Simple algebraic division

Q: What are the four components of any algebraic division problem, and what degree is the remainder when dividing by a linear divisor?

The four components are the Dividend (\f$f(x)\f$), Divisor (\f$d(x)\f$), Quotient (\f$Q(x)\f$), and Remainder (\f$R\f$). When dividing by a linear divisor (degree 1), the remainder is always a constant (degree 0).

Q: What are the four repeating steps in the polynomial long division cycle?

The four steps are: Divide , Multiply , Subtract , and Bring down (DMSB).

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Further Pure Mathematics · Pearson Edexcel IGCSE · 9 min read · Updated: 22 Oct 2026

Welcome to Simple Algebraic Division

Division is one of the four basic pillars of arithmetic that you have used since primary school. When you first learned numerical division, you divided numbers like $17$ by $5$ to get a quotient of $3$ and a remainder of $2$. In Pearson Edexcel IGCSE Further Pure Mathematics, we apply this exact same logic to algebraic expressions!

Don't worry if algebraic division seems intimidating at first glance. Once you learn the repeating four-step cycle, it becomes a predictable and reliable technique. Master this skill here, because it serves as the essential stepping stone to other key topics in this section, such as the Factor and Remainder Theorems and Factorising Cubic Equations.

1. The Core Anatomy of Division

Before jumping into the calculations, let's establish the proper mathematical terms. Every division problem has four key components:

1. Dividend: The polynomial you want to divide, often written as $f(x)$.
2. Divisor: The expression you are dividing by, written as $d(x)$. In this specification, divisors are always linear expressions: $(x + a)$, $(x - a)$, $(ax + b)$, or $(ax - b)$.
3. Quotient: The main result of the division, written as $Q(x)$.
4. Remainder: Whatever is left over at the end, written as $R$. Because our divisor is linear (degree $1$), the remainder will always be a single constant number (degree $0$).

The Fundamental Division Identity

Just as with ordinary numbers, where $\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}$, algebra follows the exact identity:

$f(x) \equiv d(x) Q(x) + R$

Alternatively, dividing both sides by the divisor $d(x)$ gives the fractional form:

$\frac{f(x)}{d(x)} \equiv Q(x) + \frac{R}{d(x)}$

Key Takeaway: If the remainder $R = 0$, then $d(x)$ divides $f(x)$ exactly, meaning $d(x)$ is a factor of $f(x)$.

2. The Long Division Method (The 4-Step Cycle)

The standard way to carry out simple algebraic division is polynomial long division. It mirrors numerical long division using a repeating four-step memory cycle:

Step 1: Divide — Divide the highest-power term of the current dividend by the first term of the divisor.
Step 2: Multiply — Multiply your new quotient term by the entire divisor.
Step 3: Subtract — Subtract this product from the current line of the dividend. Be extremely careful with minus signs!
Step 4: Bring down — Bring down the next term of the dividend, and repeat the process until the remaining term has a degree lower than the divisor.

Mnemonic Aid: Does McDonald's Sell Burgers?

A classic trick to remember the four steps in order is the phrase: Divide, Multiply, Subtract, Bring down (Does McDonald's Sell Burgers?).

Worked Example 1: Division by $(x - a)$ with Zero Remainder

Divide $x^3 + 2x^2 - 5x - 6$ by $(x - 2)$.

Step-by-step solution:

Cycle 1:
- Divide: Divide the first term of the dividend by the first term of the divisor: $\frac{x^3}{x} = x^2$. This is the first term of your quotient.
- Multiply: Multiply $x^2$ by $(x - 2)$ to get $x^3 - 2x^2$.
- Subtract: $(x^3 + 2x^2) - (x^3 - 2x^2) = 2x^2 - (-2x^2) = 4x^2$.
- Bring down: Bring down the next term, $-5x$, giving the new expression $4x^2 - 5x$.

Cycle 2:
- Divide: Divide $\frac{4x^2}{x} = +4x$.
- Multiply: Multiply $+4x$ by $(x - 2)$ to get $4x^2 - 8x$.
- Subtract: $(4x^2 - 5x) - (4x^2 - 8x) = -5x - (-8x) = 3x$.
- Bring down: Bring down the constant term, $-6$, giving $3x - 6$.

Cycle 3:
- Divide: Divide $\frac{3x}{x} = +3$.
- Multiply: Multiply $+3$ by $(x - 2)$ to get $3x - 6$.
- Subtract: $(3x - 6) - (3x - 6) = 0$.

The remainder is $0$. Therefore, the quotient is $x^2 + 4x + 3$.
We can write: $x^3 + 2x^2 - 5x - 6 \equiv (x - 2)(x^2 + 4x + 3)$.

