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Difficulty: Very hardFactoring Polynomials

Which of the following is a factor of the expression 2x4−8x2y2−2x2+8y22x^4 - 8x^2y^2 - 2x^2 + 8y^2 when it is factored completely?

  1. A
    (x−2y)2(x - 2y)^2
  2. B
    x2−8y2x^2 - 8y^2
  3. x−2yx - 2yAnswer
  4. D
    x2+2y2x^2 + 2y^2
  5. E
    x2−2yx^2 - 2y

Answer

The correct answer is x−2yx - 2y because the completely factored form of the expression is 2(x−1)(x+1)(x−2y)(x+2y)2(x - 1)(x + 1)(x - 2y)(x + 2y), which contains x−2yx - 2y as a linear factor.
The expression 2x4−8x2y2−2x2+8y22x^4 - 8x^2y^2 - 2x^2 + 8y^2 can be factored by first pulling out the greatest common factor of 2, giving 2(x4−4x2y2−x2+4y2)2(x^4 - 4x^2y^2 - x^2 + 4y^2). Grouping the terms as x2(x2−4y2)−1(x2−4y2)x^2(x^2 - 4y^2) - 1(x^2 - 4y^2) produces 2(x2−1)(x2−4y2)2(x^2 - 1)(x^2 - 4y^2). Factoring the differences of squares yields the completely factored form 2(x−1)(x+1)(x−2y)(x+2y)2(x - 1)(x + 1)(x - 2y)(x + 2y). The expression x−2yx - 2y is one of these linear factors.

Step-by-Step Solution

1
Identify and factor out the greatest common factor (GCF) of the terms in the polynomial.
The terms 2x42x^4, −8x2y2-8x^2y^2, −2x2-2x^2, and 8y28y^2 share a common factor of 2. Factoring out 2 yields: 2(x4−4x2y2−x2+4y2)2(x^4 - 4x^2y^2 - x^2 + 4y^2).
Factoring out the GCF simplifies the remaining polynomial expression, making it easier to factor further.
2
Group the terms inside the parentheses to perform factoring by grouping.
Group the terms as follows: 2[(x4−4x2y2)−(x2−4y2)]2[(x^4 - 4x^2y^2) - (x^2 - 4y^2)]. Factor out x2x^2 from the first group: 2[x2(x2−4y2)−1(x2−4y2)]2[x^2(x^2 - 4y^2) - 1(x^2 - 4y^2)]. Now, factor out the common binomial (x2−4y2)(x^2 - 4y^2) to get: 2(x2−1)(x2−4y2)2(x^2 - 1)(x^2 - 4y^2).
Grouping allows us to find common binomial factors within the terms of the polynomial.
3
Apply the difference of squares identity, a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b), to the remaining binomial factors.
For the factor (x2−1)(x^2 - 1), the difference of squares gives (x−1)(x+1)(x - 1)(x + 1). For the factor (x2−4y2)(x^2 - 4y^2), the difference of squares gives (x−2y)(x+2y)(x - 2y)(x + 2y). Substituting these back into the expression yields: 2(x−1)(x+1)(x−2y)(x+2y)2(x - 1)(x + 1)(x - 2y)(x + 2y).
Both quadratic factors are differences of squares and must be factored completely to find all linear factors.
4
Compare the complete factorization with the given choices to find the matching factor.
The linear factor x−2yx - 2y is present in the completely factored expression.
This confirms the correct option based on algebraic factorization.

Key Concept

Factoring polynomials completely using GCF, grouping, and the difference of squares identity.

Alternative Method

Instead of factoring out the GCF 2 first, you can group the terms directly: 2x4−2x2−8x2y2+8y2=2x2(x2−1)−8y2(x2−1)=(2x2−8y2)(x2−1)2x^4 - 2x^2 - 8x^2y^2 + 8y^2 = 2x^2(x^2 - 1) - 8y^2(x^2 - 1) = (2x^2 - 8y^2)(x^2 - 1). Then, factor out 2 from the first binomial to get 2(x2−4y2)(x2−1)2(x^2 - 4y^2)(x^2 - 1), and finally apply the difference of squares identity to both quadratic factors to obtain 2(x−2y)(x+2y)(x−1)(x+1)2(x - 2y)(x + 2y)(x - 1)(x + 1).
Estimated Time:1m 30s
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Topics

Factoring Polynomials

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