z\) and \(b = 1\):\[ z^3 - 1 = (z - 1)(z^2 + z + 1) \]
05
- Factor \(z^3 + 1\) using the sum of cubes
Factor \(z^3 + 1\) using the sum of cubes formula:\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]Here, let \(a = z\) and \(b = 1\):\[ z^3 + 1 = (z + 1)(z^2 - z + 1) \]
06
- Combine all factors
Combine all factors:\[ z^{12} - 1 = (z^6 - 1)(z^6 + 1) = [(z^3 - 1)(z^3 + 1)](z^6 + 1) = (z - 1)(z^2 + z + 1)(z + 1)(z^2 - z + 1)(z^6 + 1) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Difference of Squares
In algebra, the difference of squares is a useful factorization technique. It applies when you have an expression of the form \(a^2 - b^2\). This formula states that the difference of squares can be factored into:
- \((a - b)(a + b)\)
Difference of Cubes
When working with polynomials, the difference of cubes formula is another essential tool. It applies to expressions of the form \(a^3 - b^3\). To factor such an expression, use:
- \((a - b)(a^2 + ab + b^2)\)
Sum of Cubes
The sum of cubes is closely related to the difference of cubes. When you have an expression like \(a^3 + b^3\), you can factor it using:
- \((a + b)(a^2 - ab + b^2)\)
Intermediate Algebra
Intermediate Algebra serves as a bridge between basic algebra and more advanced topics like calculus. In this context, it covers a variety of factorization techniques, including the ones we've discussed: difference of squares, difference of cubes, and sum of cubes.Important concepts in Intermediate Algebra:
- Recognizing Patterns: Spotting squares and cubes within expressions makes it easy to apply specific formulas.
- Using Formulas Efficiently: Knowing when to use each formula (like \(a^2 - b^2\), \(a^3 - b^3\), and \(a^3 + b^3\)) helps simplify complex expressions.
- Combining Simplified Factors: Once you factorize parts of the expressions, you recombine them to form the fully factored expression.
