Dividing Polynomials

Divide a polynomial by (x ± a) or (ax ± b) using long division or comparing coefficients, give the quotient and remainder, and factorise a cubic fully.

In the exam for Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 14 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Division with letters

You already know how to divide numbers. 47 ÷ 5 is 9 with 2 left over, and you can check it: 5 × 9 + 2 = 47.

Polynomials divide in the same way. A polynomial is an expression built from powers of x, like $x^{3} - 4 x^{2} + x + 7$ . In this lesson you divide one by a bracket such as $x - 3$ or $2 x - 1$ .

Formula

P (x) = D (x) \times Q (x) + R

$P (x)$ — the polynomial you are dividing.
$D (x)$ — the bracket you divide by, called the divisor.
$Q (x)$ — the quotient: the answer to the division.
$R$ — the remainder: what is left over. When you divide by a bracket like x − 3, it is just a number.

Worked example

1
Use the numbers first: 47 divided by 5.
2
$47 \div 5 = 9 remainder 2$ — 5 goes into 47 nine times (that makes 45), and 2 is left.
3
$47 = 5 \times 9 + 2$ — Same shape as the formula: divisor × quotient + remainder.

Rule

Dividing a cubic by a bracket like $x - 3$ always gives a quadratic quotient and a number remainder. This fact is not on any formula sheet (0606, 4037 or 4PM1), so know it.

Why it matters: division breaks a cubic into a bracket times a quadratic. That is the first step to factorising and solving cubics, which come up often on both boards.

Key idea

A number like 1342 is secretly a polynomial: $1 \times 1 0^{3} + 3 \times 1 0^{2} + 4 \times 10 + 2$ . Put $x = 10$ into $x^{3} + 3 x^{2} + 4 x + 2$ and you get 1342. Polynomial division is the long division you learnt in primary school, with x in place of 10.

Long division

Long division of polynomials repeats the same four moves until there is nothing left to bring down.

Steps

1
1. Divide: the first term of what you have by the first term of the bracket.
2. Multiply: that answer by the whole bracket.
3. Subtract: take the product away. Change every sign in the product, then add.
4. Bring down the next term, and repeat.

Worked example

1
Divide $x^{3} - 4 x^{2} + x + 7$ by $x - 3$ .
2
$x^{3} \div x = x^{2}$ — Divide the first term by x. This is the first term of the quotient.
3
$x^{2} (x - 3) = x^{3} - 3 x^{2}$ — Multiply the whole bracket by x².
4
$(x^{3} - 4 x^{2}) - (x^{3} - 3 x^{2}) = - x^{2}$ — Subtract. The x³ terms cancel, as they always should. Bring down + x.
5
$- x^{2} \div x = - x$ — Divide again. The next term of the quotient is −x.
6
$(- x^{2} + x) - (- x^{2} + 3 x) = - 2 x$ — Multiply: −x(x − 3) = −x² + 3x. Subtract, then bring down + 7.
7
$- 2 x \div x = - 2$ — The last term of the quotient.
8
$(- 2 x + 7) - (- 2 x + 6) = 1$ — Multiply: −2(x − 3) = −2x + 6. Subtract. Nothing is left to bring down, so 1 is the remainder.
9
$x^{3} - 4 x^{2} + x + 7 = (x - 3) (x^{2} - x - 2) + 1$ — Quotient x² − x − 2, remainder 1.

Rule

Check your answer: expand $(x - 3) (x^{2} - x - 2)$ and add 1. You must get back the polynomial you started with.

Common mistake

Subtracting only the first term. The minus sign applies to the whole line: $- 4 x^{2} - (- 3 x^{2}) = - x^{2}$ , not $- 7 x^{2}$ .

Key idea

Divide, multiply, subtract, bring down. The first term should vanish every time. If it does not, a sign has gone wrong.

Gaps and plus brackets

Sometimes a power of x is missing, for example there is no $x^{2}$ term. Fill the gap with a zero term first, so that every line stays in step.

