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Chapter 5: Problem 25

Factor each difference of two squares. See Example 2. $$ 100 r^{2} s^{4}-t^{4} $$

Short Answer

Expert verified
The expression factors to \((10rs^2 - t^2)(10rs^2 + t^2)\).

Step by step solution

01

Recognizing the Difference of Squares

The expression given is \(100r^2s^4 - t^4\). The difference of squares formula is \(a^2 - b^2 = (a - b)(a + b)\). First, identify \(a^2\) and \(b^2\) such that they match the formula.
02

Determine the Squares

Recognize \(100r^2s^4\) as a square: \(a = 10rs^2\) because \((10rs^2)^2 = 100r^2s^4\). Recognize \(t^4\) as a square: \(b = t^2\) because \((t^2)^2 = t^4\).
03

Apply the Difference of Squares Formula

Using the formula \(a^2 - b^2 = (a - b)(a + b)\), substitute \(a = 10rs^2\) and \(b = t^2\) into it. The factored expression becomes \((10rs^2 - t^2)(10rs^2 + t^2)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Concept of Difference of Squares
In algebra, a difference of squares is a specific expression that involves two perfect squares separated by a subtraction sign. For example, in the expression \(a^2 - b^2\), both \(a^2\) and \(b^2\) are squares. Identifying this pattern is crucial because it allows us to apply a special factoring technique quickly. This formula is easy to remember:
  • The difference of squares formula is \(a^2 - b^2 = (a-b)(a+b)\).
  • This works because squaring numbers and then subtracting is like finding a gap between two perfect squares.
The key to using this technique is recognizing each term as a perfect square. For example, in the exercise \(100r^2s^4 - t^4\), both terms can be rewritten as squares, setting the stage to apply the difference of squares formula.
Exploring Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and operators like addition, subtraction, multiplication, and division. They can express mathematical ideas succinctly. In the expression \(100r^2s^4 - t^4\), we have variables \(r, s,\) and \(t\), as well as constants. Each part of the expression serves a role in its algebraic structure.
  • The term \(100r^2s^4\) can be broken down further as \((10rs^2)^2\).
  • Similarly, \(t^4\) can be seen as \((t^2)^2\).
Decomposing expressions into manageable parts makes it easier to manipulate and solve the equation. By identifying the component squares, we prepared these expressions for factoring using the difference of squares technique.
Diving into Factoring Techniques
Factoring is a crucial technique in algebra for simplifying expressions and solving equations. Various techniques exist, such as factoring by grouping, trial and error, and using formulas like the difference of squares. Recognizing which method to use can simplify the problem significantly. When dealing with the exercise \(100r^2s^4 - t^4\), the difference of squares formula is ideal:
  • First, identify the squares involved: \(10rs^2\) and \(t^2\).
  • Apply the formula \(a^2 - b^2 = (a-b)(a+b)\). Here, substitute \(a = 10rs^2\) and \(b = t^2\).
  • The result is the factored form \((10rs^2 - t^2)(10rs^2 + t^2)\).
Each factoring technique has its specific use cases and being familiar with them enhances problem-solving efficiency. In this problem, recognizing the structure of the given expression directly led to a quicker solution.

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Most popular questions from this chapter

Factor. Assume all variables represent natural numbers. $$ a^{3 b}-c^{3 b} $$

Look Alikes... A. \(\left(10-a b-3 a^{2} b\right)+(4-6 a b)\) B. \(\left(10-a b-3 a^{2} b\right)-(4-6 a b)\)

Factor. Assume all variables represent natural numbers. $$ \text { Factor: } x^{12}-y^{12} $$

The number of feet that a car travels before stopping depends on the driver's reaction time and the braking distance. For one driver, the stopping distance \(d(v),\) in feet, is given by the polynomial function \(d(v)=0.04 v^{2}+0.9 v\) where \(v\) is the velocity of the car in mph. Find the stopping distance at \(60 \mathrm{mph}\). (PICTURE NOT COPY)

Look Alikes... A. \(\quad 3 d^{3}-4 d^{2}-3 d+5\) \(+11 d^{3} \quad-8 d-2\) B. \(\quad 3 d^{3}-4 d^{2}-3 d+5\) \(-\left(11 d^{3} \quad-8 d-2\right)\)
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