Colour key

Yellow = definitions and key terms

Blue = rules, formulas and theorems

Green = important steps, methods and correct answers

Pink = common mistakes

Orange = exceptions, units and special cases

Lavender = results and final answers

Teal = checks and connections

How to use this compendium

• Read each section in order. Every worked example shows each step on its own line.
• Words with a small ⓘ beside them are key terms. Click the ⓘ to open a short meaning with examples. Press Escape or the cross to close it.
• The multiple-choice questions show the correct answer in green straight under each question, followed by an explanation.
• Use the Audio Teacher bar at the bottom to hear a teacher explain each part, or to have each line read aloud.
• Use the toolbar at the top to annotate, zoom, switch between dark and bright mode, or download a Word or PDF copy.
• Use the interactive deck in Section 2 to watch each division cycle appear one step at a time.

What is inside

1. Previous Knowledge
2. Topic Explanation: Long division of polynomials (explanation, interactive deck, Extra Knowledge, multiple-choice questions, worked examples, Exam Style Questions and higher-order worked examples)
3. Application
4. Tips, Tricks and Exceptions
5. Further Reading
6. Difficult Words and Vocabulary

1. Previous Knowledge

Before you start polynomialⓘ long division, make sure the words and skills below feel comfortable. Each one is used in the main explanation.

1.1 Words about polynomials

1.2 Skills you need

Index laws

When you multiply powers of the same base, add the indices: x² × x³ = x⁵.

When you divide powers of the same base, subtract the indices: x⁵ ÷ x² = x³.

This is exactly what you do when you divide leading termsⓘ in a long division. For example, 6x³ ÷ 2x = 3x², because 6 ÷ 2 = 3 and x³ ÷ x = x².

Expanding brackets

Multiply every termⓘ in the first bracket by every term in the second.

3x(x − 4) = 3x² − 12x

(x + 2)(x − 3) = x² − 3x + 2x − 6 = x² − x − 6

Subtracting polynomials

Subtracting a bracket means changing the sign of every termⓘ inside it, and then adding.

(5x² − 13x) − (5x² − 15x) = 5x² − 13x − 5x² + 15x = 2x

Common mistake: changing only the first sign in the bracket. Here that would give 5x² − 13x − 5x² − 15x, which is wrong.

Substituting a value

To evaluate P(x) = x² − 3x + 1 at x = 2, replace every x with 2 and calculate.

P(2) = 2² − 3(2) + 1 = 4 − 6 + 1 = −1

P(−2) = (−2)² − 3(−2) + 1 = 4 + 6 + 1 = 11

Take care with negatives: put brackets around a negative value so that the signs stay correct.

Factorising a quadratic

Find two numbers that multiply to the constant termⓘ and add to the coefficientⓘ of x.

x² + 5x + 6 = (x + 2)(x + 3), because 2 × 3 = 6 and 2 + 3 = 5.

1.3 Long division with whole numbers

Polynomialⓘ long division copies the method you know for whole numbers. Look at 487 ÷ 13. The dividendⓘ is 487, the divisorⓘ is 13, the quotientⓘ is the answer, and the remainderⓘ is what is left over.

Step 1: How many 13s fit into 48? There are 3, because 3 × 13 = 39.

Step 2: Subtract: 48 − 39 = 9.

Step 3: Bring down the 7 to make 97.

Step 4: How many 13s fit into 97? There are 7, because 7 × 13 = 91.

Step 5: Subtract: 97 − 91 = 6. There is nothing left to bring down, so stop.

Result: quotient 37, remainder 6.

Check: 13 × 37 + 6 = 481 + 6 = 487 ✓

The check gives a division statementⓘ: dividend = divisor × quotient + remainder. For polynomials this becomes P(x) = D(x) × Q(x) + R(x).

The four repeating actions are the same for polynomials:

1. Divide the leading termsⓘ.
2. Multiply the new quotient termⓘ by the whole divisor.
3. Subtract the product.
4. Bring down the next term.

2. Topic Explanation: Long division of polynomials

Polynomialⓘ long division lets us divide one polynomial by another when the divisorⓘ has at least one variable term, such as x − 3 or x² + x − 2. We use it to factoriseⓘ polynomials of degreeⓘ 3 or more, to solve higher-degree equations, to simplify algebraic fractionsⓘ and to find missing dimensions in real problems.

This section covers one subtopic, Long division of polynomials, in this order: 2.1.1 a detailed explanation with diagrams, 2.1.2 an interactive deck, 2.1.3 Extra Knowledge, 2.1.4 five multiple-choice questions and 2.1.5 worked examples. After the subtopic come 2.2 Exam Style Questions and 2.3 two higher-order worked examples.

The four names in a division

When we divide the polynomialⓘ P(x) by the polynomial D(x), four names are used.

• Dividendⓘ P(x): the polynomial that is being divided.
• Divisorⓘ D(x): the polynomial we divide by.
• Quotientⓘ Q(x): the answer to the division, not counting any left-over part.
• Remainderⓘ R(x): the left-over part. Its degreeⓘ must be smaller than the degree of the divisor.

Figure 1 shows the four parts for (x³ + 2x² − 13x + 10) ÷ (x − 3). The quotient is written above the dividend, the divisor is on the left of the bracket and the remainder is the last line.

