Problem 149
Question
Factor the expression completely. \((z-2)^{2}-9\)
Step-by-Step Solution
Verified Answer
The expression \((z-2)^2 - 9\) factors to \((z - 5)(z + 1)\).
1Step 1: Identify the Expression
The expression given is \((z-2)^2 - 9\). First, recognize this as a difference of squares, where \(a^2 - b^2 = (a-b)(a+b)\). In this case, \(a = (z-2)\) and \(b = 3\).
2Step 2: Apply the Difference of Squares Formula
Use the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\) to factor the expression. Substitute \(a = (z-2)\) and \(b = 3\):\[(z-2)^2 - 9 = ((z-2) - 3)((z-2) + 3)\].
3Step 3: Simplify the Factors
Simplify the factors obtained from applying the difference of squares formula:1. \((z-2) - 3 = z - 2 - 3 = z - 5\)2. \((z-2) + 3 = z - 2 + 3 = z + 1\)Thus, the expression \((z-2)^2 - 9\) factors to \((z - 5)(z + 1)\).
Key Concepts
Difference of SquaresPolynomial FactoringCollege Algebra
Difference of Squares
The difference of squares is a special type of mathematical expression. It is written in the form \(a^2 - b^2\). This expression can be factored into \((a - b)(a + b)\). In the exercise provided, we see the expression \((z-2)^2 - 9\). Here, \(a = (z-2)\) and \(b = 3\). The process of recognizing and factoring expressions using the difference of squares makes algebra less complex.
- First, identify terms that can be squared: look for a subtraction between two squared terms.
- Apply the formula \(a^2 - b^2 = (a-b)(a+b)\).
- Simplify the parentheses to arrive at the final factors.
Polynomial Factoring
Polynomial factoring is a crucial skill in algebra. It involves breaking down a polynomial into simpler components, or "factors." These factors, when multiplied together, give back the original polynomial. In this context, the factoring of \( (z-2)^2 - 9 \) involved expressing it as a product of simpler polynomials: \( (z - 5)(z + 1) \).
- Start by identifying a familiar pattern or structure, such as the difference of squares.
- Use appropriate factoring techniques, like grouping, the quadratic formula, or special products.
- Double-check by expanding the factors to ensure they match the original polynomial.
College Algebra
College Algebra serves as a foundation for higher-level mathematics courses. It covers a variety of concepts, including factoring, equations solving, and functions. Understanding how to factor polynomials, like in the exercise, is essential for success in College Algebra.
- Key skills include recognizing patterns in expressions and applying the right factoring formulas.
- It involves solving for unknowns by simplifying and reorganizing equations, beginning with foundational expressions.
- Algebraic skills form the backbone of advanced studies in calculus, statistics, and science-related fields.
Other exercises in this chapter
Problem 147
Factor the expression completely. \(9 x^{3}+6 x^{2}-3 x\)
View solution Problem 148
Factor the expression completely. \(8 x^{3}+28 x^{2}-16 x\)
View solution Problem 150
Factor the expression completely. \((y+2)^{2}-4\)
View solution Problem 151
Factor the expression completely. \(3 x^{5}-27 x^{3}+3 x^{2}-27\)
View solution