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Difficulty: Very hardEquivalent Algebraic Expressions

For all x>1x > 1, which of the following is equivalent to the expression 2x2+3x−3x−1−x3+2x2+xx2−1\frac{2x^2 + 3x - 3}{x - 1} - \frac{x^3 + 2x^2 + x}{x^2 - 1}?

  1. A
    x−3x - 3
  2. x+3x + 3Answer
  3. C
    x+4x + 4
  4. D
    x+5x + 5

Answer

x+3x + 3
The correct answer is the expression that results from simplifying both terms and combining them. Simplifying the second term of the expression yields x2+xx−1\frac{x^2+x}{x-1}. Subtracting this from the first term gives 2x2+3x−3−(x2+x)x−1=x2+2x−3x−1\frac{2x^2+3x-3 - (x^2+x)}{x-1} = \frac{x^2+2x-3}{x-1}. Factoring the numerator gives (x+3)(x−1)x−1\frac{(x+3)(x-1)}{x-1}, which simplifies to the expression representing the sum of xx and 33 since x−1x-1 cancels out.

Step-by-Step Solution

1
Factor the numerator and denominator of the second term of the expression.
The numerator is x3+2x2+x=x(x2+2x+1)=x(x+1)2x^3 + 2x^2 + x = x(x^2 + 2x + 1) = x(x+1)^2. The denominator is x2−1=(x−1)(x+1)x^2 - 1 = (x-1)(x+1).
Factoring allows for the simplification of the rational expression by identifying common factors.
2
Simplify the second term by canceling the common factor (x+1)(x+1) since x>1x > 1.
The second term simplifies to x(x+1)x−1=x2+xx−1\frac{x(x+1)}{x-1} = \frac{x^2+x}{x-1}.
Since x>1x > 1, x+1≠0x+1 \neq 0, so we can cancel it from both numerator and denominator.
3
Subtract the simplified second term from the first term.
2x2+3x−3x−1−x2+xx−1=(2x2+3x−3)−(x2+x)x−1=x2+2x−3x−1\frac{2x^2 + 3x - 3}{x - 1} - \frac{x^2 + x}{x - 1} = \frac{(2x^2 + 3x - 3) - (x^2 + x)}{x - 1} = \frac{x^2 + 2x - 3}{x - 1}.
With a common denominator of x−1x-1, we can combine the numerators directly, being careful to distribute the subtraction sign to both terms in the second numerator.
4
Factor the quadratic expression in the numerator.
x2+2x−3=(x+3)(x−1)x^2 + 2x - 3 = (x+3)(x-1).
Factoring the quadratic helps to check if there is a common factor with the denominator.
5
Cancel the common factor (x−1)(x-1) to find the final simplified expression.
(x+3)(x−1)x−1=x+3\frac{(x+3)(x-1)}{x-1} = x+3.
Since x>1x > 1, x−1≠0x-1 \neq 0, allowing us to divide out the common factor.

Key Concept

Simplifying rational expressions and combining algebraic terms over a common denominator

Alternative Method

Instead of simplifying the second term first, a common denominator can be established by multiplying the first term by x+1x+1\frac{x+1}{x+1}. This gives: (2x2+3x−3)(x+1)−(x3+2x2+x)x2−1=2x3+5x2−3−x3−2x2−xx2−1=x3+3x2−x−3x2−1\frac{(2x^2+3x-3)(x+1) - (x^3+2x^2+x)}{x^2-1} = \frac{2x^3+5x^2-3 - x^3-2x^2-x}{x^2-1} = \frac{x^3+3x^2-x-3}{x^2-1}. Factoring by grouping the numerator yields (x2−1)(x+3)x2−1=x+3\frac{(x^2-1)(x+3)}{x^2-1} = x+3.
Estimated Time:2m 0s
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Topics

Equivalent Algebraic Expressions

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