Mathematics 9709 · Pure Mathematics 3

Division of polynomials

Dividing one polynomial by another is the same long division you already do with numbers. You just carry the x's along too.
It uses one rule the whole chapter needs: dividend = divisor × quotient + remainder.

$\text{dividend}=\text{divisor}\times\text{quotient}+\text{remainder}$ . Repeat four moves — divide, multiply, subtract, bring down — until what is left has a lower degree than the divisor.

It's long division, with x's

Start at the top power and repeat four moves until the leftover has a lower degree than the divisor:

Divide the leading terms to get the next quotient term.
Multiply the whole divisor by that term.
Subtract, which kills the current leading term.
Bring down the next term and repeat.

Worked example

Divide $x^3+2x^2-11x+6$ by $x-2$ (Worked example 1.7, textbook)

The four moves give quotient $x^2+4x-3$ with nothing left over:

$x^3+2x^2-11x+6=(x-2)(x^2+4x-3)$

The remainder is 0, so $x-2$ is a factor (that is §1.5).

Answer

\text{quotient } x^2+4x-3,\ \text{remainder } 0

Mind the gaps: write 0 for a missing term

If a power is missing, hold its place with a zero so the columns stay lined up.
To divide $2x^3-x+52$ by $x+3$ , first write the dividend as $2x^3+0x^2-x+52$ .

Worked example

Divide $2x^3-x+52$ by $x+3$ (Worked example 1.8, textbook)

$2x^3+0x^2-x+52=(x+3)(2x^2-6x+17)+1$

Answer

\text{quotient } 2x^2-6x+17,\ \text{remainder } 1

Common mistake

Forgetting the

0x^2

placeholder throws every column out of line and breaks the division. Always write the dividend out in full, highest power first.

Dividing by a quadratic: same moves, stop sooner

The divisor can be a quadratic (and the dividend degree 4). Same four moves. Just stop once the leftover is degree 1 or lower (below the quadratic divisor).
So the remainder can be a linear expression like $x$ , not just a plain number.

Worked example

Real question: divide $x^3+5x^2-2x-15$ by $x^2-3$

9709/31 May/June 2025 Q10(a) — word for word from a recent paper.

Divide leading terms $x^3\div x^2=x$ , multiply $x(x^2-3)=x^3-3x$ and subtract: $(x^3+5x^2-2x-15)-(x^3-3x)=5x^2+x-15$ .
Repeat: $5x^2\div x^2=5$ , multiply $5(x^2-3)=5x^2-15$ and subtract: $(5x^2+x-15)-(5x^2-15)=x$ .
The leftover $x$ is degree 1, below the divisor, so stop: $x^3+5x^2-2x-15=(x^2-3)(x+5)+x$ .

Answer

\text{quotient } x+5,\ \text{remainder } x

Common mistake

With a quadratic divisor the remainder can still have an

x

in it. Don't force it down to a plain number. Stop the moment the leftover's degree drops below the divisor's.

A quick shortcut: match coefficients instead of dividing

For a quadratic divisor you can skip long division. The answer must look like $x^4+16=(x^2+4)(x^2+Ax+B)+(Cx+D)$ (quotient one degree lower, linear remainder). Multiply out the right side and compare coefficients for each power. Matching $x^3,x^2,x$ and the constant gives $A,B,C,D$ . For $x^4+16\div(x^2+4)$ this gives $A=0,\ B=-4,\ C=0,\ D=32$ , i.e. quotient $x^2-4$ , remainder $32$ (9709/33 Oct/Nov 2024 Q9). Useful when the long division gets messy.