The division statement

Every polynomialⓘ division can be written as one equation:

P(x) = D(x) × Q(x) + R(x), where the degreeⓘ of R(x) is less than the degree of D(x).

If we divide both sides by D(x), we also get P(x)/D(x) = Q(x) + R(x)/D(x). This is useful for simplifying algebraic fractionsⓘ.

The degree of the quotientⓘ is the degree of the dividendⓘ minus the degree of the divisorⓘ. For (x³ + 2x² − 13x + 10) ÷ (x − 3), the quotient has degree 3 − 1 = 2.

Getting ready: descending order and placeholders

Before you divide, write both polynomialsⓘ in descending orderⓘ. If a power of x is missing from the dividendⓘ, write it with a coefficientⓘ of zero. This is called a placeholderⓘ. It keeps every power in its own column.

Common mistake: leaving out the placeholder. Later terms then slide into the wrong column and the answer is wrong.

The four actions in each cycle

Figure 2 shows one cycle. Work through these four actions, then repeat.

1. Divide the leading termⓘ of the working lineⓘ by the leading term of the divisorⓘ. The answer is the next term of the quotientⓘ. Divide the coefficientsⓘ and subtract the powers.
2. Multiply that quotient term by every term of the divisor and write the product under the working line.
3. Subtract the product from the working line. Change the sign of every term in the product, then add.
4. Bring down the next term of the dividendⓘ to make a new working line.

Stop when the degreeⓘ of the working line is smaller than the degree of the divisor. That last line is the remainderⓘ.

Seeing the layout cycle by cycle

We divide x³ + 2x² − 13x + 10 by x − 3. The quotientⓘ is written above the dividendⓘ. Each quotient term sits above the column of its own power of x.

After cycle 1. x³ ÷ x = x². Then x²(x − 3) = x³ − 3x². Subtract to get 5x², then bring down −13x.

After cycle 2. 5x² ÷ x = 5x. Then 5x(x − 3) = 5x² − 15x. Subtract to get 2x, then bring down 10.

After cycle 3. 2x ÷ x = 2. Then 2(x − 3) = 2x − 6. Subtract to get 16.

The remainderⓘ 16 has degreeⓘ 0, which is less than the degree 1 of the divisorⓘ, so we stop. The quotient is x² + 5x + 2.

Subtracting carefully

Most errors happen in the subtraction step. Subtracting a row means changing every sign in that row.

(x³ + 2x²) − (x³ − 3x²) = x³ + 2x² − x³ + 3x² = 5x²

Common mistake: writing 2x² − 3x² = −x². The term being subtracted is −3x², so subtracting it adds 3x².

Writing and checking the answer

For our example the answer is quotientⓘ x² + 5x + 2, remainderⓘ 16. Write the division statementⓘ:

x³ + 2x² − 13x + 10 = (x − 3)(x² + 5x + 2) + 16

Check by multiplying: (x − 3)(x² + 5x + 2) = x³ + 5x² + 2x − 3x² − 15x − 6 = x³ + 2x² − 13x − 6. Adding the remainder 16 gives x³ + 2x² − 13x + 10, which is the dividendⓘ. ✓

As a fraction: (x³ + 2x² − 13x + 10)/(x − 3) = x² + 5x + 2 + 16/(x − 3).

When the divisor does not start with 1x

If the divisorⓘ is 2x + 1, the first quotientⓘ term is found by dividing the numbers as well as the powers. For 6x³ ÷ 2x we do 6 ÷ 2 = 3 and x³ ÷ x = x², so the term is 3x². The rest of the cycle is unchanged.

Exception: the quotient coefficientsⓘ are not always whole numbers. In Year 10 the examples are chosen so that they are.

When the divisor has degree 2

If the divisorⓘ is quadraticⓘ, each product has three terms, and the remainderⓘ can have degreeⓘ 1. Here is (x⁴ − 3x³ + 2x − 5) ÷ (x² + x − 2).

The quotientⓘ is x² − 4x + 6 and the remainder is −12x + 7. The remainder has degree 1, which is less than 2, so the division is finished. Notice the placeholderⓘ 0x² in the dividendⓘ.

Link to the remainder theorem and the factor theorem

When the divisorⓘ is x − a, the remainderⓘ is just P(a). This is the remainder theoremⓘ. In our example a = 3 and P(3) = 27 + 18 − 39 + 10 = 16, the same as the remainder. Figure 3 shows the point (3, 16) on the graph of y = P(x).

If the remainder is zero, then x − a is a factorⓘ of P(x). This is the factor theoremⓘ. The graph then crosses the x-axis at x = a. For P(x) above, P(1) = 0, P(2) = 0 and P(−5) = 0, so the factors are (x − 1), (x − 2) and (x + 5).

Link to area and volume

Figure 4 shows that x² + 5x + 6 is the area of a rectangle with sides x + 2 and x + 3. Dividing the area by one side gives the other side: (x² + 5x + 6) ÷ (x + 2) = x + 3.

Volumes work in the same way. In Figure 5, the base area is x(x + 4) = x² + 4x, so the height is (x³ + 3x² − 4x) ÷ (x² + 4x) = x − 1 metres.