3. Handling Missing Powers of $x$

One of the most frequent exam pitfalls occurs when a polynomial skips a power of $x$ (for instance, jumping from $x^3$ straight to $x$, missing the $x^2$ term). If you do not create a placeholder, your like terms will not line up vertically, leading to algebraic chaos!

The Golden Rule: Always write out the dividend in full descending powers of $x$, inserting $0x^2$ or $0x$ for any missing terms.

Worked Example 2: Missing Term and Non-Zero Remainder

Divide $2x^3 - 7x + 1$ by $(x + 3)$. Express your answer in the form $Q(x) + \frac{R}{x + 3}$.

Step-by-step solution:

First, rewrite the dividend to include the missing squared term: $2x^3 + 0x^2 - 7x + 1$.

Cycle 1:
- Divide: $\frac{2x^3}{x} = 2x^2$.
- Multiply: $2x^2(x + 3) = 2x^3 + 6x^2$.
- Subtract: $(2x^3 + 0x^2) - (2x^3 + 6x^2) = -6x^2$.
- Bring down: Bring down $-7x$ to obtain $-6x^2 - 7x$.

Cycle 2:
- Divide: $\frac{-6x^2}{x} = -6x$.
- Multiply: $-6x(x + 3) = -6x^2 - 18x$.
- Subtract: $(-6x^2 - 7x) - (-6x^2 - 18x) = -7x - (-18x) = 11x$.
- Bring down: Bring down $+1$ to obtain $11x + 1$.

Cycle 3:
- Divide: $\frac{11x}{x} = +11$.
- Multiply: $11(x + 3) = 11x + 33$.
- Subtract: $(11x + 1) - (11x + 33) = 1 - 33 = -32$.

Since $-32$ is a constant and cannot be divided further by $x$, our remainder is $R = -32$.
The quotient is $Q(x) = 2x^2 - 6x + 11$.
Writing the final result in the requested form:
$\frac{2x^3 - 7x + 1}{x + 3} \equiv 2x^2 - 6x + 11 - \frac{32}{x + 3}$.

4. Dividing by $(ax + b)$ or $(ax - b)$

When the divisor has a coefficient in front of $x$ that is not $1$, the algorithm remains completely identical. You just need to take extra care during the Divide step to divide by the leading coefficient $a$.

Worked Example 3: Divisor with Leading Coefficient Greater than 1

Find the quotient and remainder when $6x^3 + 11x^2 - x - 5$ is divided by $(2x - 1)$.

Step-by-step solution:

Cycle 1:
- Divide: $\frac{6x^3}{2x} = 3x^2$.
- Multiply: $3x^2(2x - 1) = 6x^3 - 3x^2$.
- Subtract: $(6x^3 + 11x^2) - (6x^3 - 3x^2) = 11x^2 - (-3x^2) = 14x^2$.
- Bring down: Bring down $-x$ to give $14x^2 - x$.

Cycle 2:
- Divide: $\frac{14x^2}{2x} = +7x$.
- Multiply: $7x(2x - 1) = 14x^2 - 7x$.
- Subtract: $(14x^2 - x) - (14x^2 - 7x) = -x - (-7x) = 6x$.
- Bring down: Bring down $-5$ to give $6x - 5$.

Cycle 3:
- Divide: $\frac{6x}{2x} = +3$.
- Multiply: $3(2x - 1) = 6x - 3$.
- Subtract: $(6x - 5) - (6x - 3) = -5 - (-3) = -2$.

Quotient $Q(x) = 3x^2 + 7x + 3$
Remainder $R = -2$
Expressed as an identity:
$6x^3 + 11x^2 - x - 5 \equiv (2x - 1)(3x^2 + 7x + 3) - 2$.

5. Alternative Approach: Equating Coefficients (Inspection Method)

An alternative method accepted in Pearson Edexcel examinations is finding the quotient by setting up an identity and comparing coefficients. If you find the column subtraction in long division error-prone, this method offers a neat, algebraic alternative.

When a cubic polynomial $f(x)$ is divided by a linear factor $(ax + b)$, the quotient must be a quadratic expression $Ax^2 + Bx + C$, leaving a constant remainder $R$:

$f(x) \equiv (ax + b)(Ax^2 + Bx + C) + R$

Worked Example 4: Using Equating of Coefficients

Find the quotient and remainder when $2x^3 - 3x^2 + 4x - 7$ is divided by $(x - 2)$.

