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Grade 10 / High 1 (age 15-16)

Remainder & Factor Theorem

Remainder Theorem & Factor Theorem

The remainder theorem gives the remainder without doing the division. When f(x) is divided by x − a, the remainder is exactly f(a), so one substitution settles it. If that remainder is zero, x − a is a factor, and this factor theorem unlocks higher-degree polynomials. Here you can slide a to see how the remainder changes and when it reaches zero.

Remainder Theorem — Plug In to Get the Remainder

The remainder of P(x) divided by (x − a) equals P(a). A powerful shortcut: just substitute, no division needed.

2
💡 Why It Works
①f(x) = (x−a)·Q(x) + R; substitute x = a
②f(a) = 0·Q(a) + R = R
③So R = f(a)
Remainder Theorem Formulas
Remainder Theorem
remainder of P(x) ÷ (x − a) = P(a)
The substitution value IS the remainder
Generalization
remainder of P(x) ÷ (ax + b) = P(-ba)
Substitute the root of ax + b = 0
Factor Theorem — P(a)=0 means (x−a) is a factor

Remainder zero means the graph meets the x-axis. P(a) = 0 ⇔ (x − a) is a factor of P(x).

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🔑 Heart of the Factor Theorem
①P(a) = 0 ⟺ (x − a) is a factor
②Graph: x-intercepts identify factors
③First step in factoring high-degree polys = find integer roots
Using the Factor Theorem
Factor Theorem
P(a) = 0 ⟺ (x − a) is a factor
Zero remainder = divides exactly = factor
📐 High-degree Factoring Strategy
①Try divisors of the constant (±1, ±2, ±3, ...)
②Find a value with P(a) = 0
③Use synthetic division to find the quotient
④Factor the quotient further
Work It Out
Example 1
Find the remainder when P(x) = x³ − 2x + 5 is divided by x − 2.
1
By the remainder theorem, the remainder on division by x − a is P(a) (here a = 2).
R = P(2)
2
Substitute x = 2.
P(2) = 2³ − 2·2 + 5 = 8 − 4 + 5 = 9
The remainder is 9
The remainder on division by x − a is just P(a).
Example 2
If P(x) = x³ + ax + 2 is divisible by x + 1, find the constant a.
1
By the factor theorem, divisibility by x + 1 means P(−1) = 0.
P(−1) = 0
2
Substitute x = −1 and solve.
(−1)³ + a(−1) + 2 = 1 − a = 0 ⇒ a = 1
a = 1
Divisible means remainder 0, that is, the factor theorem P(a) = 0.
Wrap-up
Remainder Theorem
R = P(a) when divided by (x − a)
One substitution gives the remainder
Factor Theorem
P(a) = 0 ⟹ P(x) = (x − a) · Q(x)
Zero remainder ⇒ 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?
0
1
2
3
4
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
🎯 Exam Points
①Remainder theorem: works for (ax+b) too — plug in the root
②Factor theorem: try divisors of the constant first
③Combine with synthetic division for high-degree factoring
④If f(x) = (x−a)(x−b)·…, then f(a)=f(b)=0
⑤Reverse: knowing R fixes coefficients
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