Extra Knowledge: why the method always works, and where it leads in Year 11 and 12

The division algorithm. For any polynomialsⓘ P(x) and D(x) with D(x) not zero, there is exactly one quotientⓘ Q(x) and exactly one remainderⓘ R(x) so that P(x) = D(x)Q(x) + R(x) with deg R < deg D. Each cycle of the method removes the leading termⓘ of the working lineⓘ, so the degreeⓘ falls every time. It cannot fall forever, so the process ends. The answer is unique, because if two different pairs worked, subtracting them would give D(x) × (difference of quotients) = (difference of remainders), and the left side has degree at least deg D while the right side has degree less than deg D. That is impossible unless both differences are zero.

Proof of the remainder theoremⓘ. Divide P(x) by x − a. The remainder has degree less than 1, so it is a constant r. Then P(x) = (x − a)Q(x) + r for every x. Put x = a. The bracket (x − a) becomes 0, so P(a) = r. The remainder theorem is therefore a direct consequence of the division statementⓘ, and the factor theoremⓘ is the special case r = 0.

Divisorsⓘ of the form ax − b. Because ax − b = a(x − b/a), the remainder on dividing by ax − b is P(b/a). For 2x − 1, use P(½).

At most n rootsⓘ. A non-zero polynomial of degree n has at most n real roots. Each root a gives a factorⓘ (x − a) by the factor theorem, and each factor lowers the degree of what is left by 1, so there is room for at most n factors.

Quadraticⓘ divisors through two linearⓘ factors. When the divisor is (x − a)(x − b) with a ≠ b, the remainder has the form px + q. Substituting x = a and x = b into P(x) = (x − a)(x − b)Q(x) + px + q gives two equations, P(a) = pa + q and P(b) = pb + q, which are solved together. This is a shortcut that avoids the long division.

Synthetic divisionⓘ and the rational root theoremⓘ. For a divisor x − a, only the coefficientsⓘ matter, and the method called synthetic division is a compact form of the same cycles. To find candidate roots of a polynomial with whole-number coefficients, the rational root theorem lists the fractions p/q to test, where p divides the constant termⓘ and q divides the leading coefficient.

Beyond polynomials. The same repeated division is the heart of the Euclidean algorithmⓘ for polynomials, of partial fractionsⓘ, and of finding oblique asymptotes of rational functions, all met in senior secondary mathematics.

Choose the best option for each question before you read the answer. The correct option is highlighted under each question, followed by a step-by-step explanation.

Worked Example 1: A cubic divided by x − 3

Divide x³ + 2x² − 13x + 10 by x − 3.

Set up. Both polynomialsⓘ are in descending orderⓘ and no power is missing.

Cycle 1 · Divide. Leading termⓘ of the working lineⓘ ÷ leading term of the divisorⓘ: x³ ÷ x = x²

Multiply. x² × (x − 3) = x³ − 3x²

Subtract. (x³ + 2x²) − (x³ − 3x²) = 5x²

Bring down −13x to give the new working line: 5x² − 13x

Cycle 2 · Divide. Leading term of the working line ÷ leading term of the divisor: 5x² ÷ x = 5x

Multiply. 5x × (x − 3) = 5x² − 15x

Subtract. (5x² − 13x) − (5x² − 15x) = 2x

Bring down 10 to give the new working line: 2x + 10

Cycle 3 · Divide. Leading term of the working line ÷ leading term of the divisor: 2x ÷ x = 2

Multiply. 2 × (x − 3) = 2x − 6

Subtract. (2x + 10) − (2x − 6) = 16

Stop. The remainderⓘ 16 has degreeⓘ 0, which is less than the divisor's degree 1, so the division is finished.

Result. Quotientⓘ = x² + 5x + 2 and remainder = 16

Division statementⓘ. x³ + 2x² − 13x + 10 = (x − 3)(x² + 5x + 2) + 16

Check. (x − 3)(x² + 5x + 2) + 16 = x³ + 2x² − 13x − 6 + 16 = x³ + 2x² − 13x + 10, which equals the dividendⓘ. ✓

Worked Example 2: A missing term and a non-zero remainder

Divide 2x³ + 3x − 5 by x + 1. State the quotientⓘ and the remainderⓘ.

Set up. The x² term is missing, so write the dividendⓘ as 2x³ + 0x² + 3x − 5.

Cycle 1 · Divide. Leading termⓘ of the working lineⓘ ÷ leading term of the divisorⓘ: 2x³ ÷ x = 2x²

Multiply. 2x² × (x + 1) = 2x³ + 2x²

Subtract. (2x³ + 0x²) − (2x³ + 2x²) = −2x²

Bring down 3x to give the new working line: −2x² + 3x

Cycle 2 · Divide. Leading term of the working line ÷ leading term of the divisor: −2x² ÷ x = −2x

Multiply. −2x × (x + 1) = −2x² − 2x

Subtract. (−2x² + 3x) − (−2x² − 2x) = 5x

Bring down −5 to give the new working line: 5x − 5

Cycle 3 · Divide. Leading term of the working line ÷ leading term of the divisor: 5x ÷ x = 5

Multiply. 5 × (x + 1) = 5x + 5

Subtract. (5x − 5) − (5x + 5) = −10

Stop. The remainder −10 has degreeⓘ 0, which is less than the divisor's degree 1, so the division is finished.

