IB Maths AA HL Topic 2 — Functions Paper 1 & 2 HL only ~8 min read

Polynomial Division

Polynomial division splits one polynomial into a quotient and a remainder — exactly like long division with whole numbers, but with x-terms instead of digits. Two methods get you there: long division (the workhorse, mechanical) and comparing coefficients (faster algebra). You’ll need both — long division when the divisor is large, comparing coefficients when the structure is simple.

📘 What you need to know

Division identity: P(x) = D(x) · Q(x) + R(x), where D is the divisor, Q the quotient, R the remainder.
Degree rules: if P has degree n and D has degree k ≤ n, then Q has degree n − k and R has degree less than k.
Long division: divide leading terms → multiply → subtract → repeat. Keep going until the working remainder has lower degree than the divisor.
Comparing coefficients: write P(x) ≡ D(x) · Q(x) + R(x) with unknown coefficients, expand, match each power of x.
Use placeholders for missing terms (write 0x², 0x etc.) — this avoids alignment mistakes.
Check by re-multiplying: D(x) · Q(x) + R(x) should give back the original polynomial.

The division identity

Polynomial division identity P(x) = D(x) · Q(x) + R(x)

This is just P/D = Q + R/D rearranged. Two facts about the degrees decide everything:

Quotient degree

deg Q = n − k

degree of P minus degree of D

Remainder degree

deg R < k

always less than the divisor’s degree

Examples: cubic ÷ linear → quadratic quotient, constant remainder. Quartic ÷ quadratic → quadratic quotient, linear (or constant) remainder. Knowing the degree of Q upfront lets you set up the comparing-coefficients method cleanly.

Long division — the workhorse

The mechanics are identical to numerical long division. At each step you handle the highest remaining power on the left.

🧭 Recipe — long division

Write the polynomial with placeholders for missing terms (0x³, 0x², etc.).
Divide the leading term of the polynomial by the leading term of the divisor — that’s the next term of the quotient.
Multiply the divisor by that term.
Subtract from the working polynomial — the leading term should cancel.
Bring down the next term and repeat from step 2.
Stop when the working remainder has lower degree than the divisor. What you have is the remainder; the running collection of terms is the quotient.

Comparing coefficients — the algebra method

When you know the degree of Q and R, write the identity with unknown coefficients, expand the right-hand side, then match each power of x on both sides.

Set-up — cubic ÷ linear ax³ + bx² + cx + d = (x − k)(px² + qx + r) + s

Expand the right-hand side, line up powers of x, then solve the system one equation at a time — start with the leading term, work down.

For exam questions, comparing coefficients usually beats long division — fewer alignment errors, cleaner working. Long division wins when the divisor is messy or you want to see the remainder appear naturally as you go.

Which method, when

Long division

divisor is large or messy

good for quartic ÷ quadratic, or any time the remainder structure is unclear

Comparing coefficients

divisor is linear or simple

faster, cleaner — works whenever you can predict the structure of Q and R

Worked examples

WE 1

Long division — cubic divided by linear

Use long division to find the quotient and remainder when 2x³ − x² − 8x + 5 is divided by (x − 2).

Step 1: 2x³ ÷ x = 2x² → first term of quotient 2x²(x − 2) = 2x³ − 4x² subtract: (2x³ − x²) − (2x³ − 4x²) = 3x² Step 2: 3x² ÷ x = 3x → next term 3x(x − 2) = 3x² − 6x subtract: (3x² − 8x) − (3x² − 6x) = −2x Step 3: −2x ÷ x = −2 → final term −2(x − 2) = −2x + 4 subtract: (−2x + 5) − (−2x + 4) = 1 quotient: 2x² + 3x − 2; remainder: 1 check: (x − 2)(2x² + 3x − 2) + 1 = 2x³ − x² − 8x + 5 ✓

WE 2

Comparing coefficients — cubic divided by linear

Find the quotient and remainder when 3x³ + 5x² − 4x + 7 is divided by (x + 2), using the comparing coefficients method.

Set up the identity 3x³ + 5x² − 4x + 7 = (x + 2)(ax² + bx + c) + r Expand RHS (x + 2)(ax² + bx + c) = ax³ + (b + 2a)x² + (c + 2b)x + 2c + r → ax³ + (b + 2a)x² + (c + 2b)x + (2c + r) Match powers of x x³: a = 3 x²: b + 2a = 5 → b = −1 x: c + 2b = −4 → c = −2 const: 2c + r = 7 → −4 + r = 7 → r = 11 quotient: 3x² − x − 2; remainder: 11 five lines of algebra — no aligning, no bringing down

WE 3

Long division — quartic divided by quadratic

Find the quotient and remainder when x⁴ + 3x³ − 2x² + 5x − 1 is divided by (x² + 1).

