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
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.
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
- 1Set x − 1 = 0, so x = 1.The remainder theorem needs the value that makes the divisor zero.
- 2f(1) = 1 − 4 + 1 + 6.Substitute.
- 3= 4.No long division needed at all.
3Factorising a cubic
- 1Try the factors of the constant 6: ±1, ±2, ±3, ±6.Any whole-number root must divide the constant term.
- 2f(2) = 8 − 16 + 2 + 6 = 0, so (x − 2) is a factor.Factor theorem.
- 3Divide: x³ − 4x² + x + 6 = (x − 2)(x² − 2x − 3).Use long division or inspection.
- 4Factorise the trinomial: x² − 2x − 3 = (x − 3)(x + 1).
- 5f(x) = (x − 2)(x − 3)(x + 1).Check the constant: (−2)(−3)(1) = 6. ✓
Quick check on any cubic factorisation: multiply the three constants together. They must give the original constant term.
- 1Test the factors of the constant 2: f(1) = 2 − 3 − 3 + 2 = −2, but f(−1) = −2 − 3 + 3 + 2 = 0, so (x + 1) is a factor.A whole-number root still has to divide the constant, even with a 2 in front of x³.
- 2Divide: 2x³ − 3x² − 3x + 2 = (x + 1)(2x² − 5x + 2).By inspection: 2x² gives the 2x³ and +2 gives the constant; the middle term must make the x² coefficient −3.
- 3Factorise the trinomial: 2x² − 5x + 2 = (2x − 1)(x − 2).Split the middle term: a × c = 4, and −4 + (−1) = −5.
- 4f(x) = (x + 1)(2x − 1)(x − 2)The roots are −1, 2 and ½ — and ½ is not a factor of 2. Testing the constant's factors found a root; the fraction came out of the quadratic.
4Solving a cubic equation
- 1(x − 2)(x − 3)(x + 1) = 0.From the factorisation above.
- 2x = 2 or x = 3 or x = −1.Zero-product rule — a cubic has up to three roots.
Start by testing factors of the constant term, not random numbers — any whole-number root has to divide it. For a monic cubic (just x³ in front) with integer coefficients and a constant of 6, the whole-number roots worth trying are ±1, ±2, ±3 and ±6. But once there is a number in front of x³, a root can be a fraction: 2x³ + x² + x − 1 = (2x − 1)(x² + x + 1) has the root ½, and neither 1 nor −1 works. A question like that normally gives you the factor to test.
Practice exercises
Work each one out, then click to reveal the answer.
- 1Determine 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 - 2Show 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. - 3Factorise f(x) = x³ − 4x² + x + 6 completely.
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(x − 2)(x − 3)(x + 1) - 4Solve for x: x³ − 4x² + x + 6 = 0
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x = 2, x = 3 or x = −1 - 5Determine the remainder when f(x) = 2x³ + x − 3 is divided by (x + 1).
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f(−1) = −2 − 1 − 3 = −6 - 6If (x − 3) is a factor of f(x), what is f(3)?
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0 - 7Which whole-number values should you test first when factorising a monic cubic with integer coefficients (just x³ in front) and constant term 12?
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The factors of 12: ±1, ±2, ±3, ±4, ±6, ±12 - 8Factorise f(x) = 2x³ + x² − 13x + 6 completely.
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f(2) = 16 + 4 − 26 + 6 = 0, so (x − 2) is a factor; 2x³ + x² − 13x + 6 = (x − 2)(2x² + 5x − 3) = (x − 2)(2x − 1)(x + 3) - 9Solve 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
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