Step-by-step solution:

Set up the identity:
$2x^3 - 3x^2 + 4x - 7 \equiv (x - 2)(Ax^2 + Bx + C) + R$

Expand the right-hand side systematically:
$(x - 2)(Ax^2 + Bx + C) + R = Ax^3 + Bx^2 + Cx - 2Ax^2 - 2Bx - 2C + R$
Grouping like powers of $x$:
$= Ax^3 + (B - 2A)x^2 + (C - 2B)x + (-2C + R)$

Now compare coefficients on both sides of the identity:

- Coefficients of $x^3$:
$A = 2$

- Coefficients of $x^2$:
$B - 2A = -3$
$B - 2(2) = -3 \implies B - 4 = -3 \implies B = 1$

- Coefficients of $x$:
$C - 2B = 4$
$C - 2(1) = 4 \implies C - 2 = 4 \implies C = 6$

- Constant terms:
$-2C + R = -7$
$-2(6) + R = -7 \implies -12 + R = -7 \implies R = 5$

Hence, the quotient is $2x^2 + x + 6$ and the remainder is $5$.

6. Common Mistakes to Avoid

Examiners routinely notice marks lost due to simple slips rather than conceptual misunderstanding. Keep an eye out for these frequent pitfalls:

1. Minus Sign Slip-ups during Subtraction: When subtracting negative terms like $-5x - (-8x)$, students often mistakenly calculate $-5x - 8x = -13x$ instead of the correct $-5x + 8x = 3x$. Put brackets around the lower line before subtracting to protect your signs!
2. Forgetting Missing Terms: Never divide into an expression like $x^3 - 4$ without first writing $x^3 + 0x^2 + 0x - 4$.
3. Incomplete Division: You must continue the division process until the remaining term has a degree strictly less than the divisor. For a linear divisor, stop only when you reach a single number (degree $0$).
4. Dividing Only by the First Letter: In $(2x - 1)$, remember that $\frac{6x^3}{2x} = 3x^2$, not $6x^2$. Always divide by the number and the variable together.

7. Quick Review & Summary Checklist

- Divisors: You are expected to divide by linear expressions of the forms $(x + a)$, $(x - a)$, $(ax + b)$, or $(ax - b)$.
- Identity Form: $\text{Dividend} \equiv \text{Divisor} \times \text{Quotient} + \text{Remainder}$.
- Degrees: Dividing a cubic (degree $3$) by a linear (degree $1$) polynomial always yields a quadratic quotient (degree $2$) and a constant remainder (degree $0$).
- Zero Remainder: If $R = 0$, the linear expression is an exact factor of the polynomial.
- Next Step: You will use this division process directly to factorise polynomials fully in the upcoming chapter on The factor and remainder theorems.

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What are the four components of any algebraic division problem, and what degree is the remainder when dividing by a linear divisor?

The four components are the Dividend (\f$f(x) $), Divisor (\f$d(x) $), Quotient (\f$Q(x) $), and Remainder (\f$R $). When dividing by a linear divisor (degree 1), the remainder is always a constant (degree 0).

What is the Fundamental Division Identity relating dividend, divisor, quotient, and remainder?

$f(x) \equiv d(x) Q(x) + R$, or in fractional form: $\frac{f(x)}{d(x)} \equiv Q(x) + \frac{R}{d(x)}$.

What are the four repeating steps in the polynomial long division cycle?

The four steps are: Divide, Multiply, Subtract, and Bring down (DMSB).

How must you handle missing powers of $x$ in a polynomial before carrying out algebraic long division?

Insert placeholder terms with a coefficient of zero (such as $0x^2$ or $0x$) so the terms are written in full descending powers of $x$.

What extra precaution must be taken during the Divide step when the divisor is of the form $(ax + b)$ or $(ax - b)$ with $a > 1$?

Divide the highest-power term of the current dividend by both the variable and the leading coefficient $a$ (divide by $ax$, not just $x$).

How does the inspection method (equating coefficients) find the quotient and remainder when dividing a cubic polynomial by a linear factor?

Assume the identity $f(x) \equiv (ax + b)(Ax^2 + Bx + C) + R$, expand the right-hand side, and equate coefficients of corresponding powers of $x$ to solve for $A$, $B$, $C$, and $R$.

What is the most effective safeguard against sign errors during the subtraction step of algebraic long division?

Place brackets around the lower line before subtracting to avoid errors like calculating $-5x - (-8x)$ as $-13x$ instead of $+3x$.

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Simple algebraic division

Welcome to Simple Algebraic Division

1. The Core Anatomy of Division

The Fundamental Division Identity

2. The Long Division Method (The 4-Step Cycle)

Mnemonic Aid: Does McDonald's Sell Burgers?

Worked Example 1: Division by \((x - a)\) with Zero Remainder

3. Handling Missing Powers of \(x\)

Worked Example 2: Missing Term and Non-Zero Remainder

4. Dividing by \((ax + b)\) or \((ax - b)\)

Worked Example 3: Divisor with Leading Coefficient Greater than 1

5. Alternative Approach: Equating Coefficients (Inspection Method)

Worked Example 4: Using Equating of Coefficients

6. Common Mistakes to Avoid

7. Quick Review & Summary Checklist

Can you answer these now?

Ready to test yourself?

More Further Pure Mathematics chapters

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