To check a factorisation, multiply the factors back together and compare with the original polynomial. If every term matches, the factorisation is right. If not, the mismatch points to the error.
This is the last step of a full cubic question, after dividing by a linear factor, and it rescues many sign slips.
How do I expand three brackets without losing terms?
Expand two brackets first, then multiply the result by the third. Write every line. A good order is to start with the two easiest brackets and keep the last multiplication for a bracket that has a coefficient.
Use quick checks before the full expansion:
- Highest-power coefficient: multiply the x coefficients of all factors. It must equal the highest-power coefficient of f(x).
- Constant term: multiply the constants of all factors. It must equal the constant of f(x).
- One substitution: try x = 1 in the original and in the factorised form. They must give the same number.
Worked example
A student claims 2x³ + 3x² − 8x + 3 = (2x − 1)(x − 1)(x + 3). Check it.
Step 1, quick checks: highest-power coefficient 2 × 1 × 1 = 2 ✓. Constant term (−1)(−1)(3) = 3 ✓.
Step 2, substitute x = 1: original 2 + 3 − 8 + 3 = 0. Factorised: (1)(0)(4) = 0 ✓.
Step 3, expand two brackets: (x − 1)(x + 3) = x² + 3x − x − 3 = x² + 2x − 3.
Step 4, multiply by (2x − 1):
2x(x² + 2x − 3) = 2x³ + 4x² − 6x
−1(x² + 2x − 3) = −x² − 2x + 3
Step 5, add: 2x³ + 3x² − 8x + 3. ✓
The factorisation is correct.
The mistake to watch for
Sign errors in the roots are the most common cause of a wrong factorisation. The roots of x³ − 2x² − 5x + 6 are 1, 3 and −2.
Mistaken factorisation: (x − 1)(x − 3)(x − 2).
The student turned the root −2 into (x − 2) instead of (x + 2).
The constant-term check catches it in seconds: (−1)(−3)(−2) = −6, but the polynomial’s constant is +6. Expanding confirms it: (x − 1)(x − 3)(x − 2) = x³ − 6x² + 11x − 6, which does not match. The correct factor is (x + 2), since (−1)(−3)(2) = 6.
Check yourself
1. Is x³ − x² − 4x + 4 equal to (x − 1)(x − 2)(x + 2)?
Show answer
(x − 2)(x + 2) = x² − 4. Then (x − 1)(x² − 4) = x³ − 4x − x² + 4 = x³ − x² − 4x + 4. Yes, it is correct.
2. Is 2x³ − 5x² − 4x + 3 equal to (x − 3)(2x + 1)(x + 1)? If not, give the correct factorisation.
Show answer
(x − 3)(x + 1) = x² − 2x − 3. Multiply by (2x + 1): 2x³ + x² − 4x² − 2x − 6x − 3 = 2x³ − 3x² − 8x − 3. This does not match, so no. Trying (2x − 1) instead: (x² − 2x − 3)(2x − 1) = 2x³ − x² − 4x² + 2x − 6x + 3 = 2x³ − 5x² − 4x + 3. The correct factorisation is (x − 3)(2x − 1)(x + 1).
3. Use expansion to show that x³ − 3x² + 4 = (x + 1)(x − 2)².
Show answer
(x − 2)² = x² − 4x + 4. Then (x + 1)(x² − 4x + 4) = x³ − 4x² + 4x + x² − 4x + 4 = x³ − 3x² + 4. ✓
Where this leads next
Return to using a known root to see where the factor came from, or test the whole module with the mixed practice set. The non-calculator working trainer can help you rehearse careful expansion by hand, and the module page polynomial factors and remainders shows the full route.
Some students find that their expansion is neat but their sign tracking is not. Spotting that pattern is part of what our teachers do in online one-to-one Additional Mathematics tuition.