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Zorluk: OrtaSimplifying and Factoring Algebraic Expressions

For all real numbers x≠2x \neq 2, which of the following expressions is equivalent to x4−16x3−2x2+4x−8\frac{x^4 - 16}{x^3 - 2x^2 + 4x - 8}?

  1. x+2x + 2Cevap
  2. B
    x−2x - 2
  3. C
    x2+4x^2 + 4
  4. D
    x+2x−2\frac{x + 2}{x - 2}
  5. E
    ±(x+2)\pm (x + 2)

Cevap

x+2x + 2
Factoring the numerator x4−16x^4 - 16 as a difference of squares yields (x−2)(x+2)(x2+4)(x - 2)(x + 2)(x^2 + 4). Factoring the denominator x3−2x2+4x−8x^3 - 2x^2 + 4x - 8 by grouping yields (x−2)(x2+4)(x - 2)(x^2 + 4). Dividing the numerator by the denominator allows the shared factor (x−2)(x2+4)(x - 2)(x^2 + 4) to cancel, leaving x+2x + 2.

Adım Adım Çözüm

1
Factor the numerator x4−16x^4 - 16
x4−16=(x2−4)(x2+4)=(x−2)(x+2)(x2+4)x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)
Apply the difference of squares factorization identity twice.
2
Factor the denominator x3−2x2+4x−8x^3 - 2x^2 + 4x - 8 by grouping
x3−2x2+4x−8=x2(x−2)+4(x−2)=(x−2)(x2+4)x^3 - 2x^2 + 4x - 8 = x^2(x - 2) + 4(x - 2) = (x - 2)(x^2 + 4)
Group the terms in pairs and factor out the common binomial factor (x−2)(x - 2).
3
Simplify the rational expression by canceling common factors
(x−2)(x+2)(x2+4)(x−2)(x2+4)=x+2\frac{(x - 2)(x + 2)(x^2 + 4)}{(x - 2)(x^2 + 4)} = x + 2
Cancel the non-zero common factor (x−2)(x2+4)(x - 2)(x^2 + 4) present in both numerator and denominator.

Anahtar Kavram

Simplifying rational expressions using difference of squares and factoring by grouping
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