In Exercises 1–68, factor completely, or state that the polynomial is prime.
x⁸ − y⁸
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1
Recognize that the expression \(x^8 - y^8\) is a difference of two powers.
Recall the formula for factoring a difference of squares: \(a^2 - b^2 = (a - b)(a + b)\).
Notice that \(x^8 - y^8\) can be rewritten as \((x^4)^2 - (y^4)^2\), which is a difference of squares.
Apply the difference of squares formula: \((x^4 - y^4)(x^4 + y^4)\).
Factor \(x^4 - y^4\) further using the difference of squares: \((x^2 - y^2)(x^2 + y^2)\).
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Difference of Squares
The difference of squares is a fundamental algebraic identity stating that a² - b² can be factored into (a - b)(a + b). This concept is crucial for factoring polynomials that can be expressed as the difference between two perfect squares, which is applicable in the given polynomial x⁸ - y⁸.
Solving Quadratic Equations by Completing the Square
Factoring Polynomials
Factoring polynomials involves rewriting a polynomial as a product of simpler polynomials or factors. Understanding how to identify common factors, apply special identities, and break down higher-degree polynomials is essential for completely factoring expressions like x⁸ - y⁸.
A prime polynomial is one that cannot be factored into the product of two non-constant polynomials with real coefficients. Recognizing when a polynomial is prime is important in algebra, as it determines whether further factorization is possible, which is relevant when analyzing the polynomial x⁸ - y⁸.