Result. Quotient = 2x² − 2x + 5 and remainder = −10

Division statementⓘ. 2x³ + 3x − 5 = (x + 1)(2x² − 2x + 5) + −10

Check. (x + 1)(2x² − 2x + 5) + −10 = 2x³ + 3x + 5 + −10 = 2x³ + 3x − 5, which equals the dividend. ✓

Remainder theoremⓘ check. x + 1 = x − (−1), so the remainder is P(−1) = 2(−1)³ + 3(−1) − 5 = −2 − 3 − 5 = −10. ✓

Worked Example 3: A divisor with leading coefficient 2

Divide 8x³ − 6x² + 5x + 1 by 2x − 1. Give the quotientⓘ and the remainderⓘ.

Set up. Both polynomialsⓘ are in descending orderⓘ. The divisorⓘ starts with 2x, so divide the coefficientsⓘ too.

Cycle 1 · Divide. Leading termⓘ of the working lineⓘ ÷ leading term of the divisor: 8x³ ÷ 2x = 4x²

Multiply. 4x² × (2x − 1) = 8x³ − 4x²

Subtract. (8x³ − 6x²) − (8x³ − 4x²) = −2x²

Bring down 5x to give the new working line: −2x² + 5x

Cycle 2 · Divide. Leading term of the working line ÷ leading term of the divisor: −2x² ÷ 2x = −x

Multiply. −x × (2x − 1) = −2x² + x

Subtract. (−2x² + 5x) − (−2x² + x) = 4x

Bring down 1 to give the new working line: 4x + 1

Cycle 3 · Divide. Leading term of the working line ÷ leading term of the divisor: 4x ÷ 2x = 2

Multiply. 2 × (2x − 1) = 4x − 2

Subtract. (4x + 1) − (4x − 2) = 3

Stop. The remainder 3 has degreeⓘ 0, which is less than the divisor's degree 1, so the division is finished.

Result. Quotient = 4x² − x + 2 and remainder = 3

Division statementⓘ. 8x³ − 6x² + 5x + 1 = (2x − 1)(4x² − x + 2) + 3

Check. (2x − 1)(4x² − x + 2) + 3 = 8x³ − 6x² + 5x − 2 + 3 = 8x³ − 6x² + 5x + 1, which equals the dividendⓘ. ✓

Remainder theoremⓘ check. 2x − 1 = 0 gives x = ½, so the remainder is P(½) = 8(⅛) − 6(¼) + 5(½) + 1 = 1 − 1.5 + 2.5 + 1 = 3. ✓

These questions are written in the style of Australian school exams. Each one joins long division with other ideas from this topic: the placeholderⓘ rule, the remainder theoremⓘ, the factor theoremⓘ, factorisingⓘ and solving, algebraic fractionsⓘ and volume. The marks are shown for every question and every part. Each solution begins with a short explanation of the idea, and then every step is on its own line.

Exam Style Question 1 (6 marks): The remainder theorem and long division

The polynomialⓘ P(x) = 2x³ − 5x² + x + 7 is divided by x − 2.

(a) Use the remainder theoremⓘ to find the remainderⓘ. (2 marks)

(b) Use long division to find the quotientⓘ and the remainder, and write P(x) in the form D(x) × Q(x) + R(x). (3 marks)

(c) Explain what the two remainders tell you about your working. (1 mark)

Idea. The remainder theorem is a shortcut, and long division is the full method. They must give the same remainder, so each one checks the other. Combining them is the best way to catch an arithmetic slip, because a slip in one method is very unlikely to produce exactly the same wrong number in the other.

(a) Remainder theorem (2 marks). The divisorⓘ is x − 2, so a = 2 and the remainder is P(2).

P(2) = 2(8) − 5(4) + 2 + 7 = 16 − 20 + 2 + 7 = 5.

(b) Long division (3 marks). No power is missing, so no placeholderⓘ is needed.

Cycle 1 · Divide. Leading termⓘ of the working lineⓘ ÷ leading term of the divisor: 2x³ ÷ x = 2x²

Multiply. 2x² × (x − 2) = 2x³ − 4x²

Subtract. (2x³ − 5x²) − (2x³ − 4x²) = −x²

Bring down x to give the new working line: −x² + x

Cycle 2 · Divide. Leading term of the working line ÷ leading term of the divisor: −x² ÷ x = −x

Multiply. −x × (x − 2) = −x² + 2x

Subtract. (−x² + x) − (−x² + 2x) = −x

Bring down 7 to give the new working line: −x + 7

Cycle 3 · Divide. Leading term of the working line ÷ leading term of the divisor: −x ÷ x = −1

Multiply. −1 × (x − 2) = −x + 2

Subtract. (−x + 7) − (−x + 2) = 5

Stop. The remainder 5 has degreeⓘ 0, which is less than the divisor's degree 1, so the division is finished.

