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Factors of Polynomials

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Cambridge (CIE) IGCSE · Additional Maths

Factors of Polynomials

The Big Idea: Any polynomial can be broken down (factorised) into a chain of simpler pieces multiplied together — and the fastest way to find those pieces is to smartly guess a root, confirm it with the factor theorem, then peel it off using polynomial division. Repeat until nothing more will factorise.

Summary — What This Chapter Covers

Operations with polynomials — what counts as a polynomial, and how to expand brackets with more than two terms/factors.
Polynomial division — dividing one polynomial by another (linear or quadratic) using the "bus stop" method, producing a quotient and a remainder.
The Factor Theorem — a fast test for whether (x − p) is a factor of a polynomial, without doing full division.
The Remainder Theorem — a way to find the remainder of a division instantly, and to solve for unknown coefficients.
Solving cubic equations — combining the factor theorem + division + quadratic formula to fully factorise and solve cubics.

1. Operations with Polynomials

What actually IS a polynomial?

A polynomial is an algebraic expression made of a finite number of terms, where every power (index) on the variable is a non-negative integer (0, 1, 2, 3, ...). That's the whole test — no fractional powers, no negative powers.

Think of it like a guest list rule at a party: only whole, non-negative numbers get an invite as exponents. Square roots (fractional powers) and "over x" terms (negative powers) are turned away at the door.

✅ These ARE polynomials 3x + 5 2x²y − 4y + 6 8 (yes — a plain number is a "constant" polynomial!)

❌ These are NOT polynomials √x (= x^½, a fractional index) 5x⁻³ (negative index) 3/x (= 3x⁻¹, negative index)

Why This Matters

Recognising polynomials matters because every rule in this chapter — division, factor theorem, remainder theorem — only applies to polynomials. If you're handed something with a √x or 1/x in it, these tools don't directly apply.

Expanding brackets with more than 2 factors

You already know FOIL for two brackets like (x + 2)(x + 3). But FOIL is really just a shortcut for a much more general rule: every term in the first bracket multiplies every term in every other bracket.

When you have three or more brackets, the trick is to only ever work with a pair at a time. Expand two of them fully first, simplify, then multiply that result by the next bracket.

General Rule (a + b)(x + y + z) = a(x + y + z) + b(x + y + z) = ax + ay + az + bx + by + bz In plain words: distribute EVERY term in bracket 1 across EVERY term in bracket 2 — nothing skipped.

Worked Example — Three Brackets

Expand (x − 4)(x + 3)(x² + x − 1)

Step 1 — pick a pair: expand (x − 4)(x + 3) first → x² − x − 12

Step 2 — multiply that result by the third bracket:
(x² − x − 12)(x² + x − 1)
= x²(x² + x − 1) − x(x² + x − 1) − 12(x² + x − 1)
= x⁴ + x³ − x² − x³ − x² + x − 12x² − 12x + 12

Step 3 — collect like terms:
= x⁴ − 14x² − 11x + 12

Common Mistake

Students often try to expand all three brackets "at once" and lose terms along the way. Always collapse to two brackets first, simplify fully, THEN bring in the next one. Rushing this step is the #1 source of sign errors in this chapter.

Practice Question 1

Expand and simplify: 4x(x − 3y)(3x + 2y − 5)

Practice Question 2

Is 6x⁻² + 5x + 1 a polynomial? Explain why or why not.

2. Polynomial Division

Polynomial division is exactly the same idea as the long division you did with numbers in primary school (the "bus stop" method) — just applied to algebraic expressions instead of digits.

The Number Analogy

Remember dividing 4836 ÷ 39? You worked out how many times 39 fits into the leading digits, subtracted, brought the next digit down, and repeated. Polynomial division is identical — except instead of "how many times does 39 fit," you ask "what do I multiply the highest power of the divisor by to match the highest power of what's left?"

What You're Building Dividend = (Divisor × Quotient) + Remainder
e.g. x³ + x² − x − 1 = (x + 1)(x² − 1)
x³ + 2x² + 3x + 4 = (x − 1)(x² + 3x + 6) + 10

The method, step by step

Divide the highest power term of what remains by the highest power term of the divisor. Write that in the answer row.
Multiply that answer by the whole divisor, and subtract it from what you had.
Bring down the next term and repeat.
Keep going until you're left with something of a lower degree than the divisor — that's your remainder.

Worked Example — Linear Divisor

Divide f(x) = x³ + 6x² − 9x − 14 by (x − 2)

Step 1: x³ ÷ x = x². Multiply: x²(x − 2) = x³ − 2x². Subtract: leaves 8x² − 9x

Step 2: Bring down −14 → 8x² − 9x − 14. Divide: 8x² ÷ x = 8x. Multiply: 8x(x−2) = 8x² − 16x. Subtract: leaves 7x − 14

Step 3: 7x ÷ x = 7. Multiply: 7(x − 2) = 7x − 14. Subtract: leaves 0

Result: x³ + 6x² − 9x − 14 = (x − 2)(x² + 8x + 7), remainder 0 — so (x − 2) IS a factor!

