Chapter 5: Problem 8
Fill in the blanks. $$ \text { a. } 36 y^{2}-49 m^{4}=(\quad)^{2}-(\quad)^{2} $$ $$ \text { b. } 125 h^{3}-27 k^{6}=(\quad)^{3}-(\quad)^{3} $$
Short Answer
Expert verified
(a) \( (6y)^2 - (7m^2)^2 \), (b) \( (5h)^3 - (3k^2)^3 \).
Step by step solution
01
Identify the Difference of Squares
For part (a), the expression \(36y^2 - 49m^4\) resembles the difference of squares form \(a^2 - b^2\). Firstly, identify what squared terms lead to 36 and 49. We recognize that \((6y)^2 = 36y^2\) and \((7m^2)^2 = 49m^4\).
02
Fill in the Difference of Squares
Based on the identified squares from Step 1, we fill in the blanks: \((6y)^2 - (7m^2)^2\). So the completed expression is \(36y^{2}-49m^{4}=(6y)^{2}-(7m^{2})^{2}\).
03
Identify the Difference of Cubes
For part (b), the expression \(125h^3 - 27k^6\) resembles the difference of cubes form \(a^3 - b^3\). Calculate the cube roots of 125 and 27. We find \((5h)^3 = 125h^3\) and \((3k^2)^3 = 27k^6\).
04
Fill in the Difference of Cubes
Using the cube roots from Step 3, complete the expression: \((5h)^3 - (3k^2)^3\). Therefore, the completed expression is \(125h^{3}-27k^{6}=(5h)^{3}-(3k^{2})^{3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Difference of Squares
The difference of squares is a fascinating algebraic concept that involves subtracting one squared term from another squared term. This type of expression often looks like this: \( a^2 - b^2 \). The unique aspect of the difference of squares is that it can always be factored as \((a-b)(a+b)\). This means you need to find two numbers whose squares make up the original terms you're dealing with.
In the exercise, we look at the expression \(36y^2 - 49m^4\). Here, you should identify the terms that are perfect squares:
So, with these identified, the expression can be rewritten as \((6y)^2 - (7m^2)^2\). This equates to a factorization using the difference of squares formula: \((6y - 7m^2)(6y + 7m^2)\). Understanding these patterns is a crucial skill in algebra, enhancing problem-solving and simplifying expressions.
In the exercise, we look at the expression \(36y^2 - 49m^4\). Here, you should identify the terms that are perfect squares:
- \(36y^2\) is the square of \(6y\).
- \(49m^4\) is the square of \(7m^2\).
So, with these identified, the expression can be rewritten as \((6y)^2 - (7m^2)^2\). This equates to a factorization using the difference of squares formula: \((6y - 7m^2)(6y + 7m^2)\). Understanding these patterns is a crucial skill in algebra, enhancing problem-solving and simplifying expressions.
Difference of Cubes
The difference of cubes involves subtracting one cubed term from another. Expressions in this form look like \( a^3 - b^3 \). Different from squares, the difference of cubes can be factored using the formula \((a-b)(a^2+ab+b^2)\). It's essential to identify the cube roots of each term correctly.
In the given problem, we have \(125h^3 - 27k^6\). To factor it, identify the terms as cubes:
In the given problem, we have \(125h^3 - 27k^6\). To factor it, identify the terms as cubes:
- \(125h^3\) is the cube of \(5h\).
- \(27k^6\) is the cube of \(3k^2\).
Algebraic Expressions
Algebraic expressions form the backbone of algebra, representing numbers and operations in a symbolic form. By combining constants and variables with mathematical operations such as addition, subtraction, multiplication, and division, these expressions are pivotal in problem-solving. Their flexible nature allows them to model real-world scenarios and abstract mathematical concepts.
When working with algebraic expressions, two special forms are frequently encountered:
When working with algebraic expressions, two special forms are frequently encountered:
- The **Difference of Squares** \( a^2 - b^2 \).
- The **Difference of Cubes** \( a^3 - b^3 \).