Result. Quotient = 2x² − x − 1 and remainder = 5

Division statementⓘ. 2x³ − 5x² + x + 7 = (x − 2)(2x² − x − 1) + 5

Check. (x − 2)(2x² − x − 1) + 5 = 2x³ − 5x² + x + 2 + 5 = 2x³ − 5x² + x + 7, which equals the dividendⓘ. ✓

(c) Comparison (1 mark). Both methods give a remainder of 5, so the working is consistent and the answer is very likely correct. ✓

Exam Style Question 2 (8 marks): The factor theorem, factorising and solving

Let P(x) = 2x³ + 3x² − 11x − 6. Figure 6 shows the graph of y = P(x).

(a) Show that x − 2 is a factorⓘ of P(x). (2 marks)

(b) Use long division to find the quotientⓘ when P(x) is divided by x − 2. (3 marks)

(c) Factoriseⓘ P(x) completely. (1 mark)

(d) Solve P(x) = 0. (1 mark)

(e) Use Figure 6 to check your solutions. (1 mark)

Idea. A cubicⓘ is hard to factorise directly. If we can find one linearⓘ factor, long division reduces the cubic to a quadraticⓘ, and we know how to factorise a quadratic. The factor theoremⓘ tells us when a linear factor exists: x − a is a factor exactly when P(a) = 0. Each factor then gives a rootⓘ, and each real root is a point where the graph meets the x-axis, so algebra and graph must agree.

(a) Test the factor (2 marks). P(2) = 2(8) + 3(4) − 11(2) − 6 = 16 + 12 − 22 − 6 = 0. So x − 2 is a factor, by the factor theorem.

(b) Long division (3 marks).

Cycle 1 · Divide. Leading termⓘ of the working lineⓘ ÷ leading term of the divisorⓘ: 2x³ ÷ x = 2x²

Multiply. 2x² × (x − 2) = 2x³ − 4x²

Subtract. (2x³ + 3x²) − (2x³ − 4x²) = 7x²

Bring down −11x to give the new working line: 7x² − 11x

Cycle 2 · Divide. Leading term of the working line ÷ leading term of the divisor: 7x² ÷ x = 7x

Multiply. 7x × (x − 2) = 7x² − 14x

Subtract. (7x² − 11x) − (7x² − 14x) = 3x

Bring down −6 to give the new working line: 3x − 6

Cycle 3 · Divide. Leading term of the working line ÷ leading term of the divisor: 3x ÷ x = 3

Multiply. 3 × (x − 2) = 3x − 6

Subtract. (3x − 6) − (3x − 6) = 0

Stop. The remainderⓘ is 0, so there is nothing left to divide and the division is exact.

Result. Quotient = 2x² + 7x + 3 and remainder = 0

Division statementⓘ. 2x³ + 3x² − 11x − 6 = (x − 2)(2x² + 7x + 3)

Check. (x − 2)(2x² + 7x + 3) = 2x³ + 3x² − 11x − 6, which equals the dividendⓘ. ✓

(c) Factorise the quotient (1 mark). 2x² + 7x + 3 = (2x + 1)(x + 3), because (2x + 1)(x + 3) = 2x² + 6x + x + 3. ✓ So P(x) = (x − 2)(2x + 1)(x + 3).

(d) Solve (1 mark). x − 2 = 0 gives x = 2. 2x + 1 = 0 gives x = −½. x + 3 = 0 gives x = −3.

(e) Check with the graph (1 mark). Figure 6 crosses the x-axis at −3, −½ and 2, which match the solutions. ✓

Exam Style Question 3 (6 marks): Long division and algebraic fractions

(a) Simplify (x³ − x² − 4x + 4)/(x − 1). (3 marks)

(b) Write (2x² + 5x − 1)/(x + 3) in the form quotientⓘ + remainderⓘ/divisorⓘ. (3 marks)

Idea. A fraction with a polynomialⓘ on top can be simplified when the top has the bottom as a factorⓘ. Dividing tells us this: a remainder of zero means the fraction simplifies completely. If there is a remainder, the fraction becomes a quotient plus a smaller fraction, and the smaller fraction has a top of lower degreeⓘ than its bottom.

(a) Divide the numerator by x − 1 (3 marks). Watch for a zero in the middle of the working.

Cycle 1 · Divide. x³ ÷ x = x²

Multiply. x²(x − 1) = x³ − x²

Subtract. (x³ − x²) − (x³ − x²) = 0x²

Bring down −4x to give the new working lineⓘ: 0x² − 4x

Cycle 2 · Divide. 0x² ÷ x = 0x. The quotient has no x term.

Multiply. 0x(x − 1) = 0x² − 0x

Subtract. (0x² − 4x) − (0x² − 0x) = −4x

Bring down 4 to give the new working line: −4x + 4

Cycle 3 · Divide. −4x ÷ x = −4

Multiply. −4(x − 1) = −4x + 4

Subtract. (−4x + 4) − (−4x + 4) = 0

Stop. The remainder is 0, so there is nothing left to divide and the division is exact.

Result. Quotient = x² − 4 and remainder = 0

Simplify. (x³ − x² − 4x + 4)/(x − 1) = x² − 4, provided x ≠ 1.

Extension. x² − 4 = (x − 2)(x + 2), so the numerator is (x − 1)(x − 2)(x + 2).

(b) Divide the numerator by x + 3 (3 marks).

