Problem 38
Question
In \(27-39,\) factor each polynomial completely. $$ 4-(y-1)^{2} $$
Step-by-Step Solution
Verified Answer
The polynomial factors as \((3-y)(y+1)\.\)
1Step 1: Recognize the Pattern
The polynomial is of the form \(4 - (y-1)^2\). This resembles the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\). Identify \(a\) and \(b\) such that \(a^2 = 4\) and \(b^2 = (y-1)^2\).
2Step 2: Identify 'a' and 'b'
For the expression \(4 - (y-1)^2\), we identify \(a^2 = 4\), which means \(a = 2\) (since \(2^2 = 4\)), and \(b = (y-1)\) since \((y-1)^2 = (y-1)^2\).
3Step 3: Apply the Difference of Squares Formula
Using the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\), substitute \(a = 2\) and \(b = (y-1)\) into the formula. This yields: \(4 - (y-1)^2 = (2 - (y-1))(2 + (y-1))\).
4Step 4: Simplify the Factors
Simplify the expressions within the parentheses: - For \( (2 - (y-1)) \), simplify to \( (2 - y + 1) = (3 - y) \). - For \( (2 + (y-1)) \), simplify to \( (2 + y - 1) = (y + 1) \).
5Step 5: Write the Final Factored Form
The fully factored form of \(4 - (y-1)^2\) is \((3-y)(y+1)\).
Key Concepts
difference of squaresalgebra 2factoring techniquesquadratic expressions
difference of squares
The "difference of squares" is a powerful algebraic identity that simplifies expressions where one square is subtracted from another. It’s given by the formula \[ a^2 - b^2 = (a-b)(a+b) \] This formula tells us that a difference of two squares is equivalent to the product of two binomials. Here, both binomials include the square roots of the original squares, distinguished merely by their signs.
- The first binomial is the difference \(a-b\),
- and the second is the sum \(a+b\).
algebra 2
Algebra 2 is a branch of mathematics that extends the concepts taught in Algebra 1. It dives deeper into equations, functions, and more complex mathematical relationships.
Here’s what you commonly explore in Algebra 2:
- Quadratic expressions and equations, which represent one of the most introduced pieces of algebraic thinking.
- Factoring techniques, like difference of squares, which are integral in simplifying and solving these expressions.
- Advanced polynomials, systems of equations, and other algebraic structures.
factoring techniques
Factoring is one of the fundamental techniques used in algebra to simplify expressions. It involves rewriting a polynomial as a product of simpler polynomials. Key factoring techniques you might encounter include:
- Difference of squares, where expressions in the form of \(a^2 - b^2\) are factored as \( (a-b)(a+b) \).
- Factoring trinomials, which splits a three-term polynomial into two binomials.
- Recognizing perfect square trinomials, using the square of a binomial pattern.
quadratic expressions
Quadratic expressions are polynomials of degree 2, usually in the form \(ax^2 + bx + c\). They hold a significant place in algebra because they appear in various scenarios, from physics to economics. Common properties of quadratic expressions include:
- They take on a parabolic shape when graphed, opening either upwards or downwards depending on the sign of \(a\).
- Factorization is a critical method for solving quadratic equations, often involving techniques like completing the square or using the quadratic formula.
- Roots or solutions of these expressions can indicate real-world quantities, like time or distance, depending on context.
Other exercises in this chapter
Problem 36
In \(27-39,\) factor each polynomial completely. $$ z^{4}-12 z^{2}+27 $$
View solution Problem 37
In \(27-39,\) factor each polynomial completely. $$ (c+2)^{2}-1 $$
View solution Problem 39
In \(27-39,\) factor each polynomial completely. $$ x^{2} y-16 y $$
View solution Problem 41
$$ 9-9(x+2)^{2} $$
View solution