Now you try

Find the quotient and remainder when $2x^3+3x^2-4$ is divided by $x-1$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Use long division (and remember $+0x$ for the missing x term):

Answer

\text{quotient } 2x^2+5x+5,\ \text{remainder } 1

Now a real one with a quadratic divisor: find the quotient and remainder when $x^4-3x^3+9x^2-12x+27$ is divided by $x^2+5$ . (9709/32 Feb/March 2024 Q1)

Your turn— tap to reveal the worked answer (9709/32 Feb/March 2024 Q1)

Same four moves; stop once the leftover drops below degree 2:

Answer

\text{quotient } x^2-3x+4,\ \text{remainder } 3x+7

PreviousSolving modulus inequalities NextThe factor theorem

Division of polynomials

Dividing one polynomial by another is the same long division you already do with numbers. You just carry the x's along too.
It uses one rule the whole chapter needs: dividend = divisor × quotient + remainder.

$\text{dividend}=\text{divisor}\times\text{quotient}+\text{remainder}$ . Repeat four moves — divide, multiply, subtract, bring down — until what is left has a lower degree than the divisor.

It's long division, with x's

Start at the top power and repeat four moves until the leftover has a lower degree than the divisor:

Divide the leading terms to get the next quotient term.
Multiply the whole divisor by that term.
Subtract, which kills the current leading term.
Bring down the next term and repeat.

Worked example

Divide $x^3+2x^2-11x+6$ by $x-2$ (Worked example 1.7, textbook)

The four moves give quotient $x^2+4x-3$ with nothing left over:

$x^3+2x^2-11x+6=(x-2)(x^2+4x-3)$

The remainder is 0, so $x-2$ is a factor (that is §1.5).

Answer

\text{quotient } x^2+4x-3,\ \text{remainder } 0

Mind the gaps: write 0 for a missing term

If a power is missing, hold its place with a zero so the columns stay lined up.
To divide $2x^3-x+52$ by $x+3$ , first write the dividend as $2x^3+0x^2-x+52$ .

Worked example

Divide $2x^3-x+52$ by $x+3$ (Worked example 1.8, textbook)

$2x^3+0x^2-x+52=(x+3)(2x^2-6x+17)+1$

Answer

\text{quotient } 2x^2-6x+17,\ \text{remainder } 1

Common mistake

Forgetting the

0x^2

placeholder throws every column out of line and breaks the division. Always write the dividend out in full, highest power first.

Dividing by a quadratic: same moves, stop sooner

The divisor can be a quadratic (and the dividend degree 4). Same four moves. Just stop once the leftover is degree 1 or lower (below the quadratic divisor).
So the remainder can be a linear expression like $x$ , not just a plain number.

Worked example

Real question: divide $x^3+5x^2-2x-15$ by $x^2-3$

9709/31 May/June 2025 Q10(a) — word for word from a recent paper.

Divide leading terms $x^3\div x^2=x$ , multiply $x(x^2-3)=x^3-3x$ and subtract: $(x^3+5x^2-2x-15)-(x^3-3x)=5x^2+x-15$ .
Repeat: $5x^2\div x^2=5$ , multiply $5(x^2-3)=5x^2-15$ and subtract: $(5x^2+x-15)-(5x^2-15)=x$ .
The leftover $x$ is degree 1, below the divisor, so stop: $x^3+5x^2-2x-15=(x^2-3)(x+5)+x$ .

Answer

\text{quotient } x+5,\ \text{remainder } x

Common mistake

With a quadratic divisor the remainder can still have an

x

in it. Don't force it down to a plain number. Stop the moment the leftover's degree drops below the divisor's.

A quick shortcut: match coefficients instead of dividing

Now you try

Find the quotient and remainder when $2x^3+3x^2-4$ is divided by $x-1$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Use long division (and remember $+0x$ for the missing x term):

Answer

\text{quotient } 2x^2+5x+5,\ \text{remainder } 1

Now a real one with a quadratic divisor: find the quotient and remainder when $x^4-3x^3+9x^2-12x+27$ is divided by $x^2+5$ . (9709/32 Feb/March 2024 Q1)

Your turn— tap to reveal the worked answer (9709/32 Feb/March 2024 Q1)

Same four moves; stop once the leftover drops below degree 2:

Answer

\text{quotient } x^2-3x+4,\ \text{remainder } 3x+7