Worked example

1
Divide $2 x^{3} - 5 x + 4$ by $x + 2$ .
2
$2 x^{3} + 0 x^{2} - 5 x + 4$ — There is no x² term, so write 0x² as a placeholder.
3
$2 x^{3} \div x = 2 x^{2}$ — The first term of the quotient.
4
$(2 x^{3} + 0 x^{2}) - (2 x^{3} + 4 x^{2}) = - 4 x^{2}$ — Multiply: 2x²(x + 2) = 2x³ + 4x². Subtract, then bring down −5x.
5
$- 4 x^{2} \div x = - 4 x$ — The next term of the quotient.
6
$(- 4 x^{2} - 5 x) - (- 4 x^{2} - 8 x) = 3 x$ — Multiply: −4x(x + 2) = −4x² − 8x. Minus a minus is plus: −5x + 8x = 3x. Bring down + 4.
7
$3 x \div x = 3$ — The last term of the quotient.
8
$(3 x + 4) - (3 x + 6) = - 2$ — Multiply: 3(x + 2) = 3x + 6. Subtract to leave −2.
9
$2 x^{3} - 5 x + 4 = (x + 2) (2 x^{2} - 4 x + 3) - 2$ — Quotient 2x² − 4x + 3, remainder −2.

Common mistake

Skipping the gap. Without $0 x^{2}$ , the $- 5 x$ lands under the $x^{2}$ terms and every later line is wrong.

Common mistake

Sign slips with a plus bracket. Subtracting $- 8 x$ means adding $8 x$ .

Key idea

A remainder can be negative. It is simply the number left after the last subtraction.

Dividing by ax + b

Both boards also divide by brackets like $2 x - 1$ or $3 x + 2$ . The four moves do not change. The only difference: you now divide by $2 x$ , the first term of the bracket, not by $x$ .

Worked example

1
Divide $6 x^{3} + 7 x^{2} - 9 x + 5$ by $2 x - 1$ .
2
$6 x^{3} \div 2 x = 3 x^{2}$ — Divide the first term by 2x.
3
$(6 x^{3} + 7 x^{2}) - (6 x^{3} - 3 x^{2}) = 10 x^{2}$ — Multiply: 3x²(2x − 1) = 6x³ − 3x². Subtract: 7x² + 3x² = 10x². Bring down −9x.
4
$10 x^{2} \div 2 x = 5 x$ — The next term of the quotient.
5
$(10 x^{2} - 9 x) - (10 x^{2} - 5 x) = - 4 x$ — Multiply: 5x(2x − 1) = 10x² − 5x. Subtract, then bring down + 5.
6
$- 4 x \div 2 x = - 2$ — The last term of the quotient.
7
$(- 4 x + 5) - (- 4 x + 2) = 3$ — Multiply: −2(2x − 1) = −4x + 2. Subtract to leave 3.
8
$6 x^{3} + 7 x^{2} - 9 x + 5 = (2 x - 1) (3 x^{2} + 5 x - 2) + 3$ — Quotient 3x² + 5x − 2, remainder 3.

Common mistake

Dividing by $x$ instead of $2 x$ . The quotient would start with $6 x^{2}$ , and every later line would be wrong.

Key idea

Look at the first term of the bracket. That is what you divide by, every single time.

Comparing coefficients

When the bracket is a factor, the remainder is 0. Then there is a quicker way: write the cubic as the bracket times an unknown quadratic, and match the terms. Examiners accept this in place of long division.

Formula

P (x) \equiv (x - a) (A x^{2} + B x + C)

$\equiv$ — identical: the two sides are equal for every value of x, so matching terms must be equal.
$A$ — found from the x³ terms.
$C$ — found from the number terms.
$B$ — found from the x² terms. The x terms then give a free check.

