Problem 37
Question
Factor each completely. $$ (y+2)^{2}-49 $$
Step-by-Step Solution
Verified Answer
The expression \((y+2)^2 - 49\) factors to \((y-5)(y+9)\).
1Step 1: Recognize the form
The given expression \((y+2)^2 - 49\) is a difference of squares. The difference of squares formula is \(a^2 - b^2 = (a-b)(a+b)\). Identify \(a\) and \(b\) in this expression.
2Step 2: Identify terms as squares
In the expression \((y+2)^2 - 49\), recognize \((y+2)^2\) as \(a^2\) and 49 as \(b^2\), where \(a = (y+2)\) and \(b = 7\). This sets up the expression as \((y+2)^2 - 7^2\).
3Step 3: Apply difference of squares formula
Utilize the difference of squares formula: \((a-b)(a+b)\) on the expression \((y+2)^2 - 7^2\). This results in \(((y+2) - 7)((y+2) + 7)\).
4Step 4: Simplify the expression
Carry out the simplification: \((y+2) - 7 = y - 5\) and \((y+2) + 7 = y + 9\). Therefore, the factorized form of the expression is \((y-5)(y+9)\).
Key Concepts
FactoringPolynomialsAlgebraic Expressions
Factoring
Factoring is a fundamental process in algebra that simplifies expressions by finding equivalent expressions that can be quickly evaluated. At its heart, factoring involves breaking down expressions into products of simpler components, called factors.
Why is factoring important? Well, it makes solving equations much easier and offers insight into the behavior of functions represented by those equations. When dealing with polynomials, factoring allows us to identify roots or zeros, which are crucial for sketching graphs and understanding the function's nature.
The difference of squares is a specific scenario in factoring. It deals with expressions of the form \(a^2 - b^2\). This type of expression can be cleverly factored using the formula \((a-b)(a+b)\). This shortcut results in substantial simplification, since two squares are often easier to visualize and work with.
Why is factoring important? Well, it makes solving equations much easier and offers insight into the behavior of functions represented by those equations. When dealing with polynomials, factoring allows us to identify roots or zeros, which are crucial for sketching graphs and understanding the function's nature.
The difference of squares is a specific scenario in factoring. It deals with expressions of the form \(a^2 - b^2\). This type of expression can be cleverly factored using the formula \((a-b)(a+b)\). This shortcut results in substantial simplification, since two squares are often easier to visualize and work with.
- Recognize perfect squares within the expression.
- Identify the difference of squares pattern \(a^2 - b^2\).
- Apply the formula \((a-b)(a+b)\) for quick factoring.
Polynomials
Polynomials are mathematical expressions that involve variables raised to whole number exponents, along with coefficients. They are central to algebra and higher mathematics because they can describe an array of behaviors and scenarios. A simple polynomial might look like \(2x^2 - 3x + 5\).
Polynomials can be classified based on their degree, which is the highest exponent of the variable in the expression. The example above is a second-degree polynomial because the highest power of \(x\) is 2. When factoring polynomials, we often seek ways to express them as products of their simplest components—either other polynomials or numbers.
The difference of squares, as illustrated in the original exercise, is a special case of polynomial factorization. By recognizing that the expression fits the difference of squares pattern, you can apply the relevant formula and break it down efficiently into linear components, like \((y-5)\) and \((y+9)\). This ability to simplify complex expressions is why understanding polynomial factoring is so invaluable in mathematics.
Polynomials can be classified based on their degree, which is the highest exponent of the variable in the expression. The example above is a second-degree polynomial because the highest power of \(x\) is 2. When factoring polynomials, we often seek ways to express them as products of their simplest components—either other polynomials or numbers.
The difference of squares, as illustrated in the original exercise, is a special case of polynomial factorization. By recognizing that the expression fits the difference of squares pattern, you can apply the relevant formula and break it down efficiently into linear components, like \((y-5)\) and \((y+9)\). This ability to simplify complex expressions is why understanding polynomial factoring is so invaluable in mathematics.
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and arithmetic operations like addition, subtraction, multiplication, and division. These expressions form the building blocks of algebra, helping us model and solve real-world problems.
The original exercise dealt with an algebraic expression in the form of a difference of squares, \((y+2)^2 - 49\), which includes both a variable and constants. In algebra, understanding how to manipulate and simplify these expressions is crucial.
Recognizing patterns, such as the difference of squares, empowers you to approach problems confidently and efficiently. It relies on identifying underlying structures within expressions to simplify and solve them.
The original exercise dealt with an algebraic expression in the form of a difference of squares, \((y+2)^2 - 49\), which includes both a variable and constants. In algebra, understanding how to manipulate and simplify these expressions is crucial.
Recognizing patterns, such as the difference of squares, empowers you to approach problems confidently and efficiently. It relies on identifying underlying structures within expressions to simplify and solve them.
- Combine numbers, variables, and operations into cohesive structures.
- Identify opportunities to simplify using algebraic identities, like \(a^2 - b^2 = (a-b)(a+b)\).
- Employ factoring for efficiency in solving algebraic problems.
Other exercises in this chapter
Problem 36
Factor each trinomial completely. See Examples 1 through 5 . \(3 n^{2}+20 n+5\)
View solution Problem 36
Solve. $$ 36 x^{2}+x-21=0 $$
View solution Problem 37
At the end of 2 years, \(P\) dollars invested at an interest rate \(r\) compounded annually increases to an amount, \(A\) dollars, given by $$ A=P(1+r)^{2} $$ F
View solution Problem 37
Factor each trinomial by grouping. Exercises 9 through 12 are broken into parts to help you get started. $$ 12 x^{3}-27 x^{2}-27 x $$
View solution