Step 1: x⁴ ÷ x² = x² x²(x² + 1) = x⁴ + x² subtract: (x⁴ + 3x³ − 2x²) − (x⁴ + x²) = 3x³ − 3x² Step 2: 3x³ ÷ x² = 3x 3x(x² + 1) = 3x³ + 3x subtract: (3x³ − 3x² + 5x) − (3x³ + 3x) = −3x² + 2x Step 3: −3x² ÷ x² = −3 −3(x² + 1) = −3x² − 3 subtract: (−3x² + 2x − 1) − (−3x² − 3) = 2x + 2 quotient: x² + 3x − 3; remainder: 2x + 2 remainder is degree 1 because the divisor is degree 2 — matches the rule deg R < deg D

WE 4

Express in identity form

Express 4x³ + 5x − 8 in the form (x − 1)Q(x) + R, where R is a constant.

Insert placeholder for missing x² term 4x³ + 0x² + 5x − 8 Long division by (x − 1) 4x³ ÷ x = 4x²; 4x²(x − 1) = 4x³ − 4x² remainder so far: 0x² − (−4x²) = 4x² 4x² ÷ x = 4x; 4x(x − 1) = 4x² − 4x remainder so far: 5x − (−4x) = 9x 9x ÷ x = 9; 9(x − 1) = 9x − 9 remainder so far: −8 − (−9) = 1 4x³ + 5x − 8 = (x − 1)(4x² + 4x + 9) + 1 forgetting the 0x² placeholder is the most common mistake — it shifts every subsequent column

WE 5

Division with multiple missing terms

Find the quotient and remainder when x⁴ + 2x³ − 5 is divided by (x² − 3).

Insert placeholders for missing x² and x terms x⁴ + 2x³ + 0x² + 0x − 5; divisor x² + 0x − 3 Step 1: x⁴ ÷ x² = x² x²(x² − 3) = x⁴ − 3x² subtract: (x⁴ + 2x³ + 0x²) − (x⁴ − 3x²) = 2x³ + 3x² Step 2: 2x³ ÷ x² = 2x 2x(x² − 3) = 2x³ − 6x subtract: (2x³ + 3x² + 0x) − (2x³ − 6x) = 3x² + 6x Step 3: 3x² ÷ x² = 3 3(x² − 3) = 3x² − 9 subtract: (3x² + 6x − 5) − (3x² − 9) = 6x + 4 quotient: x² + 2x + 3; remainder: 6x + 4 writing all the placeholders keeps the columns aligned — small effort, big payoff

WE 6

Find an unknown coefficient from the remainder

When the polynomial 2x³ + 5x² − 4x + c is divided by (x² + 3x − 1), the remainder is (x − 7). Find the value of c.

Step 1: Set up the identity — quotient is linear (degree 1) 2x³ + 5x² − 4x + c = (x² + 3x − 1)(ax + b) + (x − 7) Step 2: Expand RHS (x² + 3x − 1)(ax + b) = ax³ + (b + 3a)x² + (3b − a)x − b + (x − 7) → ax³ + (b + 3a)x² + (3b − a + 1)x + (−b − 7) Step 3: Match coefficients x³: a = 2 x²: b + 3a = 5 → b + 6 = 5 → b = −1 const: −b − 7 = c → 1 − 7 = c → c = −6 c = −6 the x-coefficient gives a consistency check: 3(−1) − 2 + 1 = −4 ✓

💡 Top tips

Always insert placeholders for missing terms (0x³, 0x², 0x). Five seconds of writing prevents most alignment errors.
Use comparing coefficients for linear divisors — it’s faster and the algebra is cleaner.
Use long division for quadratic or higher divisors — predicting the unknown structure gets messy fast.
Predict the degree of Q and R before starting. Knowing “Q will be quadratic, R will be linear” makes the comparing-coefficients set-up trivial.
Always verify by re-multiplying: D(x) · Q(x) + R(x) should give back the original polynomial. Catches arithmetic slips immediately.
Watch the signs when subtracting. The leading term must cancel — if it doesn’t, the multiplication step is wrong.
Stop when the working remainder has lower degree than the divisor. Going further by mistake is a common slip.

⚠ Common mistakes

Skipping the placeholders for missing terms. Misalignment cascades through every later step.
Sign errors in the subtraction step. Always rewrite the subtraction as “add the negative” — much harder to slip.
Stopping too early: if the working remainder still has degree ≥ deg D, the division isn’t finished.
Stopping too late: continuing once the working remainder is lower-degree than the divisor produces nonsense.
Forgetting the remainder structure when setting up comparing coefficients. For divisor of degree k, the remainder has at most k unknown coefficients.
Not multiplying out to verify. The 30 seconds it takes to check is much less than the time lost finding the slip later.
Using long division when comparing coefficients would be cleaner. Reach for the algebra method whenever the divisor is small.

Polynomial division is the engine for the next two notes. Factor and remainder theorem uses division by linear factors to find roots and remainders without doing the full division. Then graphs and roots brings everything together — once you know how to factorise a polynomial, you can sketch its graph quickly.

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