Just substitute x = a: Dividing P(x) by (x − a)? The remainder is P(a) (remainder theorem).
And (x − a) is a factor exactly when P(a) = 0 (factor theorem). No long division needed — just substitute.
IB-style question — find a remainder
Find the remainder when P(x) = x³ − 2x² + 5x − 1 is divided by (x − 2).
Step by step
- Remainder theorem: the remainder is P(2). Substitute x = 2.
- Evaluate.
Final answer
The remainder is 9.
IB-style question — confirm a factor
Show that (x − 1) is a factor of x³ − 6x² + 11x − 6.
Step by step
- Factor theorem: (x − 1) is a factor iff P(1) = 0.
- Evaluate.
Final answer
P(1) = 0, so (x − 1) is a factor.
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Each clue gives an equation: A given factor means P(value) = 0; a given remainder means P(value) = remainder. Each clue becomes an equation — solve them together for the unknowns.
IB-style question — find p and q
P(x) = x³ + px + q has (x − 1) as a factor, and leaves a remainder of −12 when divided by (x + 2).
Find p and q.
Step by step
- (x − 1) a factor ⇒ P(1) = 0.
- Remainder −12 on ÷(x + 2) ⇒ P(−2) = −12.
- Subtract the equations to eliminate q.
- So p = 1, then q = −1 − p.
Final answer
p = 1, q = −2.