Substitute one number and you know the remainder — no division needed.
Given P(x) = x³ − 4x² + 5x − 2
Find the remainder when it is divided by (x − 3).
The remainder theorem replaces the whole division with one substitution: P(3) = 27 − 36 + 15 − 2 = 4. B would mean (x − 3) is a factor, and a remainder of 4 says it is not.
divide P(x) by (x − a) → remainder = P(a)
divisor (x − 3), so a = 3
the value that makes it zero
P(3) = 27 − 36 + 15 − 2
substitute
= 4
the remainder
For (x − 3) you substitute +3. For (x + 3) you substitute −3. Use whichever value makes the bracket zero.
Remainder zero means it divides exactly — which is what a factor does.
P(2) = 8 − 16 + 10 − 2
test a = 2
= 0
nothing left over
so (x − 2) is a factor
and x = 2 is a root
Factor, root and zero all describe the same thing: (x − 2) is a factor, x = 2 is a root, and the graph crosses the axis at 2.
Only the factors of the constant term can be whole-number roots.
the factors of −2
| Try a | P(a) | Factor? |
|---|---|---|
| 1 | 1 − 4 + 5 − 2 = 0 | yes |
| 2 | 8 − 16 + 10 − 2 = 0 | yes |
| −1 | −1 − 4 − 5 − 2 = −12 | no |
Both 1 and 2 work, so (x − 1) and (x − 2) are both factors.
Synthetic division with the root 2.
| 2 | | 1 | −4 | 5 | −2 |
|---|---|---|---|---|
| bring down / multiply | — | 2 | −4 | 2 |
| add | 1 | −2 | 1 | 0 |
quotient = x² − 2x + 1
one degree lower
= (x − 1)²
a perfect square trinomial
P(x) = (x − 2)(x − 1)²
fully factored
The final 0 in the bottom row is the remainder — confirming (x − 2) divides exactly.
A favourite exam use of the theorem.
x³ + kx − 6 has factor (x − 2)
find k
P(2) = 0
the factor theorem
8 + 2k − 6 = 0
substitute
k = −1
2k = −2
Same move if the question gives a remainder instead: set P(a) equal to that remainder rather than to zero.
The rule never changes: substitute the value that makes the divisor zero. For 2x + 1 that value is a fraction. Find the remainder when 4x³ + 2x² − x + 3 is divided by 2x + 1.
2x + 1 = 0 → x = −12
not −1, and not +12
P(−12) = 4(−18) + 2(14) − (−12) + 3
substitute in brackets
= −12 + 12 + 12 + 3 = 72
the remainder
Given a remainder, set P(a) equal to it. x³ + kx² − x + 5 leaves remainder 3 when divided by x − 1. Find k.
P(1) = 1 + k − 1 + 5 = k + 5
x − 1 is zero at 1
k + 5 = 3 → k = −2
equal to the remainder, not to 0
P(x) = x³ + ax² + bx − 6 has x − 1 as a factor and leaves remainder −8 when divided by x + 1. Find a and b. Each condition gives one equation.
P(1) = 1 + a + b − 6 = 0 → a + b = 5
factor: remainder 0
P(−1) = −1 + a − b − 6 = −8 → a − b = −1
remainder −8
2a = 4 → a = 2, b = 3
add the two equations
P(x) = x³ + 2x² + 3x − 6
P(1) = 0 ✓, P(−1) = −8 ✓
The same two-equation idea finds the remainder on a quadratic divisor. Dividing by (x − 1)(x − 3) leaves a remainder of the form px + q. If P(1) = 2 and P(3) = 8, then p + q = 2 and 3p + q = 8, so p = 3, q = −1: the remainder is 3x − 1.
Every distractor in these questions comes from substituting +1 for x + 1, or −12 for 2x − 1. Say the zero aloud before you substitute: “x + 1 is zero at −1”.
Remainder questions are never long division. Find k so that something is a factor, find a remainder (often with huge powers like x40, where (−1)even = 1), or solve two conditions together. Zero the divisor, substitute, and set the result equal to 0 or to the given remainder.
Given P(x) = x³ + 2x² − 5x − 6
Find the remainder when divided by (x + 1).
x + 1 is zero at x = −1, so the remainder is P(−1) = −1 + 2 + 5 − 6 = 0. A zero remainder means (x + 1) is a factor — the factor theorem is just this special case. C evaluates at +1.