Dividing by a quadratic divisor

Sometimes you divide by a quadratic like (x² + qx + r) instead of a linear term. The process is identical — but since x² can't divide neatly into a single x term, the leftover remainder can be of linear form (rx + s) rather than just a constant. (It's still possible for r = 0, meaning the remainder ends up being just a constant.)

Worked Example — Quadratic Divisor

Find the remainder when x⁴ + 4x³ − x + 1 is divided by x² − 2x

Write the missing x² term as 0x² to keep place-value alignment: x⁴ + 4x³ + 0x² − x + 1

x⁴ ÷ x² = x². Multiply & subtract → 6x³ + 0x²

6x³ ÷ x² = 6x. Multiply & subtract → 12x² − x

12x² ÷ x² = 12. Multiply & subtract → 23x + 1

Result: quotient = x² + 6x + 12, remainder = 23x + 1

Common Mistake

Forgetting to write in "placeholder" terms with a coefficient of 0 (like 0x³ or 0x²) for missing powers. If you skip a power, your columns misalign and every subsequent subtraction goes wrong. Always list every power from highest to lowest, even if the coefficient is zero.

Practice Question

Divide f(x) = x⁴ + 11x² − 1 by (x + 3), and state the remainder.

3. Factor & Remainder Theorem

The Factor Theorem

The factor theorem is the single most useful shortcut in this whole chapter. Instead of doing a full polynomial division to check whether (x − p) is a factor, you just plug x = p into the polynomial. If the result is zero, you've found a factor — instantly.

Factor Theorem (i) If f(p) = 0, then (x − p) is a factor of f(x)
(ii) If (x − p) is a factor of f(x), then f(p) = 0 In plain words: substituting the root gives zero — it's a two-way street, both directions are true.

Worked Example

Show that (x − 2) is a factor of f(x) = x³ + 6x² − 9x − 14

By the factor theorem, if (x − 2) is a factor, then f(2) must equal 0.

f(2) = (2)³ + 6(2)² − 9(2) − 14 = 8 + 24 − 18 − 14 = 0

Since f(2) = 0, (x − 2) is confirmed to be a factor.

Which Values of p Do I Try?

In exams, the value of p that gives f(p) = 0 is almost always a small integer close to zero. Try p = 1 and p = −1 first, then 2 and −2, then 3 and −3. It's very unlikely you'll ever need to go beyond that. Also — if there's no constant term (d = 0), then x = 0 is automatically a root, since f(0) = 0 trivially!

The Remainder Theorem

The factor theorem is actually just a special case of a bigger, more general rule called the remainder theorem. It tells you the leftover remainder of ANY polynomial division — even when the divisor ISN'T a factor — without doing the division at all.

Remainder Theorem When f(x) is divided by (x − a), the remainder is f(a).
Formally: f(x) = (x − a)Q(x) + f(a) Q(x) is the quotient (the "answer on top"). f(a) is what's left over. When f(a) = 0, there's no remainder — that's exactly the factor theorem!

Extended Version (for divisors like ax − b) If f(x) is divided by (ax − b), the remainder is f(b/a) Just find the x-value that makes the divisor zero (set ax − b = 0), then substitute — it doesn't have to be a whole number.

Worked Example — Finding Unknown Coefficients

Given: p(x) = 8x⁴ + ax² + bx − 1. When p(x) ÷ (x − 1), remainder = 9. When p(x) ÷ (2x − 1), remainder = 1. Find a and b.

Using x − 1 = 0 → x = 1:
p(1) = 9 → 8 + a + b − 1 = 9 → a + b = 2

Using 2x − 1 = 0 → x = ½:
p(½) = 1 → 8(1/16) + a(1/4) + b(1/2) − 1 = 1 → a + 2b = 6

Solving simultaneously:
a + b = 2
a + 2b = 6
Subtracting: b = 4, so a = 2 − 4 = −2

Answer: a = −2, b = 4

Common Mistake

Mixing up the sign: for divisor (x − a), you substitute x = a (positive a), NOT x = −a. Students frequently flip this sign under exam pressure. Double check: if the divisor is (x + 3), that's (x − (−3)), so you substitute x = −3.

Practice Question

Find the remainder when g(x) = 2x³ − 7x² + 9 is divided by (x − 3). Is (x − 3) a factor of g(x)?

