Polynomials and Factoring
The Remainder and Factor Theorems
Why the remainder on dividing by x - a is exactly f(a), how that turns into a factor test, and how the two together factor a cubic without guesswork.
What this page covers
- State and derive the remainder theorem from the division algorithm
- Use the factor theorem to test a candidate root in one substitution
- Choose sensible candidate roots rather than guessing
- Factor a cubic completely by combining the theorem with long division
The remainder theorem
Dividing a polynomial by a linear factor produces a constant remainder, and there is a way to obtain that constant without doing the division at all.
When f(x) is divided by (x - a), the remainder is f(a).
The derivation is three lines and the source gives it in full. Start from the division algorithm with divisor x - a. Because the divisor has degree 1, the remainder is a constant r:
The whole argument turns on substituting the one value that makes the divisor vanish. Whatever the quotient q(x) happens to be, it is multiplied by zero and disappears.
Worked example — the source's case
Find the remainder when f(x) = x3 - 5x2 + 7x - 9 is divided by x - 4.
The remainder is 3. No division was performed. Doing the long division would take five lines and produce the same 3, along with a quotient that was not asked for.
Watch the sign convention. Dividing by x + 2 means dividing by x - (-2), so the value to substitute is a = -2, not +2. Getting this backwards is the most common error with the theorem.
The factor theorem
A remainder of zero means the division came out exactly, which is what 'is a factor' means. Setting r = 0 in the remainder theorem gives a test.
(x - a) is a factor of f(x) if and only if f(a) = 0.
Both directions are useful. Forwards: if a factor is known, a root is known. Backwards, which is the direction actually used: if a root can be found, a factor has been found, and long division extracts it.
Choosing candidates rather than guessing
The source notes list, for x3 - 5x2 + 7x - 12 = 0, the values worth trying: ±1, ±2, 3, 4, 6. That list is not arbitrary, and knowing where it comes from turns trial and error into a finite search.
Suppose f(x) = anxn + … + a0 has integer coefficients and an integer root p. Then substituting gives a0 = -p(anpn-1 + … + a1), so p must divide a0. Any integer root divides the constant term.
For x3 - 5x2 + 7x - 12 the constant term is -12, whose divisors are ±1, ±2, ±3, ±4, ±6, ±12. That is twelve substitutions at worst, and in practice far fewer.
| a | f(a) | Root? |
|---|---|---|
| 1 | 1 - 5 + 7 - 12 = -9 | No |
| 2 | 8 - 20 + 14 - 12 = -10 | No |
| 3 | 27 - 45 + 21 - 12 = -9 | No |
| 4 | 64 - 80 + 28 - 12 = 0 | Yes |
With f(4) = 0, the factor theorem gives (x - 4) as a factor. Long division then supplies the rest:
The quadratic has discriminant 1 - 12 = -11 < 0, so it contributes no further real roots. The equation has exactly one real solution, x = 4.
When the leading coefficient is not 1, fractional roots become possible. For 2x3 + 5x2 - 14x + 3 the candidates are ±pq with p dividing 3 and q dividing 2 — that is ±1, ±3, ±12, ±32. The root turns out to be 32, which no search over integers alone would ever have found.
Worked example: factoring a cubic completely
Solve x3 + 4x2 + 2x - 4 = 0
The constant term is -4, so the candidates are ±1, ±2, ±4.
So (x + 2) is a factor. Dividing:
The remaining quadratic does not factor over the integers. The source completes the square rather than reaching for the formula, which is neat here:
The three roots should sum to -4, the negative of the x2 coefficient: -2 + (-1 + √3) + (-1 - √3) = -4. The surds cancel, as they must for a conjugate pair.
That last check is worth adopting as a habit. For a cubic x3 + bx2 + cx + d the roots sum to -b and multiply to -d, so both can be verified in a few seconds without re-doing any of the work.
Why irrational roots arrive in pairs
In the example above the two surd roots were -1 ± √3 — a conjugate pair. That is not a coincidence.
A quadratic with rational coefficients has roots -b ± √b2 - 4ac2a. The only irrational part is the radical, and it enters with both signs. So an irrational root of the form p + q√n is always accompanied by p - q√n.
If a cubic with rational coefficients has one rational root, the other two are either both rational or a conjugate surd pair or a conjugate complex pair. There is no way to have exactly one irrational root and one rational one among the remaining two.
The same reasoning applies to complex roots, which is why the notes' cubic x3 - 8 = 0 produces x = 2 together with the conjugate pair -1 ± i√3.
When the method does not apply
| Situation | What happens | What to do instead |
|---|---|---|
| No rational root exists | Every candidate fails | Use a numerical method, or the cubic formula if exactness is required |
| Coefficients are not integers | The divisor-of-the-constant argument does not apply | Clear denominators first to get integer coefficients |
| Divisor is quadratic or higher | The theorem says nothing about the remainder | Do the long division |
| The quotient is wanted, not just the remainder | The theorem gives only the remainder | Do the long division |
It is also worth knowing that failure to find a rational root proves nothing about real roots. Every cubic with real coefficients has at least one real root, because its graph runs from -∞ to +∞ and must cross the axis. That root may simply be irrational.
Common mistakes
| Mistake | Correct | Why |
|---|---|---|
| Testing f(2) for the divisor x + 2 | Test f(-2) | x + 2 = x - (-2), so a = -2 |
| Guessing candidates at random | Test divisors of the constant term | An integer root must divide the constant |
| Ignoring fractional candidates when an ≠ 1 | Also test ±pq | 2x3 + 5x2 - 14x + 3 has root 32 |
| Stopping after finding one root | Divide out and factor the quotient | A cubic has three roots counted with multiplicity |
| Concluding 'no real roots' after a failed search | Every real cubic has a real root | The root may be irrational and invisible to this method |
| Using the theorem with a quadratic divisor | It applies only to linear divisors | The remainder need not be constant |
Frequently asked questions
Why is the remainder a constant when dividing by x - a?
Because the remainder must have degree strictly less than the divisor. x - a has degree 1, so the remainder has degree 0 — a constant.
What if the divisor is 2x - 3 rather than x - a?
Rewrite it as 2(x - 32). The root is x = 32, so test f(32). The factor theorem is about the root, and the leading coefficient of the divisor only rescales the quotient.
Which numbers should I test?
The integer factors of the constant term, positive and negative. For x3 - 5x2 + 7x - 12 the constant is -12, so try ±1, ±2, ±3, ±4, ±6, ±12. If the leading coefficient is not 1, also try fractions whose numerator divides the constant and whose denominator divides the leading coefficient.
Does every cubic have a nice root?
No. Many have no rational root at all, and then this method finds nothing. It is worth a few substitutions because it is cheap, but a failure to find one is not evidence that the cubic has no real roots — every real cubic has at least one.
Source. Handwritten teaching notes, Week 6, pages 7-8, and Week 7, page 1.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.