Factor x^4 - 5x^2 + 6 into irreducible factors over the real numbers.

• A. (x^2-1)(x^2-6)
• B. (x^2-2)(x^2-3)
• C. (x-√2)(x+√2)(x-√3)(x+√3)
• D. x^4 -5x^2 +6

Answer: (B)

Treat the expression as a quadratic in x^2. Let y = x^2. Then x^4 - 5x^2 + 6 becomes y^2 - 5y + 6, which factors as (y - 2)(y - 3). Replacing y with x^2 gives the factorization (x^2 - 2)(x^2 - 3).

This is a natural real-factorization because you’re factoring a quadratic in x^2. Each quadratic factor can be further broken over the real numbers into linear factors: x^2 - 2 = (x - √2)(x + √2) and x^2 - 3 = (x - √3)(x + √3). So the fully factored form over reals is (x - √2)(x + √2)(x - √3)(x + √3).

If you’re specifically asked for irreducible factors in terms of x, the complete irreducible factorization over the real numbers is the four linear factors above. The two-quadratic form (x^2 - 2)(x^2 - 3) is a standard intermediate factorization, often used because it’s simpler and still valid over the reals.