Cycle 1 · Divide. Leading termⓘ of the working line ÷ leading term of the divisor: 2x² ÷ x = 2x

Multiply. 2x × (x + 3) = 2x² + 6x

Subtract. (2x² + 5x) − (2x² + 6x) = −x

Bring down −1 to give the new working line: −x − 1

Cycle 2 · Divide. Leading term of the working line ÷ leading term of the divisor: −x ÷ x = −1

Multiply. −1 × (x + 3) = −x − 3

Subtract. (−x − 1) − (−x − 3) = 2

Stop. The remainder 2 has degree 0, which is less than the divisor's degree 1, so the division is finished.

Result. Quotient = 2x − 1 and remainder = 2

Write the fraction. (2x² + 5x − 1)/(x + 3) = 2x − 1 + 2/(x + 3), provided x ≠ −3.

Check. (x + 3)(2x − 1) + 2 = 2x² + 5x − 3 + 2 = 2x² + 5x − 1, which equals the dividendⓘ. ✓

Exam Style Question 4 (7 marks): A rainwater tank

Figure 7 shows a rectangular rainwater tank. Its volume is V = 2x³ + 7x² + 3x cubicⓘ metres, its width is x metres and its length is (x + 3) metres.

(a) Find the height in terms of x. (3 marks)

(b) The tank holds 12 cubic metres. Show that x = 1 is the only real solution and state the dimensions of the tank. (4 marks)

Idea. Volume = length × width × height, so the height is the volume divided by the base area. This is polynomialⓘ division in a real setting. Part (b) then uses the factor theoremⓘ to solve a cubic equation, and the discriminantⓘ of the remaining quadraticⓘ shows that no other real solution exists.

(a) Base area (1 mark). x(x + 3) = x² + 3x.

Divide V by the base area (2 marks). Write V as 2x³ + 7x² + 3x + 0 and divide by x² + 3x + 0.

Cycle 1 · Divide. 2x³ ÷ x² = 2x. Multiply. 2x(x² + 3x) = 2x³ + 6x². Subtract. (2x³ + 7x²) − (2x³ + 6x²) = x². Bring down 3x to give x² + 3x.

Cycle 2 · Divide. x² ÷ x² = 1. Multiply. 1(x² + 3x) = x² + 3x. Subtract. (x² + 3x) − (x² + 3x) = 0.

Result. The height is h = 2x + 1 metres. Check: x(x + 3)(2x + 1) = x(2x² + 7x + 3) = 2x³ + 7x² + 3x. ✓

(b) Form the equation (1 mark). 2x³ + 7x² + 3x = 12, so 2x³ + 7x² + 3x − 12 = 0.

Find a factorⓘ (1 mark). Try x = 1: 2 + 7 + 3 − 12 = 0. So x − 1 is a factor.

Cycle 1 · Divide. Leading termⓘ of the working lineⓘ ÷ leading term of the divisorⓘ: 2x³ ÷ x = 2x²

Multiply. 2x² × (x − 1) = 2x³ − 2x²

Subtract. (2x³ + 7x²) − (2x³ − 2x²) = 9x²

Bring down 3x to give the new working line: 9x² + 3x

Cycle 2 · Divide. Leading term of the working line ÷ leading term of the divisor: 9x² ÷ x = 9x

Multiply. 9x × (x − 1) = 9x² − 9x

Subtract. (9x² + 3x) − (9x² − 9x) = 12x

Bring down −12 to give the new working line: 12x − 12

Cycle 3 · Divide. Leading term of the working line ÷ leading term of the divisor: 12x ÷ x = 12

Multiply. 12 × (x − 1) = 12x − 12

Subtract. (12x − 12) − (12x − 12) = 0

Stop. The remainderⓘ is 0, so there is nothing left to divide and the division is exact.

Result. Quotientⓘ = 2x² + 9x + 12 and remainder = 0

Factoriseⓘ (1 mark). 2x³ + 7x² + 3x − 12 = (x − 1)(2x² + 9x + 12).

Test the quadratic. Discriminant = 9² − 4(2)(12) = 81 − 96 = −15. This is negative, so 2x² + 9x + 12 = 0 has no real solutions.

Conclusion (1 mark). The only real solution is x = 1. The tank is 1 m wide, 4 m long and 3 m high, and 1 × 4 × 3 = 12 m³. ✓

Higher-Order Example 1: Unknown coefficients and a complete factorisation

The polynomialⓘ P(x) = 2x³ + ax² + bx − 12 has x − 2 and x + 3 as factorsⓘ. Find a and b, and then factoriseⓘ P(x) completely.

Idea. Each known factor gives one equation, because the factor theoremⓘ says P(2) = 0 and P(−3) = 0. Two unknowns need two equations, so we have enough. Once a and b are known, we can divide by the two known factors together, because (x − 2)(x + 3) = x² + x − 6 is a quadraticⓘ divisorⓘ, and the quotientⓘ is the missing third factor.

Use P(2) = 0. 2(8) + 4a + 2b − 12 = 0, so 4a + 2b = −4, which simplifies to 2a + b = −2. (1)

Use P(−3) = 0. 2(−27) + 9a − 3b − 12 = 0, so 9a − 3b = 66, which simplifies to 3a − b = 22. (2)

Add (1) and (2). 5a = 20, so a = 4.

Substitute into (1). 2(4) + b = −2, so b = −10.

Write P(x). P(x) = 2x³ + 4x² − 10x − 12.

