Question

Difficulty: MediumSimplifying and Factoring Algebraic Expressions

For all real numbers xx and yy such that x2yx \neq 2y and x2yx \neq -2y, which of the following expressions is equivalent to x3+2x2y4xy28y3x24y2\frac{x^3 + 2x^2y - 4xy^2 - 8y^3}{x^2 - 4y^2}?

  1. A
    x2yx - 2y
  2. B
    x+4yx + 4y
  3. x+2yx + 2yAnswer
  4. D
    x+2yx2y\frac{x + 2y}{x - 2y}
  5. E
    x2+2yx^2 + 2y

Answer

The simplified expression is x+2yx + 2y.
Grouping terms in the numerator gives x2(x+2y)4y2(x+2y)=(x24y2)(x+2y)x^2(x + 2y) - 4y^2(x + 2y) = (x^2 - 4y^2)(x + 2y). Dividing this by the denominator (x24y2)(x^2 - 4y^2) cancels out the identical non-zero factor (x24y2)(x^2 - 4y^2), leaving the linear binomial x+2yx + 2y.

Step-by-Step Solution

1
Group the four terms in the numerator in pairs to factor by grouping
x3+2x2y4xy28y3=x2(x+2y)4y2(x+2y)x^3 + 2x^2y - 4xy^2 - 8y^3 = x^2(x + 2y) - 4y^2(x + 2y)
Grouping the first two terms and the last two terms allows factoring out x2x^2 and 4y2-4y^2 respectively.
2
Factor out the common binomial factor (x+2y)(x + 2y) from the numerator
x2(x+2y)4y2(x+2y)=(x24y2)(x+2y)x^2(x + 2y) - 4y^2(x + 2y) = (x^2 - 4y^2)(x + 2y)
Both terms share the common factor (x+2y)(x + 2y).
3
Divide the factored numerator by the denominator (x24y2)(x^2 - 4y^2)
(x24y2)(x+2y)x24y2=x+2y\frac{(x^2 - 4y^2)(x + 2y)}{x^2 - 4y^2} = x + 2y
Since x±2yx \neq \pm 2y, x24y20x^2 - 4y^2 \neq 0, allowing the common polynomial factor (x24y2)(x^2 - 4y^2) to be canceled.

Key Concept

Factoring Four-Term Polynomials by Grouping and Rational Expression Simplification
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