The factor theorem says that if substituting x = a into a polynomial f(x) gives 0, then (x − a) is a factor. It is the fastest way to start factorising a cubic in Additional Mathematics. It appears whenever a question says “show that (x − 2) is a factor” or “given that one root is 3”.
This lesson opens polynomial factors and remainders and leads straight into the remainder theorem.
Why does a zero give a factor?
If f(a) = 0, then dividing f(x) by (x − a) leaves no remainder, so f(x) = (x − a) × (something). The value a is a root of the equation f(x) = 0, and the corresponding factor is (x − a).
Watch the sign. A root of 3 gives the factor (x − 3). A root of −2 gives the factor (x + 2), because x − (−2) = x + 2.
How do I find a root to test?
- Write down the constant term. For a polynomial with whole-number coefficients, any whole-number root divides the constant term.
- List its factors, positive and negative. For 6 that is ±1, ±2, ±3, ±6.
- Substitute each one in turn until f(a) = 0. Start with the small values, which are quicker to calculate.
- Turn the root into a factor by writing (x − a) and clearing any fractions.
When the highest-power coefficient is not 1, a root can be a fraction. For 2x³ + 3x² − 8x + 3, try ½ as well, because 2 × (½) − 1 = 0 gives the factor (2x − 1).
Worked example
Given f(x) = x³ − 2x² − 5x + 6, find all the linear factors.
Step 1, candidates: the constant is 6, so test ±1, ±2, ±3, ±6.
Step 2, test x = 1: f(1) = 1 − 2 − 5 + 6 = 0. So (x − 1) is a factor.
Step 3, keep testing: f(2) = 8 − 8 − 10 + 6 = −4, not zero. f(3) = 27 − 18 − 15 + 6 = 0, so (x − 3) is a factor. f(−2) = −8 − 8 + 10 + 6 = 0, so (x + 2) is a factor.
Step 4, stop at three: a cubic has at most three linear factors, so we are done.
Step 5, check the constant: (−1) × (−3) × 2 = 6, which matches the constant term of f(x), and the highest-power coefficient 1 matches too.
f(x) = (x − 1)(x − 3)(x + 2)
The mistake to watch for
A slip that appears when the factor has a coefficient: testing the wrong sign. Suppose a question asks whether (2x − 1) is a factor of 2x³ + 3x² − 8x + 3.
Mistaken working: f(−½) = 2(−⅛) + 3(¼) − 8(−½) + 3 = −0.25 + 0.75 + 4 + 3 = 7.5, so not a factor.
The student solved 2x + 1 = 0 instead of 2x − 1 = 0.
The correction is to set the factor equal to zero first. 2x − 1 = 0 gives x = ½, so test f(½) = 2(⅛) + 3(¼) − 8(½) + 3 = 0.25 + 0.75 − 4 + 3 = 0. (2x − 1) is a factor.
Check yourself
1. Show that (x − 2) is a factor of x³ − 3x² + x + 2.
Show answer
f(2) = 8 − 12 + 2 + 2 = 0, so (x − 2) is a factor.
2. Is (x + 1) a factor of x³ + 2x² − x − 3?
Show answer
Solve x + 1 = 0 to get x = −1. f(−1) = −1 + 2 + 1 − 3 = −1, which is not zero. (x + 1) is not a factor.
3. Find the value of k so that (x − 3) is a factor of x³ + kx² − x − 6.
Show answer
f(3) = 27 + 9k − 3 − 6 = 18 + 9k = 0, so k = −2. Check: 27 − 18 − 3 − 6 = 0.
Where this leads next
Once a root gives you a factor, the natural question is what is left over. Dividing a cubic by a linear factor shows how to finish the factorisation, and the mixed practice set tests both skills. The non-calculator working trainer and the quadratic structure explorer help with the arithmetic and with the quadratic that remains.
If the first move on an unfamiliar cubic is where you stall, that is a pattern our teachers work on in online one-to-one Additional Mathematics tuition.