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Ch. P - Fundamental Concepts of Algebra
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 1, Problem 86

In Exercises 65–92, factor completely, or state that the polynomial is prime. x2−10x+25−36y2

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Identify the structure of the polynomial: The given expression is \( x^2 - 10x + 25 - 36y^2 \). Notice that \( x^2 - 10x + 25 \) is a quadratic trinomial, and \( 36y^2 \) is a perfect square.
Factor \( x^2 - 10x + 25 \): Recognize that \( x^2 - 10x + 25 \) is a perfect square trinomial. It can be factored as \( (x - 5)^2 \), since \( (x - 5)(x - 5) = x^2 - 10x + 25 \).
Rewrite the expression: Substitute \( (x - 5)^2 \) for \( x^2 - 10x + 25 \). The expression becomes \( (x - 5)^2 - 36y^2 \).
Recognize the difference of squares: The expression \( (x - 5)^2 - 36y^2 \) is a difference of squares, which can be factored using the formula \( a^2 - b^2 = (a - b)(a + b) \). Here, \( a = (x - 5) \) and \( b = 6y \).
Factor the difference of squares: Apply the formula to factor \( (x - 5)^2 - 36y^2 \) as \( ((x - 5) - 6y)((x - 5) + 6y) \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Factoring Polynomials

Factoring polynomials involves rewriting a polynomial expression as a product of simpler polynomials. This process is essential for simplifying expressions and solving equations. Common techniques include identifying common factors, using the difference of squares, and applying the quadratic formula when necessary.
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Difference of Squares

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