1.02fPureStretch

Algebraic Division

Q: Does a remainder of zero always mean a factor?

Yes. Zero remainder means f(x) = d(x) q(x) exactly, which is the definition of d(x) being a factor.

Once the factor theorem has handed you one factor of a cubic, something has to divide it out. OCR calls this "simple algebraic division", and there are two accepted routes: formal long division, or comparing coefficients. Both are examined; knowing when each is faster is most of the skill.

35 min Polynomials

The big picture

Division is what turns a degree-3 problem into a degree-2 problem, and degree 2 is where all your existing tools live. That reduction is the engine behind solving cubics, sketching their graphs, and — later in the course — splitting a rational function into partial fractions before integrating it. There is also a conceptual payoff: dividing polynomials behaves exactly like dividing whole numbers, quotient and remainder included. Recognising that $f (x) = divisor \times quotient + remainder$ is the same statement as $17 = 5 \times 3 + 2$ makes the whole topic feel less like a new procedure.

What you'll be able to do

Divide a polynomial by a linear expression using long division
Achieve the same result by comparing coefficients
Express a result in the form $f (x) = (x - a) q (x) + r$
Choose the more efficient method for a given question

What division is asking for

Dividing $f (x)$ by $d (x)$ means finding a quotient $q (x)$ and a remainder $r (x)$ with $f (x) = d (x) q (x) + r (x)$ , where the remainder has lower degree than the divisor.

Dividing by a linear expression therefore always leaves a $constant$ remainder, because anything of lower degree than 1 is a number. When that constant is zero, the divisor was a factor — which is the factor theorem restated.

f (x) = (x - a) q (x) + r

Linear divisor ⇒ constant remainder.

Why this works

Degrees are a free check. Dividing a cubic by a linear expression must leave a quadratic quotient. If your quotient comes out cubic or linear, something has gone wrong before the arithmetic is worth checking.

Method 1: long division

The layout mirrors numerical long division. At each stage, divide the leading term of what remains by the leading term of the divisor, multiply the divisor by that result, and subtract.

The one non-negotiable is including $zero placeholders$ for missing powers. Dividing $x^{3} - 7 x + 6$ means writing $x^{3} + 0 x^{2} - 7 x + 6$ ; without the $0 x^{2}$ the columns misalign and every later term is wrong.

x^{3} \div x = x^{2}

. Multiply:

x^{2} (x - 1) = x^{3} - x^{2}

. Subtract, leaving

- 5 x^{2} + 11 x - 6

- 5 x^{2} \div x = - 5 x

. Multiply:

- 5 x (x - 1) = - 5 x^{2} + 5 x

. Subtract, leaving

6 x - 6

6 x \div x = 6

. Multiply:

6 (x - 1) = 6 x - 6

. Subtract, leaving

0

AnswerQuotient

x^{2} - 5 x + 6

, remainder

0

Tip — Subtraction is where this method goes wrong. Changing both signs of the line below and adding is more reliable than subtracting in place.

Method 2: comparing coefficients

Write the answer in the shape you know it must take, with unknown coefficients, then multiply out and match. For a cubic divided by a linear factor: $f (x) = (x - a) (p x^{2} + q x + r)$ .

Two of the unknowns come immediately. The $x^{3}$ coefficient forces $p$ , and the constant term forces $r$ . Only the middle needs an actual comparison, so a cubic division reduces to one small equation.

1Compare

x^{3}

: the only source is

x \times p x^{2}

, so

p = 2

2Compare constants:

- 2 \times r = - 6

, so

r = 3

3Compare

x^{2}

: from

x \times q x

and

- 2 \times p x^{2}

, giving

q - 2 p = 3

, so

q - 4 = 3

and

q = 7

4Check with the

x

terms:

r - 2 q = 3 - 14 = - 11

. It matches.

Answer

(x - 2) (2 x^{2} + 7 x + 3)

Why this works

The unused comparison is a free verification. Having found $p$ , $q$ and $r$ from three of the four coefficients, checking against the fourth confirms the whole division in one line — long division offers nothing equivalent.