Worked example

1
Given that $x + 3$ is a factor, write $x^{3} + 2 x^{2} - 5 x - 6$ as $x + 3$ times a quadratic.
2
$x^{3} + 2 x^{2} - 5 x - 6$ $\equiv (x + 3) (A x^{2} + B x + C)$ — Set up the product.
3
$A = 1$ — x³ terms: x × Ax² must give 1x³.
4
$3 C = - 6, C = - 2$ — Number terms: 3 × C must give −6.
5
$B + 3 A = 2, B = - 1$ — x² terms: x × Bx and 3 × Ax² together give 2x².
6
$C + 3 B = - 2 - 3 = - 5$ — Check the x terms: −5x matches the cubic. Good.
7
$x^{3} + 2 x^{2} - 5 x - 6 = (x + 3) (x^{2} - x - 2)$ — A linear factor times a quadratic factor.

Common mistake

Forgetting a product. The $x^{2}$ terms come from two places: $x \times B x$ and $3 \times A x^{2}$ . Miss one and B is wrong.

Key idea

Top term gives A. Bottom term gives C. The middle terms give B, and then check it.

Factorise completely

A favourite exam question gives you one factor and asks you to factorise the cubic completely. Completely means that no bracket can be factorised any further.

Steps

1
1. Divide by the given factor, by long division or by comparing coefficients.
2. Write the cubic as the factor times a quadratic.
3. Factorise the quadratic, if it can be done.
4. Write the answer as one product of all the brackets.

Worked example

1
Given that $2 x + 1$ is a factor of $2 x^{3} - 3 x^{2} - 8 x - 3$ , factorise it completely.
2
$2 x^{3} \div 2 x = x^{2}$ — Divide by 2x, the first term of the bracket.
3
$(2 x^{3} - 3 x^{2}) - (2 x^{3} + x^{2}) = - 4 x^{2}$ — Multiply: x²(2x + 1). Subtract, then bring down −8x.
4
$(- 4 x^{2} - 8 x) - (- 4 x^{2} - 2 x) = - 6 x$ — Divide: −4x² ÷ 2x = −2x. Multiply: −2x(2x + 1). Subtract, then bring down −3.
5
$(- 6 x - 3) - (- 6 x - 3) = 0$ — Divide: −6x ÷ 2x = −3. Multiply and subtract. Remainder 0, as it must be for a factor.
6
$2 x^{3} - 3 x^{2} - 8 x - 3 = (2 x + 1) (x^{2} - 2 x - 3)$ — Quotient x² − 2x − 3.
7
$x^{2} - 2 x - 3 = (x - 3) (x + 1)$ — Two numbers that multiply to −3 and add to −2: −3 and 1.
8
$2 x^{3} - 3 x^{2} - 8 x - 3 = (2 x + 1) (x - 3) (x + 1)$ — The final answer: a product of three linear factors.

Mark schemes give a method mark for trying to find the quadratic factor, an accuracy mark for the correct quadratic, and the last mark for the full product. So show each stage.

Common mistake

Stopping at $(2 x + 1) (x^{2} - 2 x - 3)$ . The question says completely, so the quadratic must be split too.

Rule

If the quadratic factor will not factorise, leave it whole. For example, $x^{2} + x + 1$ has no simpler factors.

Key idea

One division turns a cubic into a quadratic problem, and quadratics you already know how to handle.

Which method? Read the question

Before you start, read what the question asks. The wording tells you which tool to use.

You see	You do
“Divide … by $x - a$ ”	Long division: divide, multiply, subtract, bring down
A power of x is missing	Put in a zero term first, such as $0 x^{2}$
Divide by $a x + b$	Same four moves; divide each time by $a x$
“Find the quotient and the remainder” (4PM1)	Give both, or write $P (x) = D (x) \times Q (x) + R$
“Given that … is a factor”	Divide or compare coefficients; the remainder must be 0
“Factorise completely”	Divide, then factorise the quadratic too

Common mistake

Giving only the quotient. If the question asks for the remainder, state it, even when it is 0.

Key idea

Every division checks itself: divisor × quotient + remainder must give back the polynomial you started with.

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Real past-paper questions on dividing polynomials

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