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Zorluk: Çok zorSimplifying and Factoring Algebraic Expressions

For all real numbers xx and yy such that x≠yx \neq y and x≠−yx \neq -y, which of the following is equivalent to the algebraic expression x3(x+2y)−y3(y+2x)x2−y2\frac{x^3(x + 2y) - y^3(y + 2x)}{x^2 - y^2}?

  1. A
    x2+y2x^2 + y^2
  2. (x+y)2(x + y)^2Cevap
  3. C
    (x−y)2(x - y)^2
  4. D
    x2+2xy−y2x^2 + 2xy - y^2
  5. E
    x+yx + y

Cevap

The simplified expression is (x+y)2(x + y)^2.
Expanding and grouping the terms in the numerator gives (x4−y4)+2xy(x2−y2)=(x2−y2)(x2+2xy+y2)=(x2−y2)(x+y)2(x^4 - y^4) + 2xy(x^2 - y^2) = (x^2 - y^2)(x^2 + 2xy + y^2) = (x^2 - y^2)(x + y)^2. Canceling the common factor (x2−y2)(x^2 - y^2) from both the numerator and the denominator leaves (x+y)2(x + y)^2.

Adım Adım Çözüm

1
Expand the numerator terms
x3(x+2y)−y3(y+2x)=x4+2x3y−y4−2xy3x^3(x + 2y) - y^3(y + 2x) = x^4 + 2x^3y - y^4 - 2xy^3
Apply the distributive property to remove parentheses in the numerator.
2
Group the terms in the numerator to factor
(x4−y4)+(2x3y−2xy3)=(x2−y2)(x2+y2)+2xy(x2−y2)(x^4 - y^4) + (2x^3y - 2xy^3) = (x^2 - y^2)(x^2 + y^2) + 2xy(x^2 - y^2)
Use difference of squares on x4−y4x^4 - y^4 and factor out the greatest common factor 2xy2xy from the remaining terms.
3
Factor out the common term (x2−y2)(x^2 - y^2) from the numerator
(x2−y2)(x2+y2+2xy)=(x2−y2)(x+y)2(x^2 - y^2)(x^2 + y^2 + 2xy) = (x^2 - y^2)(x + y)^2
Recognize that x2+2xy+y2x^2 + 2xy + y^2 is the perfect square binomial (x+y)2(x + y)^2.
4
Simplify the rational expression by canceling common factors
\frac{(x^2 - y^2)(x + y)^2}{x^2 - y^2} = (x + y)^2
Divide numerator and denominator by (x2−y2)(x^2 - y^2), which is non-zero since x≠±yx \neq \pm y.

Anahtar Kavram

Factoring high-degree algebraic expressions by grouping terms, recognizing difference of squares, and applying perfect square binomial identities.
Tahmini Süre:2m 0s
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Simplifying and Factoring Algebraic Expressions

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