Divide by the quadratic (x − 2)(x + 3) = x² + x − 6.

Cycle 1 · Divide. Leading termⓘ of the working lineⓘ ÷ leading term of the divisor: 2x³ ÷ x² = 2x

Multiply. 2x × (x² + x − 6) = 2x³ + 2x² − 12x

Subtract. (2x³ + 4x² − 10x) − (2x³ + 2x² − 12x) = 2x² + 2x

Bring down −12 to give the new working line: 2x² + 2x − 12

Cycle 2 · Divide. Leading term of the working line ÷ leading term of the divisor: 2x² ÷ x² = 2

Multiply. 2 × (x² + x − 6) = 2x² + 2x − 12

Subtract. (2x² + 2x − 12) − (2x² + 2x − 12) = 0

Stop. The remainderⓘ is 0, so there is nothing left to divide and the division is exact.

Result. Quotient = 2x + 2 and remainder = 0

Third factor. The quotient is 2x + 2 = 2(x + 1).

Complete factorisation. P(x) = 2(x − 2)(x + 3)(x + 1).

Check. P(−1) = −2 + 4 + 10 − 12 = 0, which agrees with the factor x + 1. ✓

Higher-Order Example 2: A quadratic divisor and exact solutions

Show that x² + x − 2 divides x⁴ + x³ − 5x² − 3x + 6 exactly. Hence factoriseⓘ the quarticⓘ completely and solve x⁴ + x³ − 5x² − 3x + 6 = 0, giving exact answersⓘ.

Idea. A quartic has degreeⓘ 4, so it is much harder to factorise by guessing. A quadraticⓘ divisorⓘ removes degree 2 at once. If the remainderⓘ is zero then the quadratic is a factorⓘ, and both the divisor and the quotientⓘ are quadratics that we can factorise further. Exact answers mean we keep square rootsⓘ instead of using decimals.

Set up. Write the dividendⓘ as x⁴ + x³ − 5x² − 3x + 6. No power is missing. The divisor is x² + x − 2.

Cycle 1 · Divide. x⁴ ÷ x² = x²

Multiply. x²(x² + x − 2) = x⁴ + x³ − 2x²

Subtract. (x⁴ + x³ − 5x²) − (x⁴ + x³ − 2x²) = 0x³ − 3x²

Bring down −3x to give the new working lineⓘ: 0x³ − 3x² − 3x

Cycle 2 · Divide. 0x³ ÷ x² = 0x. The quotient has no x term.

Multiply. 0x(x² + x − 2) = 0x³ + 0x² − 0x

Subtract. (0x³ − 3x² − 3x) − (0x³ + 0x² − 0x) = −3x² − 3x

Bring down 6 to give the new working line: −3x² − 3x + 6

Cycle 3 · Divide. −3x² ÷ x² = −3

Multiply. −3(x² + x − 2) = −3x² − 3x + 6

Subtract. (−3x² − 3x + 6) − (−3x² − 3x + 6) = 0

Stop. The remainder is 0, so there is nothing left to divide and the division is exact.

Result. Quotient = x² − 3 and remainder = 0

Conclusion. The remainder is 0, so x² + x − 2 divides the quartic exactly. x⁴ + x³ − 5x² − 3x + 6 = (x² + x − 2)(x² − 3).

Factorise the divisor. x² + x − 2 = (x + 2)(x − 1), because 2 × (−1) = −2 and 2 + (−1) = 1.

Factorise the quotient. x² − 3 = (x − √3)(x + √3), a difference of two squares with 3 = (√3)².

Complete factorisation. (x + 2)(x − 1)(x − √3)(x + √3).

Solve. Set each factor to zero: x = −2, x = 1, x = √3, x = −√3.

Check one root. At x = 1: 1 + 1 − 5 − 3 + 6 = 0. ✓

3. Application

This section shows where polynomialⓘ long division is used, how it connects to what you are expected to learn in Year 10, and where it leads next.

3.1 Real-life applications in Australian contexts

Polynomialsⓘ describe lengths, areas and volumes, so polynomial division is a tool for finding a missing measurement. Here are some Australian settings where the same skills appear.

3.2 Links to the school mathematics curriculum

Year 10 is not a NAPLAN year, because NAPLAN is held in Years 3, 5, 7 and 9. This section therefore explains the links to the Year 10 mathematics curriculum, to school assessment and to later study. The points below describe the content in general terms. Your teacher will confirm exactly which parts your school teaches in Year 10 and which are extension content.

Content descriptions

• Algebra strand. Year 10 content asks students to expand, factoriseⓘ and simplify algebraic expressions and to solve equations. Polynomialⓘ long division builds directly on these skills.
• Extension content. Work with polynomials of degreeⓘ 3 or more, including the factorⓘ and remainderⓘ theorems, is usually met in Year 10 extension or at the start of Year 11. Long division is the method that connects them.
• Functions and graphs. Factors of a polynomial match the x-interceptsⓘ of its graph. Dividing helps you find all the intercepts of a cubicⓘ once one is known.
• Working mathematically. Long division gives practice in carrying out multi-step procedures accurately, checking results and explaining each step in correct notation.

