IB Maths AA HL Topic 2 — Functions Paper 1 & 2 HL only ~7 min read

Factor & Remainder Theorem

Two shortcuts that save you the trouble of doing a full polynomial division. The remainder theorem tells you the remainder when P(x) is divided by (ax − b) just by evaluating P(b/a). The factor theorem is the special case where that remainder is zero — meaning (ax − b) is a factor. Together they let you find roots, factorise polynomials, and solve for unknown coefficients in seconds.

📘 What you need to know

Remainder theorem: when P(x) is divided by (ax − b), the remainder is P(b/a).
Factor theorem: (ax − b) is a factor of P(x) ⟺ P(b/a) = 0. (The factor theorem is just the remainder theorem with remainder 0.)
Roots and factors are paired: x = b/a is a root ⟺ (ax − b) is a factor.
Finding unknown coefficients: substitute b/a into P(x) and set equal to the remainder (or 0 for a factor).
Rational root candidates: for P(x) with integer coefficients, possible rational roots are p/q, where p divides the constant term and q divides the leading coefficient.
Fully factorising: find one factor with the factor theorem, divide it out, then factorise the resulting quotient (often a quadratic you can solve directly).

The factor theorem — testing for factors

Factor theorem (ax − b) is a factor of P(x) ⟺ P(ba) = 0

To test whether (ax − b) is a factor: set the bracket equal to zero, solve for x, and substitute that value into P(x). Zero means yes; anything else means no.

Monic linear: (x − k)

test by computing P(k)

simplest case — substitute the root

Non-monic: (ax − b)

test by computing P(b/a)

always set the bracket to zero first

The remainder theorem — finding remainders fast

Remainder theorem remainder when P(x) is divided by (ax − b) is P(ba)

🤔 Why does this work?

Write the division identity: P(x) = (ax − b) · Q(x) + r. The remainder is a constant because the divisor is linear. Substitute x = b/a: the bracket becomes zero, the entire (ax − b) · Q(x) part vanishes, and you’re left with r. So P(b/a) = r.

One substitution gives you the remainder — no long division required. The factor theorem is just this with r = 0.

Finding unknown coefficients

The most common Paper 1 question type: a polynomial with one or two unknowns, plus information about its factors or remainders. Each piece of information gives you one equation; solve the system.

🧭 Recipe — finding unknowns from factor/remainder info

For each linear divisor (ax − b), compute x = b/a.
Substitute that value into P(x) — this gives an expression in the unknowns.
Set it equal to the given remainder (or 0 if it’s a factor).
Solve the resulting equation(s). One unknown needs one equation; two unknowns need two.

Fully factorising a polynomial

If you can find one factor, you can usually find the rest. The trick is knowing where to look.

Rational root theorem (mini version): if P(x) has integer coefficients, any rational root p/q must satisfy: p divides the constant term, q divides the leading coefficient. So for 2x³ − 5x² + 1, candidates are ±1, ±1/2.

🧭 Recipe — fully factorise a cubic (or higher)

List rational candidates ±p/q.
Test each by computing P(±p/q). The first one that gives 0 is a root.
Write the corresponding factor (use (qx − p) for root p/q).
Divide the polynomial by that factor (long division or comparing coefficients).
Factorise the resulting quotient — usually a quadratic you can handle directly.
Write the polynomial as a product of all factors found.

Worked examples

WE 1

Test whether (x + 4) is a factor

Determine whether (x + 4) is a factor of P(x) = x³ + 6x² + 5x − 12.

Step 1: Set x + 4 = 0 → x = −4 Step 2: Compute P(−4) P(−4) = (−4)³ + 6(−4)² + 5(−4) − 12 = −64 + 96 − 20 − 12 = 0 P(−4) = 0, so (x + 4) is a factor a “yes” verdict for the factor theorem means the remainder is zero — equivalently, x = −4 is a root

WE 2

Find an unknown coefficient using the factor theorem

Given that (x − 3) is a factor of P(x) = x³ + ax² − 7x + 12, find the value of a.

Step 1: x − 3 = 0 → x = 3 Step 2: Apply factor theorem — P(3) = 0 P(3) = 27 + 9a − 21 + 12 = 0 9a + 18 = 0 a = −2 a = −2 verify: P(x) = x³ − 2x² − 7x + 12; P(3) = 27 − 18 − 21 + 12 = 0 ✓

WE 3

Find a remainder using the remainder theorem

Find the remainder when P(x) = 4x³ − 3x² + 5x − 8 is divided by (x − 2).

