Math Problem Statement

Simpler way

Simplify \( \frac{x - 1}{5(x - 1)^2} \) and factorize fully \( 50 - 2y^2 \).

Solution

Let's break down and solve each part of the question.

Part (a): Simplify the expression

x15(x1)2\frac{x - 1}{5(x - 1)^2}

Step-by-step solution:

  1. Cancel common terms: The term (x1)(x - 1) appears in both the numerator and denominator.

    • The numerator has x1x - 1.
    • The denominator has (x1)2(x - 1)^2, which is (x1)×(x1)(x - 1) \times (x - 1).

    You can cancel one x1x - 1 term from both the numerator and the denominator: (x1)5(x1)2=15(x1)\frac{(x - 1)}{5(x - 1)^2} = \frac{1}{5(x - 1)}

Thus, the simplified expression is: 15(x1)\frac{1}{5(x - 1)}


Part (b): Factorize fully

502y250 - 2y^2

Step-by-step solution:

  1. Factor out the greatest common factor (GCF): The GCF of 5050 and 2y22y^2 is 22. Factor out 22: 502y2=2(25y2)50 - 2y^2 = 2(25 - y^2)

  2. Recognize a difference of squares: The expression inside the parentheses, 25y225 - y^2, is a difference of squares: 25y2=(5y)(5+y)25 - y^2 = (5 - y)(5 + y)

Thus, the fully factorized form is: 2(5y)(5+y)2(5 - y)(5 + y)


Do you want further details on any of the steps or have any questions?

5 Related Questions:

  1. What are the general rules for canceling terms in algebraic expressions?
  2. How can you identify a difference of squares quickly?
  3. Why is it important to factor out the greatest common factor first in factoring problems?
  4. Can the process of simplifying and factoring be applied to higher-order polynomial expressions?
  5. How do you approach expressions with more complex denominators when simplifying?

Tip:

Always check for common factors before moving on to more complex factoring techniques; this can make the process much easier.

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Math Problem Analysis

Mathematical Concepts

Algebra
Simplification
Factoring
Difference of Squares

Formulas

Simplification by canceling common factors
Difference of squares formula: \( a^2 - b^2 = (a - b)(a + b) \)

Theorems

Properties of exponents
Difference of squares theorem

Suitable Grade Level

Grades 8-10