Handling a non-zero remainder

When the divisor is not a factor, both methods still work; the leftover constant is the remainder. The result is then written $f (x) = (x - a) q (x) + r$ , or equivalently as a quotient plus a fraction.

For comparing coefficients, simply include the remainder as an extra unknown from the start and match as before.

1Constant term:

- 1 \times c + d = - 4

2Compare

x^{2}

b - 1 = 2

, so

b = 3

3Compare

x

c - b = 0

, so

c = 3

4Back to the constant:

- 3 + d = - 4

, so

d = - 1

Answer

(x - 1) (x^{2} + 3 x + 3) - 1

Tip — Cross-check the remainder with the remainder theorem: $f (1) = 1 + 2 - 4 = - 1$ , matching $d = - 1$ . Two independent routes agreeing is as close to certainty as exam work gets.

Think like an examiner

Insert zero coefficients for missing powers before starting long division. Misalignment here corrupts everything after it.
If the question specifies a form such as $(x - a) (p x^{2} + q x + r)$ , give the answer in that exact form and state the constants if asked.
Either method scores full marks. Comparing coefficients is usually quicker for a cubic and carries a built-in check.
Verify a non-zero remainder against $f (a)$ from the remainder theorem — it takes one substitution.

Common misconceptions

Dividing

x^{3} - 7 x + 6

without writing the

0 x^{2}

term.

Missing powers need zero placeholders or the columns shift and every subsequent term is wrong.

Adding the line below instead of subtracting it.

Each stage subtracts. Flipping both signs and then adding is the safer way to execute it.

Expecting a linear quotient when dividing a cubic by a linear expression.

Degrees subtract: cubic ÷ linear gives a quadratic quotient.

Ignoring the leftover constant when the divisor is not a factor.

That constant is the remainder and belongs in the answer as

+ r

Division facts

f (x) = d (x) q (x) + r (x)

The division identity.

de g q = de g f - de g d

Degrees subtract.

linear divisor \Rightarrow r constant

Remainder is a number.

r = f (a) when dividing by (x - a)

Cross-check via the remainder theorem.

Stretch yourself

When $f (x) = x^{3} + p x^{2} + q x + 6$ is divided by $(x - 1)$ the remainder is $12$ , and $(x + 2)$ is a factor. Find $p$ and $q$ , then factorise $f (x)$ completely.

Hint — Two conditions give two equations. Use the remainder theorem for the first and the factor theorem for the second.

Questions students ask

Which method should I use in the exam?

Comparing coefficients for a cubic divided by a linear factor — it is quicker and self-checking. Long division earns its place with higher degrees or a quadratic divisor, where the unknowns multiply up.

Why do missing terms need a zero?

Long division works by aligning like powers in columns. A missing

x^{2}

term leaves a gap that shifts everything after it into the wrong column, so the placeholder keeps the arithmetic honest.

Can I divide by a quadratic?

Yes, and the same identity applies — but the remainder may then be linear rather than constant. H240 describes the requirement as "simple algebraic division", so linear divisors are what you should expect.

Does a remainder of zero always mean a factor?

Yes. Zero remainder means

f (x) = d (x) q (x)

exactly, which is the definition of

d (x)

being a factor.

Key takeaways

Division seeks $f (x) = d (x) q (x) + r (x)$ with the remainder of lower degree.
A linear divisor always leaves a constant remainder.
Long division needs zero placeholders for missing powers.
Comparing coefficients fixes the outer terms instantly and leaves one comparison to check against.
The remainder theorem independently verifies any non-zero remainder.

How this fits the course

Build on

The Factor Theorem

Factorising Expressions

Leads to

Straight Lines

Test yourself

Ready to lock in Algebraic Division? Pick a mode and earn XP & Dobloons.

Next step

Practise this topic

Real past-paper questions on Algebraic Division, marked mark-by-mark. How you do feeds straight into your weak-topic list, so your revision keeps targeting what actually needs work.

Practise this topic

Algebraic Division

What division is asking for

Method 1: long division

Method 2: comparing coefficients

Handling a non-zero remainder

Practise this topic

Related topics

Algebraic Division

What division is asking for

Method 1: long division

Method 2: comparing coefficients

Handling a non-zero remainder

Practise this topic

Related topics