Soru

Zorluk: OrtaFactoring Polynomials

Which of the following expressions represents the complete factorization of 2x4322x^4 - 32?

  1. A
    2(x24)(x2+4)2(x^2 - 4)(x^2 + 4)
  2. B
    (2x28)(x2+4)(2x^2 - 8)(x^2 + 4)
  3. C
    2(x2)(x+2)(x+2)22(x - 2)(x + 2)(x + 2)^2
  4. 2(x2)(x+2)(x2+4)2(x - 2)(x + 2)(x^2 + 4)Cevap
  5. E
    2(x4)(x+4)2(x - 4)(x + 4)

Cevap

2(x2)(x+2)(x2+4)2(x - 2)(x + 2)(x^2 + 4)
The correct answer is found by first factoring out the greatest common factor of 22, resulting in 2(x416)2(x^4 - 16). Next, recognize that x416x^4 - 16 is a difference of squares, (x2)242(x^2)^2 - 4^2, which factors into (x24)(x2+4)(x^2 - 4)(x^2 + 4). Finally, factor the remaining difference of squares, x24x^2 - 4, into (x2)(x+2)(x - 2)(x + 2). The sum of squares, x2+4x^2 + 4, cannot be factored further. Combining these parts gives the complete factorization.

Adım Adım Çözüm

1
Factor out the greatest common factor from the polynomial.
2(x416)2(x^4 - 16)
Both terms of 2x4322x^4 - 32 are divisible by 22, so we factor it out to simplify the remaining expression.
2
Factor the difference of squares inside the parentheses.
2(x24)(x2+4)2(x^2 - 4)(x^2 + 4)
The expression x416x^4 - 16 is a difference of squares because it can be written as (x2)242(x^2)^2 - 4^2.
3
Factor the remaining difference of squares.
2(x2)(x+2)(x2+4)2(x - 2)(x + 2)(x^2 + 4)
The binomial x24x^2 - 4 is also a difference of squares (x222x^2 - 2^2), which factors into (x2)(x+2)(x - 2)(x + 2). The sum of squares x2+4x^2 + 4 cannot be factored further using real numbers.

Anahtar Kavram

Factoring polynomials completely by extracting the greatest common factor and repeatedly applying the difference of squares formula.
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