Problem 24
Question
For the following exercises, factor the polynomial. \(25 y^{2}-196\)
Step-by-Step Solution
Verified Answer
The factored form is \((5y-14)(5y+14)\).
1Step 1: Identify the type of polynomial
The polynomial provided is \(25y^2 - 196\). This is a binomial, meaning it has two terms. Noticing the structure, this resembles a difference of squares situation.
2Step 2: Express each term as a square
Recognize that \(25y^2\) is a perfect square, as \((5y)^2 = 25y^2\), and \(196\) is also a perfect square, as \(14^2 = 196\). Therefore, we can express the polynomial as \((5y)^2 - (14)^2\).
3Step 3: Apply the difference of squares formula
The difference of squares formula is \(a^2 - b^2 = (a-b)(a+b)\). Here, \(a = 5y\) and \(b = 14\). Substitute these into the formula to get:\[(5y - 14)(5y + 14)\].
4Step 4: Verify the factorization
To confirm correctness, multiply \((5y - 14)(5y + 14)\) and ensure it yields \(25y^2 - 196\). Calculating:\[(5y - 14)(5y + 14) = 5y(5y) + 5y(14) - 14(5y) - 14(14) = 25y^2 + 70y - 70y - 196 = 25y^2 - 196\]The factorization is correct.
Key Concepts
Understanding BinomialsExploring the Difference of SquaresRecognizing Perfect Squares
Understanding Binomials
The concept of a binomial is fundamental in algebra. A binomial consists of two terms combined together, typically connected by a plus or minus sign. In mathematical terms, a binomial expression can look like this: \(a + b\) or \(a - b\). Let's take the example from our exercise: \(25y^2 - 196\). This expression has two distinct terms: \(25y^2\) and \(-196\). The presence of only two terms is what makes it a binomial.Binomials are important because they often serve as the basic building blocks for more complex algebraic expressions. When factoring or expanding polynomials, recognizing binomials helps to apply specific algebraic methods more effectively.When working with binomials, always look for opportunities to simplify or factor them. This could involve recognizing patterns like the difference of squares or applying the distributive property to factor expression further.
- Simplifying binomials can make complex problems easier to solve.
- Knowing whether an expression is a binomial helps in choosing the right factoring method.
Exploring the Difference of Squares
The difference of squares is a specific algebraic pattern that is highly useful in factoring polynomials. When we talk about the difference of squares, we mean the expression \(a^2 - b^2\). This expression can always be factored into \((a - b)(a + b)\). In our exercise, we identified the polynomial \(25y^2 - 196\) as a difference of squares.To apply this concept, we need to verify:
- Both terms are perfect squares.
- The terms are connected by subtraction.
- \((5y)^2 = 25y^2\)
- \((14)^2 = 196\)
- Substitute \(a = 5y\) and \(b = 14\).
- Factor the polynomial as \((5y - 14)(5y + 14)\).
Recognizing Perfect Squares
Perfect squares are numbers or expressions that are the square of a whole number or another expression. They are pivotal in algebra because recognizing them makes factoring easier and simplifies calculus computations.For example, we say \(x^2\) is a perfect square because it is the square of \(x\). In addition, numbers like 25 or 196 are perfect squares because:
- \(5^2 = 25\)
- \(14^2 = 196\)
Other exercises in this chapter
Problem 23
For the following exercises, simplify the given expression. \(25 \div 5^{2}-7\)
View solution Problem 24
For the following exercises, divide the rational expressions. \(\frac{3 y^{2}-7 y-6}{2 y^{2}-3 y-9} \div \frac{y^{2}+y-2}{2 y^{2}+y-3}\)
View solution Problem 24
For the following exercises, expand the binomial \((4 x+5)^{2}\)
View solution Problem 24
For the following exercises, simplify each expression. \(12 \sqrt{3}-4 \sqrt{75}\)
View solution