Polynomials: Remainder & Factor Theorem (Grade 12)

Using the remainder theorem to find a remainder without dividing, the factor theorem to find the first factor of a cubic, and then factorising and solving the cubic completely.

1The two theorems

Remainder theorem
If f(x) is divided by (x − a), the remainder is f(a).

Factor theorem
(x − a) is a factor of f(x) if and only if f(a) = 0.

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The sign flips: for the factor (x − 2) you substitute x = +2; for (x + 1) you substitute x = −1. Set the bracket equal to zero and solve.

2Finding a remainder

Worked ExampleDetermine the remainder when f(x) = x³ − 4x² + x + 6 is divided by (x − 1)

1
Set x − 1 = 0, so x = 1.
The remainder theorem needs the value that makes the divisor zero.
2
f(1) = 1 − 4 + 1 + 6.
Substitute.
3
= 4.
No long division needed at all.

3Factorising a cubic

Worked ExampleFactorise f(x) = x³ − 4x² + x + 6 completely

1
Try the factors of the constant 6: ±1, ±2, ±3, ±6.
Any whole-number root must divide the constant term.
2
f(2) = 8 − 16 + 2 + 6 = 0, so (x − 2) is a factor.
Factor theorem.
3
Divide: x³ − 4x² + x + 6 = (x − 2)(x² − 2x − 3).
Use long division or inspection.
4
Factorise the trinomial: x² − 2x − 3 = (x − 3)(x + 1).
5
f(x) = (x − 2)(x − 3)(x + 1).
Check the constant: (−2)(−3)(1) = 6. ✓

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Quick check on any cubic factorisation: multiply the three constants together. They must give the original constant term.

4Solving a cubic equation

Worked ExampleSolve x³ − 4x² + x + 6 = 0

1
(x − 2)(x − 3)(x + 1) = 0.
From the factorisation above.
2
x = 2 or x = 3 or x = −1.
Zero-product rule — a cubic has up to three roots.

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Start by testing factors of the constant term, not random numbers. For a constant of 6 you only ever need to try ±1, ±2, ±3 and ±6.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Determine the remainder when f(x) = x³ − 4x² + x + 6 is divided by (x − 1).
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f(1) = 1 − 4 + 1 + 6 = 4
2
Show that (x − 2) is a factor of f(x) = x³ − 4x² + x + 6.
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f(2) = 8 − 16 + 2 + 6 = 0, so by the factor theorem (x − 2) is a factor.
3
Factorise f(x) = x³ − 4x² + x + 6 completely.
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(x − 2)(x − 3)(x + 1)
4
Solve for x: x³ − 4x² + x + 6 = 0
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x = 2, x = 3 or x = −1
5
Determine the remainder when f(x) = 2x³ + x − 3 is divided by (x + 1).
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f(−1) = −2 − 1 − 3 = −6
6
If (x − 3) is a factor of f(x), what is f(3)?
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0
7
Which values should you test first when factorising a cubic with constant term 12?
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The factors of 12: ±1, ±2, ±3, ±4, ±6, ±12
8
Solve for x: x³ − 7x + 6 = 0
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f(1) = 0 → (x−1)(x²+x−6) = (x−1)(x+3)(x−2) → x = 1, 2 or −3