Apply the remainder theorem with x = 2 P(2) = 4(8) − 3(4) + 5(2) − 8 = 32 − 12 + 10 − 8 = 22 remainder = 22 no long division needed — one substitution gives the answer

WE 4

Remainder when divisor is non-monic

Find the remainder when P(x) = 3x³ + 4x² − 6x + 1 is divided by (3x + 2).

Step 1: Set 3x + 2 = 0 → x = −2/3 Step 2: Compute P(−2/3) P(−2/3) = 3(−8/27) + 4(4/9) − 6(−2/3) + 1 = −8/9 + 16/9 + 4 + 1 = 8/9 + 5 = 8/9 + 45/9 = 53/9 remainder = 53/9 always set the bracket to zero first — substituting x = 2/3 instead of x = −2/3 is the most common slip

WE 5

Find an unknown using the remainder theorem

The polynomial P(x) = 2x³ + ax² − 5x + 6 leaves a remainder of 4 when divided by (x + 1). Find the value of a.

Step 1: Set x + 1 = 0 → x = −1 Step 2: P(−1) = remainder = 4 P(−1) = 2(−1)³ + a(−1)² − 5(−1) + 6 = −2 + a + 5 + 6 = a + 9 a + 9 = 4 → a = −5 a = −5 verify: P(−1) = −2 + (−5) + 5 + 6 = 4 ✓

WE 6

Fully factorise a cubic

Fully factorise f(x) = 2x³ − 3x² − 11x + 6.

Step 1: Rational candidates ±p/q p ∈ {1, 2, 3, 6} (factors of 6); q ∈ {1, 2} candidates: ±1, ±2, ±3, ±6, ±1/2, ±3/2 Step 2: Test until f = 0 f(−2) = −16 − 12 + 22 + 6 = 0 ✓ → (x + 2) is a factor Step 3: Divide f(x) by (x + 2) 2x³ − 3x² − 11x + 6 = (x + 2)(2x² + bx + c) compare: 2x²: 2 ✓; x²: b + 4 = −3 → b = −7; const: 2c = 6 → c = 3 quotient: 2x² − 7x + 3 Step 4: Factorise the quadratic 2x² − 7x + 3 = (2x − 1)(x − 3) f(x) = (x + 2)(2x − 1)(x − 3) cubic with three real roots — confirm by expanding back, or check f(3) = 54 − 27 − 33 + 6 = 0 ✓

💡 Top tips

Always set the bracket equal to zero first: for (3x − 2), the test value is x = 2/3, not 3/2 or −2/3.
The factor theorem is one-way only when remainder ≠ 0: (ax − b) is not a factor. Don’t conclude anything stronger.
Pair facts with equations: each “factor” or “remainder” piece of information gives exactly one equation in the unknowns.
Check rational roots before resorting to other methods: ±p/q with p dividing the constant term, q dividing the leading coefficient.
Test small candidates first: ±1, ±2, ±1/2 catch most exam-friendly polynomials.
After finding one factor, divide and factorise the quotient. The quotient is usually a quadratic — finish with the quadratic formula or factoring.
Verify by expanding: a fully factorised polynomial should expand back to the original. Quick check, big confidence boost.

⚠ Common mistakes

Substituting the wrong sign: for (x + 5), the test value is x = −5. Always solve the bracket = 0 first.
Treating non-monic divisors as monic: for (2x − 3), use x = 3/2, not x = 3.
Concluding “factor” from a small non-zero result. The remainder must be exactly zero — anything else means it’s not a factor.
Stopping after finding one root when fully factorising. Cubics have three roots (counting multiplicity); keep going until you’ve factorised everything.
Missing repeated roots. The factor theorem only confirms a root is present — it doesn’t tell you the multiplicity. Check by re-substituting after dividing.
Sign errors with negative roots: (−2)³ = −8, not 8. Be especially careful with odd powers.
Forgetting to write “is/is not a factor” as the conclusion. The question often asks for a verbal answer, not just a number.

The factor theorem turns root-finding into a substitution problem — and root-finding is the gateway to graph-sketching. The next note, Graphs & Roots of Polynomial Functions, uses everything you’ve learned here to sketch cubics, quartics, and beyond from their factorised form.

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