Chapter 6: Problem 42
Divide. $$ \frac{y^{3}-8}{y-2} $$
Short Answer
Expert verified
The result of dividing is \(y^2 + 2y + 4\), with \(y \neq 2\).
Step by step solution
01
Recognize the Form
The expression \( \frac{y^3 - 8}{y-2} \) is a division of two polynomials. Notice that the numerator is a difference of cubes, which is in the form of \(a^3 - b^3\), where \(a = y\) and \(b = 2\).
02
Apply the Difference of Cubes Formula
The difference of cubes formula is \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \). Apply this to \(y^3 - 8\):\[y^3 - 8 = (y-2)(y^2 + 2y + 4)\]
03
Simplify the Expression
Substitute the factorized form from Step 2 into the original expression:\[\frac{y^3 - 8}{y-2} = \frac{(y-2)(y^2 + 2y + 4)}{y-2}\]Since \(y-2\) is a factor in both the numerator and the denominator, they cancel each other out, resulting in the simplified expression:\[ y^2 + 2y + 4\]
04
Verify the Cancellation
Ensure that \(y eq 2\), as this would make the denominator zero and therefore the expression undefined. Since \(y eq 2\), the cancellation is valid.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Difference of Cubes
When looking at the expression \( y^3 - 8 \), it helps to recognize it as a difference of cubes. This is a specific algebraic pattern that can be factored using a very helpful formula:
- \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \)
- \( a = y \)
- \( b = 2 \)
- \( y^3 - 8 = (y - 2)(y^2 + 2y + 4) \)
Polynomial Simplification
After factoring the expression using the difference of cubes, we substitute it back into our original division problem. We start with:
- \( \frac{y^3 - 8}{y-2} = \frac{(y-2)(y^2 + 2y + 4)}{y-2} \)
- The term \( (y-2) \) appears in both the numerator and the denominator.
- We can cancel \( (y-2) \) from both parts.
- This leaves us with \( y^2 + 2y + 4 \).
Rational Expressions
Rational expressions are fractions in which the numerator and the denominator are polynomials. In our problem, we dealt with the rational expression \( \frac{y^3 - 8}{y-2} \). The goal was to simplify this expression by canceling common factors.It's crucial to recognize potential simplifications:
- Identify common factors in the numerator and denominator.
- Use factoring methods like the difference of cubes to uncover these factors.