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 1Test 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 2Find 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 3Find 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 4Remainder 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 5Find 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 6Fully 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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