Achievement standard links

• The achievement standard expects Year 10 students to use algebraic techniques fluently and to apply them in unfamiliar problems. Long division helps you factoriseⓘ and solve higher-degree expressions, which supports this.
• The standard expects clear reasoning and justification. Writing each step on its own line and checking by multiplication shows this reasoning.
• The standard expects students to connect algebra with graphs. The link between the remainderⓘ, the factor theoremⓘ and x-interceptsⓘ is a strong example.

School-based assessment

• Topic tests often include a long division question worth several marks. Marks are given for the method as well as the answer, so show every step.
• Problem-solving and investigation tasks may use volume or area models like the planter box and rainwater tank examples.
• Checking your work with the remainder theoremⓘ or by multiplication shows teachers that you can verify your own results.

Later senior secondary pathways

• Methods-style senior mathematics courses: polynomialⓘ functions, the factorⓘ and remainderⓘ theorems and solving polynomial equations are used in the early units and throughout the course.
• Specialist-style senior mathematics courses: polynomial and rational functions, partial fractionsⓘ and oblique asymptotes all rely on polynomial division.
• General senior mathematics and other studies: the algebraic fluency you build here supports modelling and problem solving.
• Beyond school: engineering, science, information technology, economics and building design all use polynomial models.

These pathway descriptions are a general guide. The official course documents give the exact requirements for each course.

4. Tips, Tricks and Exceptions

4.1 Popular tips

4.2 Exceptions and special cases

A remainder of zero

When the remainderⓘ is 0 the division is exact and the divisorⓘ is a factorⓘ. Example: (x³ − 8) ÷ (x − 2) = x² + 2x + 4 with remainder 0, so x³ − 8 = (x − 2)(x² + 2x + 4).

A divisor with a higher degree than the dividend

If the divisor has a larger degreeⓘ, no division is possible. The quotientⓘ is 0 and the remainder is the dividendⓘ. Example: (x + 2) ÷ (x² + 1) has quotient 0 and remainder x + 2.

Dividing by a number or by x alone

If the divisor is just a number or just x, divide each term separately. No long division is needed. Example: (6x³ − 4x²) ÷ 2x = 3x² − 2x.

A quotient with a zero term

Sometimes a subtraction makes the next leading termⓘ vanish. Example: (x⁴ − 1) ÷ (x² + 1) = x² − 1. After the first cycle the x³ term is 0, so the quotient has no x term. Write the placeholdersⓘ 0x³, 0x² and 0x in the dividend and keep the columns aligned so this is easy to see.

The remainder theorem with 2x − 1 or similar

For a divisor ax − b, substitute x = b/a. For 2x − 1, substitute x = ½. Example: for 8x³ − 6x² + 5x + 1 the remainder is P(½) = 3.

Restrictions in algebraic fractions

When you simplify (x³ − x² − 4x + 4)/(x − 1) to x² − 4, remember that the original expression is not defined at x = 1. We write x ≠ 1.

4.3 Common mistakes with examples

5. Further Reading

These ideas build on polynomialⓘ long division and appear in Year 11 and 12 or at university.

5.1 Synthetic division

When the divisorⓘ is x − a, you can divide using only the coefficientsⓘ. For (x³ + 2x² − 13x + 10) ÷ (x − 3), use a = 3 and the coefficients 1, 2, −13, 10.

Bring down the first coefficient: 1.

Multiply by 3 and add to the next coefficient: 1 × 3 = 3, and 2 + 3 = 5.

Multiply by 3 and add: 5 × 3 = 15, and −13 + 15 = 2.

Multiply by 3 and add: 2 × 3 = 6, and 10 + 6 = 16.

The last number is the remainderⓘ, 16. The other numbers 1, 5, 2 are the quotientⓘ x² + 5x + 2.

5.2 Theorems for polynomials of any degree

The remainder theoremⓘ and the factor theoremⓘ work for polynomialsⓘ of every degreeⓘ. They are the basis of solving polynomial equations in senior secondary mathematics.

5.3 The rational root theorem

If a polynomial with whole-number coefficientsⓘ has a rational rootⓘ p/q in simplest form, then p divides the constant termⓘ and q divides the leading coefficient. This gives a short list of numbers to test with the factor theorem.

5.4 Partial fractions

A fraction such as (5x + 7)/((x + 1)(x + 2)) can be split into simpler fractions. If the top has a degree at least as large as the bottom, long division is done first.

5.5 Oblique asymptotes

Writing (x² + 3x + 5)/(x + 1) as x + 2 + 3/(x + 1) shows that for large x the graph gets close to the line y = x + 2. This line is called an oblique asymptoteⓘ.

5.6 The Euclidean algorithm for polynomials

Repeated polynomial division finds the highest common factorⓘ of two polynomials, in the same way as repeated whole-number division finds a highest common factor.

5.7 The fundamental theorem of algebra

A polynomial of degree n has exactly n roots when complex numbers are allowed and repeated roots are counted. For example x² + 1 has no real roots but has two complex roots.

5.8 Polynomials in coding and science

Computer scientists use polynomial division to design error-correcting codes. Scientists use polynomial models to describe curves in data.

6. Difficult Words and Vocabulary

Every key term marked with ⓘ in this compendium is defined here, in alphabetical order. The definitions are the same ones that open when you click the ⓘ.