4. Solving Cubic Equations

A cubic equation has the form ax³ + bx² + cx + d = 0. Solving it means fully factorising the cubic first, then setting each factor to zero. This is where everything from this chapter comes together: factor theorem to find the first root, polynomial division to peel it off, then the quadratic formula (or factorising) to finish the job.

The 5-Step Method 1. Use the factor theorem: find p such that f(p) = 0
2. Use polynomial division: divide f(x) by (x − p)
3. Write f(x) = (x − p)(ax² + bx + c)
4. Factorise the quadratic if possible (or use the quadratic formula)
5. Set every factor equal to zero to find all solutions

Worked Example — Full Cubic Solve

Solve: x³ − 10x² + 12x + 8 = 0

STEP 1 — Factor theorem:
f(1) = 11 ≠ 0 f(−1) = −15 ≠ 0 f(2) = 0 ✓
So (x − 2) is a factor.

STEP 2 & 3 — Divide (by inspection):
f(x) = (x − 2)(ax² + bx + c). Since x³ only comes from x × ax², a = 1. Since the constant term only comes from −2 × c, c = −4. Matching x-coefficients: 12 = c − 2b → b = (−4 − 12)/2 = −8
So f(x) = (x − 2)(x² − 8x − 4)

STEP 4 — the quadratic doesn't factorise nicely, so use the quadratic formula:
x² − 8x − 4 = 0 → x = (8 ± √(64+16))/2 = 4 ± 2√5

STEP 5 — All solutions:
x = 2, x = 4 + 2√5, x = 4 − 2√5

How Many Solutions Can a Cubic Have?

A cubic can have 1, 2, or 3 real solutions. It has 3 if the quadratic factor splits into two more real roots; 1 if the quadratic has no real roots (negative discriminant); and cubics can also have repeated roots — e.g. (x − 2)²(x + 1) = 0 gives x = 2 (repeated, "double root") and x = −1.

Common Mistake

Stopping after finding ONE root and forgetting the cubic has up to two more hiding inside the quadratic factor! Always push through to Step 4/5 — a fully factorised cubic should give you (up to) THREE roots, not one.

Practice Question

Fully factorise x³ + 4x² − 11x − 30, then solve x³ + 4x² − 11x − 30 = 0.

What to Memorise

Polynomial

Expression with a finite number of terms, all indices non-negative integers.

Factor Theorem

f(p) = 0 ↔ (x − p) is a factor of f(x).

Remainder Theorem

Dividing f(x) by (x − a) leaves remainder f(a). If divisor is (ax − b), remainder is f(b/a).

Division Identity

f(x) = (x − a)·Q(x) + f(a), where Q(x) is the quotient.

Quadratic Divisor Rule

Dividing by x² + qx + r can leave a remainder of linear form (rx + s), not just a constant.

Cubic Solving Steps

Factor theorem → polynomial division → factorise/quadratic formula on remaining quadratic → set every factor to zero.

First Values to Try

p = 1, −1, 2, −2, 3, −3 (small integers, in that order) when hunting for roots.

Repeated Roots

A factor like (x − 2)² gives a repeated (double) root — the cubic can still have up to 3 solutions total, counting repeats.

Concepts Checklist

I can identify whether an expression is a polynomial (checking for non-negative integer indices).

I can expand three or more brackets by working pair-by-pair.

I can perform polynomial long division ("bus stop" method) with a linear divisor.

I can perform polynomial division with a quadratic divisor, including missing-term placeholders.

I can use the factor theorem to test and confirm factors of a polynomial.

I can use the remainder theorem to find a remainder without full division.

I can use the remainder theorem to solve for unknown coefficients using simultaneous equations.

I can fully factorise a cubic polynomial using factor theorem + division.

I can solve a cubic equation and find all its real roots (including repeated roots).

I understand the extended remainder theorem for divisors of the form (ax − b).

Exam Tips

Give yourself room. Polynomial division is easy to mess up when cramped. Write it out large, keep columns aligned by power of x, and if it gets messy — restart rather than trying to patch errors mid-way.

Watch your signs when subtracting. The most common error in polynomial division is a sign slip when subtracting the multiplied divisor from the current line. Distribute the minus sign across every term before combining.

Zero placeholders are not optional. If a power of x is missing in the polynomial (e.g. no x² term), you must still write it in as 0x² so your division columns stay aligned.

Examiners reward the factor theorem shortcut. Don't jump straight into long division to "find" a factor — that wastes time. Test small integer values with the factor theorem first to identify one root, THEN divide.

A cubic isn't "solved" until you've found every root. Mark schemes usually award marks for each solution found — don't stop at the linear factor; push through the resulting quadratic too.

Sign check on substitution. For divisor (x + p), that's really (x − (−p)) — so you substitute x = −p, not x = p. This trips up more students than any other single step in this chapter.

Also in the full note

3. Factor & Remainder Theorem

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