Factor x⁴−5x²+4: Biquadratic, Step by Step

Factor x⁴ − 5x² + 4. A degree-4 polynomial looks intimidating, and students either freeze or reach for the rational root theorem and test ±1, ±2, ±4 one by one. There is a much better move hiding in plain sight: x⁴ is (x²)², so this polynomial is secretly a quadratic wearing a disguise.

Substitute u = x², factor the easy quadratic, then translate back. The whole problem collapses in three moves.

Worked solution

Spot the pattern: the exponents are 4, 2, 0 (even numbers, descending by 2). Set u = x², so x⁴ = u². The polynomial becomes u² − 5u + 4.
Factor the quadratic. Find two numbers that multiply to 4 and add to −5: they are −1 and −4. So u² − 5u + 4 = (u − 1)(u − 4).
Substitute back with u = x²: (x² − 1)(x² − 4). Do not stop here; the answer is not finished.
Each factor is a difference of squares: x² − 1 = (x − 1)(x + 1) and x² − 4 = (x − 2)(x + 2).
Full factorization: x⁴ − 5x² + 4 = (x − 1)(x + 1)(x − 2)(x + 2).

Why this method works

Substitution renames the complexity away: with u = x², a quartic in x becomes a quadratic in u, and every factoring technique for quadratics applies unchanged. Translating back restores x, and the difference-of-squares step splits each quadratic factor into linears. It works because the original exponents (4, 2, 0) are all multiples of 2, which is exactly what makes the disguise possible.

Common mistakes

Forgetting to substitute back, answering (u − 1)(u − 4). The question asked about x. The u was scaffolding, and scaffolding comes down at the end.
Sign errors in step 2. (u + 1)(u + 4) expands to u² + 5u + 4, which has the wrong middle sign. Since the constant 4 is positive and the middle coefficient −5 is negative, both numbers must be negative.
Stopping at (x² − 1)(x² − 4). Each factor splits further into a difference of squares; leaving them whole is an incomplete answer.
Missing the substitution entirely and trying to factor x⁴ directly, which burns minutes for nothing.

Check (never skip)

Expand back to prove the factorization. (x − 1)(x + 1) = x² − 1 and (x − 2)(x + 2) = x² − 4. Then (x² − 1)(x² − 4) = x⁴ − 4x² − x² + 4 = x⁴ − 5x² + 4. ✓ Exactly the original polynomial, so the factorization is complete and correct. Expanding back is the universal factoring check.

A second example: x⁴ − 13x² + 36

Same three moves, new numbers. Substitute u = x²: u² − 13u + 36. Find two numbers that multiply to 36 and add to −13: −4 and −9, so (u − 4)(u − 9). Substitute back: (x² − 4)(x² − 9). Split each difference of squares: (x − 2)(x + 2)(x − 3)(x + 3). Check by expanding the pairs: (x² − 4)(x² − 9) = x⁴ − 9x² − 4x² + 36 = x⁴ − 13x² + 36. ✓ Once the u-substitution clicks, every biquadratic is the same problem in different clothes. Try one yourself before reading on: x⁴ − 10x² + 9 factors as (x − 1)(x + 1)(x − 3)(x + 3); the u-quadratic is u² − 10u + 9 = (u − 1)(u − 9).

When the disguise is not there

The substitution works only because every exponent is a multiple of 2. For something like x⁴ + 3x³ − 4, no single substitution reduces it to a quadratic, and you fall back to the rational root theorem or numerical methods. Before substituting, always glance at the exponents: 4, 2, 0 means go; anything else means stop and pick a different tool. Recognizing when a trick does not apply is as valuable as knowing the trick itself, and it is what separates a method you understand from a recipe you memorized.

FAQ

Q: What is a biquadratic? A: A quartic (degree 4) containing only even powers of x, of the form ax⁴ + bx² + c. The u = x² substitution always reduces it to a quadratic.

Q: What if the quadratic in u does not factor nicely? A: Use the quadratic formula on u. And if a factor like x² + 1 appears after substituting back, it is irreducible over the reals, so it stays as is.

Q: How do I get the roots from the factored form? A: Set each factor to zero: x = 1, −1, 2, −2. Each linear factor hands you a root directly, which is the whole point of factoring.

Solve it with steps on our factoring calculator.