# Simple algebraic division

> Edexcel International GCSE Further Pure Mathematics · 4PM1
> Source: https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s3-simple-algebraic-division/

This guide teaches you how to divide polynomials by linear divisors as required for Edexcel IGCSE Further Pure Maths (4PM1) S3. You will practice finding both the quotient and remainder using two standard, exam-approved methods.

**Prerequisites:** [Understanding of polynomial degree and basic algebra operations](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s1-algebraic-expressions/); [Ability to expand binomial products](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s1-binomial-expansion/)

## Learning objectives

- Divide any polynomial by linear divisors of the form (x ± a) or (ax ± b)
- Correctly identify and write the quotient and remainder of algebraic divisions
- Apply both long division and coefficient matching methods to solve exam-style algebraic division questions

## Key Definitions & Core Division Principle

**Algebraic Division Identity** — For any polynomial dividend $P(x)$ and linear divisor $D(x)$: $P(x) = D(x) \times Q(x) + R$, where $Q(x)$ is the quotient polynomial and $R$ is the constant remainder.

*Example:* Dividing $3x^3 + 2x^2 - x + 4$ by $(x - 2)$ follows the identity: $3x^3 + 2x^2 - x + 4 = (x-2)Q(x) + R$

When dividing a polynomial of degree $n$ by a linear (degree 1) divisor, the quotient will always be a polynomial of degree $n-1$, and the remainder will be a constant. This rule lets you check the validity of your result before you finish calculations.

> **tip**
>
> Always confirm the degree of your quotient first to catch errors early. For example, a cubic polynomial divided by a linear divisor will always produce a quadratic quotient.

## Method 1: Algebraic Long Division

Algebraic long division works identically to numerical long division, except you operate on terms of $x$ instead of digits. Follow the cycle: divide leading terms, multiply divisor by the result, subtract from the current dividend segment, bring down the next term, repeat.

**Worked example:** Divide $2x^3 + 5x^2 - x - 6$ by $(x + 2)$. State the quotient and remainder.

1. 1. Divide the leading term of the dividend by the leading term of the divisor: $2x^3 \div x = 2x^2$. Write this as the first term of the quotient.
2. 2. Multiply the entire divisor by $2x^2$: $2x^2(x + 2) = 2x^3 + 4x^2$. Subtract this from the first two terms of the dividend: $(2x^3 + 5x^2) - (2x^3 + 4x^2) = x^2$.
3. 3. Bring down the next term $-x$ to get $x^2 - x$. Divide leading terms: $x^2 \div x = x$. Add $x$ to the quotient.
4. 4. Multiply the divisor by $x$: $x(x+2) = x^2 + 2x$. Subtract: $(x^2 -x) - (x^2 + 2x) = -3x$.
5. 5. Bring down the final term $-6$ to get $-3x -6$. Divide leading terms: $-3x \div x = -3$. Add $-3$ to the quotient.
6. 6. Multiply the divisor by $-3$: $-3(x+2) = -3x -6$. Subtract: $(-3x -6) - (-3x -6) = 0$.
7. Final result: Quotient = $2x^2 + x - 3$, Remainder = 0

> **Exam tip:** Show all long division steps in your exam to earn method marks even if you make a small arithmetic error.

*Calculator:* allowed

## Method 2: Equating Coefficients

This method uses the algebraic division identity directly. You first write the general form of the quotient based on its expected degree, expand the right-hand side of the identity, then match coefficients of like powers of $x$ to solve for unknown constants in the quotient and remainder.

**Worked example:** Divide $3x^3 - 4x^2 + 2x - 1$ by $(2x - 1)$. State the quotient and remainder.

1. 1. The dividend is degree 3, so the quotient is quadratic: let $Q(x) = Ax^2 + Bx + C$, remainder = $R$ (constant).
2. 2. Write the identity: $3x^3 - 4x^2 + 2x - 1 = (2x - 1)(Ax^2 + Bx + C) + R$
3. 3. Expand the right-hand side: $2Ax^3 + (2B - A)x^2 + (2C - B)x + (R - C)$
4. 4. Equate coefficients of like terms:
5. $$x^3: 2A = 3 \implies A = \frac{3}{2}$$
6. $$x^2: 2B - A = -4 \implies 2B - \frac{3}{2} = -4 \implies B = -\frac{5}{4}$$
7. $$x: 2C - B = 2 \implies 2C + \frac{5}{4} = 2 \implies C = \frac{3}{8}$$
8. $$constant: R - C = -1 \implies R = -1 + \frac{3}{8} = -\frac{5}{8}$$
9. Final result: Quotient = $\frac{3}{2}x^2 - \frac{5}{4}x + \frac{3}{8}$, Remainder = $-\frac{5}{8}$

> **tip**
>
> Use fractions instead of decimals for coefficients to avoid rounding errors in follow-up calculations.

