The remainder theorem is a shortcut that gives the remainder of a polynomial division without doing the division.
The remainder when P(x) is divided by x − a equals P(a), so putting a in place of x once gives the remainder.
If that remainder is 0, P(x) is exactly divisible by x − a, which means x − a is a factor of P(x). This is the factor theorem, and it is the first step in factorising cubics and beyond.
In the first picture, as you change a, the height of the point (a, P(a)) on the graph reads as the remainder; in the second, factors can be read off where the graph meets the x-axis.
Remainder theorem: substitute to get the remainder
The remainder when P(x) is divided by x − a is equal to P(a). You can find it just by substituting a value, without dividing.
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The graph of P(x) = x³ − 2x² + x − 2, with the height P(a) at x = a read as the remainder on dividing by x − a
💡 Why the remainder theorem works
①Dividing P(x) by x − a gives P(x) = (x − a)·Q(x) + R
②Put x = a: P(a) = 0·Q(a) + R = R
③So the remainder R equals P(a)
④In the picture, a = 2 gives P(2) = 0, so this polynomial is exactly divisible by x − 2
⑤For division by ax + b, the remainder is P(−b/a), as in the formula below
The remainder theorem
Remainder theorem
remainder of P(x) ÷ (x − a) = P(a)
The value you substitute gives the remainder
Dividing by ax + b
remainder of P(x) ÷ (ax + b) = P(-ba)
Substitute the solution of ax + b = 0
Factor theorem: if P(a) = 0, it is a factor
A remainder of 0 means the graph meets the x-axis at that point. If P(a) = 0, then x − a is a factor of P(x).
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The graph of P(x) = (x − a)(x² + 1) meeting the x-axis at x = a, showing that x − a is a factor
🔑 The heart of the factor theorem
①P(a) = 0 and 'x − a is a factor of P(x)' say the same thing
②On a graph, the x-coordinate of a point where it meets the x-axis is the a of a factor x − a
③The polynomial in the picture meets the x-axis only once, because x² + 1 is never 0
④Factorising a cubic or higher starts with finding one root like this
Using the factor theorem
Factor theorem
P(a) = 0 ⟺ x − a is a factor of P(x)
Remainder 0 means exact division, and exact division means a factor
📐 Steps for factorising higher-degree polynomials
①Take the divisors of the constant term as candidate roots (±1, ±2, ±3, …)
②Substitute them to find a value with P(a) = 0
③Divide P(x) by x − a with synthetic division to get the quotient
④Factorise the quotient further
Worked examples and unit test
Example 1
Find the remainder when P(x) = x³ − 2x + 5 is divided by x − 2.
1
By the remainder theorem, the remainder on dividing by x − 2 is P(2).
R = P(2)
2
Substitute x = 2.
P(2) = 2³ − 2·2 + 5 = 8 − 4 + 5 = 9
▸ 9
Dividing with synthetic division also leaves 9 in the last cell. The remainder theorem shortens that work to a single substitution.
Example 2
P(x) = x³ + ax + 2 is exactly divisible by x + 1. Find the constant a.
1
x + 1 = x − (−1), so by the factor theorem P(−1) = 0.
P(−1) = 0
2
Substitute x = −1 and solve.
(−1)³ + a(−1) + 2 = 1 − a = 0 ⇒ a = 1
▸ a = 1
With a = 1, P(x) = x³ + x + 2 = (x + 1)(x² − x + 2), so x + 1 really is a factor.
Summary
Remainder theorem
remainder of P(x) ÷ (x − a): R = P(a)
One substitution fixes the remainder
Factor theorem
P(a) = 0 ⟹ P(x) = (x − a) · Q(x)
A remainder of 0 means x − a is a factor
Grade-10 education-office assessment type
P(x) leaves remainder 3 on division by x − 1 and −3 on division by x + 2. If the remainder on division by (x − 1)(x + 2) is R(x) = ax + b, what is ab?
①0
②1
③2
④3
⑤4
▸ ③ 2
1
By the remainder theorem, P(1) = R(1) and P(−2) = R(−2) give two equations.
a + b = 3, −2a + b = −3
2
Solving the system: 3a = 6, so a = 2, b = 1.
a = 2, b = 1 ⇒ ab = 2
🎯 Points that often come up
①For division by x − a substitute a; for ax + b substitute −b/a
②When hunting for roots with the factor theorem, start with divisors of the constant term
③Combine with synthetic division to factorise higher-degree polynomials
④If P(x) is divisible by (x − a)(x − b), then P(a) = 0 and P(b) = 0 both hold
⑤Use a given remainder in reverse to find an unknown coefficient