Problem 55
Question
Multiply. $$ (2 m+5 n)(2 m-5 n) $$
Step-by-Step Solution
Verified Answer
\(4m^2 - 25n^2\)
1Step 1: Recognize the Pattern
Notice that the expression \((2m + 5n)(2m - 5n)\) follows the difference of squares pattern \((a + b)(a - b) = a^2 - b^2\). Here, \(a = 2m\) and \(b = 5n\).
2Step 2: Apply the Difference of Squares Formula
Using the difference of squares formula, we substitute \(a = 2m\) and \(b = 5n\) into \(a^2 - b^2\).
3Step 3: Square the First Term
Calculate \((2m)^2\). This gives: \[(2m)^2 = 4m^2.\]
4Step 4: Square the Second Term
Calculate \((5n)^2\). This gives: \[(5n)^2 = 25n^2.\]
5Step 5: Subtract the Squares
Now subtract the second squared term from the first squared term to get the final answer: \[4m^2 - 25n^2.\]
Key Concepts
Algebraic ExpressionsPolynomial MultiplicationFactoring Patterns
Algebraic Expressions
Algebraic expressions are a foundational concept in mathematics. They consist of variables, constants, and operations that together create phrases that express algebraic relationships. In the exercise, we have an expression involving the terms \(2m\) and \(5n\), which are both algebraic expressions on their own. Here:
Expressions such as (2m + 5n) and (2m - 5n) are seen in various algebraic operations including polynomial multiplication.Understanding these expressions is key to performing operations like the difference of squares. By recognizing the components, we effectively manipulate and simplify expressions in equations.
- \(2m\) is the product of 2 and the variable \(m\).
- \(5n\) is the product of 5 and the variable \(n\).
Expressions such as (2m + 5n) and (2m - 5n) are seen in various algebraic operations including polynomial multiplication.Understanding these expressions is key to performing operations like the difference of squares. By recognizing the components, we effectively manipulate and simplify expressions in equations.
Polynomial Multiplication
Polynomial multiplication involves distributing each term in one polynomial by every term in another polynomial. Here, we use a specific case, known as the difference of squares. The expression offered in the exercise (2m + 5n)(2m - 5n)can be multiplied using this method.
When multiplying such binomials, each term in the first binomial is multiplied by each term in the second binomial. However, the difference of squares formula allows an easier approach.
When multiplying such binomials, each term in the first binomial is multiplied by each term in the second binomial. However, the difference of squares formula allows an easier approach.
- The first term from each binomial, \(2m\), is squared: \((2m)^2 = 4m^2\).
- The second term from each binomial, \(5n\), is also squared: \((5n)^2 = 25n^2\).
Factoring Patterns
Factoring is a method used to simplify mathematical expressions by finding factors that multiply to form the original expression. One important pattern in factoring is the difference of squares. Our initial expression \((2m + 5n)(2m - 5n)\)exemplifies this pattern. In the difference of squares, we use the formula:
Such factoring patterns make solving complex problems easier. They allow for quick simplification and insight into the structure of polynomial expressions. Becoming familiar with these patterns is a practical tool in algebra, giving students a powerful shortcut when working through exercises.
- \((a + b)(a - b) = a^2 - b^2\)
Such factoring patterns make solving complex problems easier. They allow for quick simplification and insight into the structure of polynomial expressions. Becoming familiar with these patterns is a practical tool in algebra, giving students a powerful shortcut when working through exercises.
Other exercises in this chapter
Problem 55
Use the quotient rule and simplify each expression. $$ \frac{x^{3}}{x} $$
View solution Problem 55
Multiply vertically. \((x+3)\left(2 x^{2}+4 x-1\right)\)
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Subtract \(\left(y^{2}-9\right)\) from the sum of \(\left(3 y^{2}+y+4\right)\) and \(\left(2 y^{2}-6 y-10\right)\)
View solution Problem 56
Simplify each polynomial by combining any like terms. See Examples 13 and 14. $$ 17 a^{2} b-16 a b^{2}+3 a^{3}+4 b a^{3}-b^{2} a $$
View solution