> **Exam tip:** This method is often faster for linear divisors and reduces the risk of subtraction errors common in long division.

*Calculator:* allowed

## Exam-Style Practice

**Check your understanding**

1. When dividing $4x^4 + 2x^3 - x + 5$ by $(3x + 2)$, what is the degree of the quotient?

   - A) 3
   - B) 4
   - C) 2
   - D) 1

   *Why:* The quotient degree is always equal to the dividend degree minus 1 for linear divisors: 4 - 1 = 3.

2. If $(x - 2)(x^2 + 3x + 1) + R = x^3 + x^2 - 5x + 2$, what is the value of $R$?

   *Why:* Expand the left-hand side: $(x-2)(x^2+3x+1) = x^3 + x^2 -5x -2$. Rearranging gives $R = (x^3 +x^2 -5x +2) - (x^3 +x^2 -5x -2) = 4$.

**Worked example:** Divide $x^4 - 3x^2 + x + 7$ by $(x - 3)$. State the quotient and remainder.

1. 1. First rewrite the dividend with all powers of $x$ present, inserting 0 coefficients for missing terms: $x^4 + 0x^3 - 3x^2 + x + 7$.
2. 2. Use the long division method, aligning like terms correctly throughout the process.
3. 3. After completing all division steps, the resulting quotient is $x^3 + 3x^2 + 6x + 19$, and the remainder is 64.
4. 4. Verify using the identity: $(x-3)(x^3 +3x^2 +6x +19) + 64 = x^4 -3x^2 +x +7$, which matches the original dividend.

## Common pitfalls

- **Wrong:** Forgetting to include missing terms (e.g., writing $x^3 + 2x +1$ instead of $x^3 + 0x^2 + 2x +1$) when dividing.
  - Why it fails: Missing terms lead to incorrect alignment of like terms during calculation, resulting in wrong quotient coefficients.
  - Correct: Always rewrite the dividend with all powers of $x$ present, inserting 0 coefficients for any missing terms before starting division.
- **Wrong:** Making sign errors when subtracting products during long division.
  - Why it fails: Sign errors are the most common mistake in algebraic long division, and often lead to lost method and answer marks even if your process is otherwise correct.
  - Correct: Write brackets around the product you are subtracting, then distribute the negative sign before combining terms to avoid sign errors.
- **Wrong:** Using an incorrect degree for the quotient when using the equating coefficients method.
  - Why it fails: An incorrectly sized quotient will have missing or extra terms, making it impossible to match coefficients correctly.
  - Correct: Always calculate the expected quotient degree as (dividend degree - 1) before writing the general form of the quotient.
- **Wrong:** Omitting the remainder when presenting your final answer.
  - Why it fails: Exam questions explicitly ask for both quotient and remainder, so omitting either will cost you answer marks even if your quotient is correct.
  - Correct: Always present your final answer in the form $P(x) = D(x)Q(x) + R$, or clearly label and state both the quotient and remainder separately.

## Cheatsheet

| Divisor Type | Quotient Degree (for degree n dividend) | General Identity | Approved Methods |
| --- | --- | --- | --- |
| $(x \pm a)$ | $n-1$ | $P(x) = (x\pm a)Q(x) + R$ | Long division, Equating coefficients |
| $(ax \pm b)$ | $n-1$ | $P(x) = (ax\pm b)Q(x) + R$ | Long division, Equating coefficients |

## What's next

Now that you have mastered simple algebraic division, you are ready to move on to the factor and remainder theorems, which build directly on the quotient and remainder rules you learned here. This topic forms the foundation for solving higher-degree polynomial equations and simplifying algebraic expressions that appear across the Edexcel IGCSE Further Pure Maths syllabus, including in coordinate geometry and calculus questions. Practice both division methods regularly to build speed and accuracy for your exam, as they are frequently tested in both short and long answer questions.

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