3. Factors of polynomials
- Syllabus
- 0606–2028–2029
- Topic
- 3
- Level
- —
Learning objectives
3.1Use remainder and factor theorems• Know and use the remainder theorem and the factor theorem.3.2Find polynomial factors• Find factors of polynomials.• For a cubic polynomial, first obtain a product of a linear factor and a quadratic factor, for example by observation or algebraic long division.3.3Solve cubic equations• Solve cubic equations.
Use evaluation to find remainders and test factors
The remainder theorem replaces polynomial division by one substitution. When p(x) is divided by x−a, the constant remainder is p(a).
| Divisor | Value to substitute | Conclusion |
|---|---|---|
| x−a | x=a | remainder p(a) |
| mx−n | x=n/m | remainder p(n/m) |
| either linear divisor | its zero gives p(x)=0 | the divisor is a factor |
p(x)=6x3+x2−12x+5,p(21)=0⇒2x−1 is a factor
If coefficients are unknown, translate each statement into an equation: ‘x+2 is a factor’ gives p(−2)=0, while ‘remainder 40 on division by x−3’ gives p(3)=40. Solve the resulting simultaneous equations.
For the divisor mx−n, substitute n/m, not n. A zero remainder proves a factor; any non-zero value is only the remainder and does not prove a factor.
Reduce a cubic to find all its factors
Once one linear factor of a cubic is known, dividing by it leaves a quadratic. Factorising that quadratic can reveal the complete product of linear factors.
Write the cubic in descending powers, inserting a zero coefficient for any missing term. Divide by the known factor using algebraic long division, synthetic division when appropriate, or coefficient comparison. Multiply the divisor by the quotient to check that every coefficient is recovered.
2x3−3x2−8x−3=(x−3)(2x2+3x+1)
2x2+3x+1=(2x+1)(x+1)⇒p(x)=(x−3)(2x+1)(x+1)
A known factor is only the first step. Do not stop at ‘linear factor × quadratic’ when the quadratic can be factorised further, and do not lose a missing-power zero during division.
Solve a cubic from its factors
To solve p(x)=0, express the cubic as a product and use the zero-product rule: if a product is zero, at least one factor is zero.
p(x)=(x−2)(3x+2)(2x−5)=0⇒x=2,−32,25
| Remaining factor | Real solutions contributed |
|---|---|
| two distinct linear factors | two real roots |
| a repeated linear factor | one repeated real root |
| quadratic with D<0 | no additional real roots |
For (3x−1)(2x2+3x+5)=0, the quadratic has D=32−4(2)(5)=−31<0. It contributes no real roots, so the only real solution is x=1/3.
List roots from every factor and verify them in the original cubic. A cubic need not have three distinct real roots: repeated factors or an irreducible quadratic change the number of distinct real solutions.