high school 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. Slide a and how the remainder changes and when it reaches zero appear.

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.

value of a2

💡 Why It Works

①f(x) = (x−a)·Q(x) + R; substitute x = a

②f(a) = 0·Q(a) + R = R

③So R = f(a)

④This sketch is (x − a) only. The remainder P(−b/a) for (ax + b) is in the formula box

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).

root a3

🔑 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.

By the remainder theorem, the remainder on division by x − a is P(a) (here a = 2).

R = P(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.

By the factor theorem, divisibility by x + 1 means P(−1) = 0.

P(−1) = 0

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?

①0

②1

③2

④3

⑤4

▸ ③ 2

By the remainder theorem, P(1) = R(1) and P(−2) = R(−2) give two equations.

a + b = 3, −2a + b = −3

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