Once a root has given you a factor, you divide the cubic by that factor to get a quadratic quotient, then factorise the quadratic. This completes a cubic’s factorisation and lets you solve the cubic equation.
It follows directly from using a known root and is needed in most full cubic questions.
How does the division work?
If (x − a) is a factor of f(x), then f(x) = (x − a)(quadratic). Dividing recovers that quadratic. Two layouts are common: long division, and synthetic division, which is the same process written with coefficients only.
For synthetic division by (x − a):
- Write the coefficients of f(x) in order, including a 0 for any missing power.
- Bring down the first coefficient.
- Multiply it by a, write the result under the next coefficient, and add.
- Repeat across the row. The last total is the remainder. It should be 0 if (x − a) is a factor.
- Read the quotient from the other totals, starting with x².
Worked example
Factorise f(x) = 2x³ − x² − 13x − 6 fully, given that f(3) = 0.
Step 1, factor: f(3) = 54 − 9 − 39 − 6 = 0, so (x − 3) is a factor.
Step 2, synthetic division with a = 3: coefficients 2, −1, −13, −6.
- Bring down 2.
- −1 + 3 × 2 = 5.
- −13 + 3 × 5 = 2.
- −6 + 3 × 2 = 0, so the remainder is 0.
Step 3, quotient: 2x² + 5x + 2.
Step 4, factorise the quadratic: 2x² + 5x + 2 = (2x + 1)(x + 2).
Step 5, answer: f(x) = (x − 3)(2x + 1)(x + 2), so the roots are x = 3, x = −½ and x = −2.
Check: (x − 3)(2x² + 5x + 2) = 2x³ + 5x² + 2x − 6x² − 15x − 6 = 2x³ − x² − 13x − 6. ✓
The mistake to watch for
A cubic with a missing power is the classic trap. Take x³ − 7x + 6 divided by (x − 2).
Mistaken working: coefficients 1, −7, 6 (the x² term was skipped). −7 + 2 = −5, then 6 + 2(−5) = −4. The remainder is −4, so the student thinks (x − 2) is not a factor.
But f(2) = 8 − 14 + 6 = 0, so it is a factor. The missing x² term needed a coefficient of 0.
The correct row is 1, 0, −7, 6.
Bring down 1, then 0 + 2 = 2, then −7 + 4 = −3, then 6 − 6 = 0. The quotient is x² + 2x − 3 = (x + 3)(x − 1). A remainder that disagrees with f(a) is a signal to recheck the row.
Check yourself
1. Divide x³ − 6x² + 11x − 6 by (x − 1).
Show answer
Coefficients 1, −6, 11, −6 with a = 1: 1; −6 + 1 = −5; 11 − 5 = 6; −6 + 6 = 0. The quotient is x² − 5x + 6, remainder 0.
2. Factorise x³ + x² − 4x − 4 fully, given that (x + 1) is a factor.
Show answer
a = −1. Coefficients 1, 1, −4, −4: 1; 1 − 1 = 0; −4 − 0 = −4; −4 + 4 = 0. The quotient is x² − 4 = (x − 2)(x + 2). So (x + 1)(x − 2)(x + 2).
3. Factorise 2x³ + 3x² − 8x + 3, given that (2x − 1) is a factor.
Show answer
Long division: 2x³ ÷ 2x = x², subtract x²(2x − 1) = 2x³ − x², leaving 4x² − 8x + 3. Then 4x² ÷ 2x = 2x, subtract 4x² − 2x, leaving −6x + 3. Then −3 gives −6x + 3, remainder 0. The quotient is x² + 2x − 3 = (x + 3)(x − 1). So (2x − 1)(x − 1)(x + 3).
Where this leads next
The skill pairs with checking a factorisation by expansion, which is the quickest way to confirm your quotient. You can also look at the quadratic part with the quadratic structure explorer, and then try the mixed practice set.
When a full cubic question has connected steps, it helps to have someone read your working line by line, which is what our teachers do in online one-to-one Additional Mathematics tuition.