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Find the complete factorisation of the polynomial x^4 + 64 over real numbers. Choose the correct factorised form.

Difficulty: Medium

Correct Answer: (x^2 + 4x + 8)(x^2 - 4x + 8)

Explanation:


Introduction / Context:
This problem tests a famous algebraic factorisation called Sophie Germain identity, used for expressions of the form x^4 + 4a^4. It helps factor a fourth-degree polynomial into two quadratics over real numbers.


Given Data / Assumptions:

  • Polynomial: x^4 + 64
  • We want complete factorisation over reals.


Concept / Approach:
Use Sophie Germain identity: x^4 + 4a^4 = (x^2 + 2ax + 2a^2)(x^2 - 2ax + 2a^2). Rewrite 64 in the form 4a^4 to match the pattern.


Step-by-Step Solution:

Write 64 as 4a^4: 64 = 4*16, so a^4 = 16. Thus a = 2 (since 2^4 = 16). Apply the identity with a = 2: x^4 + 4*(2^4) = (x^2 + 2*2*x + 2*(2^2)) (x^2 - 2*2*x + 2*(2^2)). Simplify coefficients: 2*2*x = 4x and 2*(2^2) = 2*4 = 8. So x^4 + 64 = (x^2 + 4x + 8)(x^2 - 4x + 8).


Verification / Alternative check:
Multiply the factors: (x^2 + 8 + 4x)(x^2 + 8 - 4x) = (x^2 + 8)^2 - (4x)^2 = (x^4 + 16x^2 + 64) - 16x^2 = x^4 + 64.


Why Other Options Are Wrong:

(x^2 + 8)^2 expands to x^4 + 16x^2 + 64, which has an extra 16x^2 term. (x^2 + 8)(x^2 - 8) equals x^4 - 64, wrong sign. The forms with -8 or different middle terms do not cancel the x^2 term correctly.


Common Pitfalls:
Confusing x^4 + 64 with x^4 - 64, or trying to use difference of squares directly without the Sophie Germain structure.


Final Answer:
(x^2 + 4x + 8)(x^2 - 4x + 8)

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