Mathematics/Lessons/Algebraic Methods/Algebraic Division
Pure · Algebraic Methods · Year 2

Algebraic Division

When a numerator's degree catches up with the denominator's, no amount of factorising helps — divide first, and everything that follows gets simpler.

10 min read 4 worked examples

Dividing polynomials by hand

Algebraic long division divides one polynomial by another the same way long division divides numbers — matching the leading term, multiplying back, subtracting, and repeating with what's left. Dividing a polynomial $f(x)$ by a linear divisor $(x-a)$ produces a quotient $q(x)$, one degree lower than $f(x)$, and a constant remainder $r$:

$f(x) = (x-a)\,q(x) + r$

At each step: divide the leading term of what's left by the leading term of the divisor to get the next term of the quotient, multiply that term by the whole divisor, and subtract the result from what's left. Stop once the degree of what remains is lower than the divisor's degree — that's the remainder.

A shortcut for checking the remainder

By the remainder theorem (from Year 1), dividing $f(x)$ by $(x-a)$ always leaves remainder $f(a)$. This doesn't replace doing the division — the quotient still has to be found by the algorithm — but it's a fast, independent check that the remainder came out right.

Improper fractions need division first

An algebraic fraction is improper if its numerator's degree is greater than or equal to its denominator's — the case partial fractions can't handle directly. Dividing first rewrites it as a polynomial plus a proper fraction, which can then be split into partial fractions if the denominator factorises:

$\dfrac{f(x)}{g(x)} = q(x) + \dfrac{r(x)}{g(x)}, \qquad \deg r < \deg g$

Worked example 1

Exact division (remainder zero)
Divide $x^3+x^2-10x+8$ by $(x-2)$, and use the result to fully factorise the cubic.
Bus-stop long division of x cubed plus x squared minus 10x plus 8 by x minus 2, showing each subtraction row and ending with remainder 0.
The division laid out bus-stop style: $(x-2)$ outside the bracket, $x^3+x^2-10x+8$ inside, with the quotient $x^2+3x-4$ building up on top and each subtraction row shown underneath.
MethodMatch, multiply, subtract — one term at a time
1
$x^3\div x = x^2$. Multiply: $x^2(x-2)=x^3-2x^2$. Subtract: $(x^3+x^2)-(x^3-2x^2)=3x^2$, bringing down $-10x$ gives $3x^2-10x$.
First term of the quotient.
2
$3x^2\div x = 3x$. Multiply: $3x(x-2)=3x^2-6x$. Subtract: $(3x^2-10x)-(3x^2-6x)=-4x$, bringing down $+8$ gives $-4x+8$.
Second term of the quotient.
3
$-4x\div x = -4$. Multiply: $-4(x-2)=-4x+8$. Subtract: $(-4x+8)-(-4x+8)=0$.
Remainder is zero — an exact factor.
4
Quotient $x^2+3x-4=(x+4)(x-1)$, remainder $0$.
Factorise the quadratic quotient.
5
So $x^3+x^2-10x+8 = (x-2)(x+4)(x-1)$.
Combine the divisor with the factorised quotient.
▪ solved
Examiner's eye

A remainder of zero means the divisor is a genuine factor — exactly what the factor theorem would have predicted by checking $f(2)=0$ first. Long division goes further than the factor theorem by finding the full quotient, not just confirming a factor exists.

Worked example 2

Division with a non-zero remainder
Divide $2x^3-3x^2+4x-1$ by $(x-1)$, giving the quotient and remainder.
Bus-stop long division of 2x cubed minus 3x squared plus 4x minus 1 by x minus 1, ending with remainder 2.
Same bus-stop layout, but this time the last row doesn't cancel: quotient $2x^2-x+3$ with remainder $2$ left over.
MethodSame algorithm — this time it doesn't reach zero
1
$2x^3\div x=2x^2$. Multiply: $2x^2(x-1)=2x^3-2x^2$. Subtract: $-x^2$, bring down $4x$: $-x^2+4x$.
First term of the quotient.
2
$-x^2\div x=-x$. Multiply: $-x(x-1)=-x^2+x$. Subtract: $3x$, bring down $-1$: $3x-1$.
Second term of the quotient.
3
$3x\div x=3$. Multiply: $3(x-1)=3x-3$. Subtract: $(3x-1)-(3x-3)=2$.
Degree of what's left (0) is now below the divisor's degree (1) — stop.
4
Quotient $2x^2-x+3$, remainder $2$: $2x^3-3x^2+4x-1 = (x-1)(2x^2-x+3) + 2$.
Assemble the final identity.
5
Check: $f(1) = 2-3+4-1=2$, matching the remainder exactly.
Remainder theorem confirms the answer independently.
▪ verified

Worked example 3

Rewriting an improper fraction
Express $\dfrac{x^2+2x-1}{x+1}$ in the form $q(x) + \dfrac{r}{x+1}$.
Bus-stop long division of x squared plus 2x minus 1 by x plus 1, ending with remainder negative 2.
Dividing $x^2+2x-1$ by $(x+1)$ bus-stop style: quotient $x+1$ on top, remainder $-2$ at the bottom.
MethodDivide first, then read off the polynomial-plus-fraction form
1
$x^2\div x=x$. Multiply: $x(x+1)=x^2+x$. Subtract: $x$, bring down $-1$: $x-1$.
First term of the quotient.
2
$x\div x=1$. Multiply: $1(x+1)=x+1$. Subtract: $(x-1)-(x+1)=-2$.
Second term of the quotient; remainder is a constant.
3
Quotient $x+1$, remainder $-2$.
The numerator's degree (2) matched the denominator's (1) plus one — improper.
4
So $\dfrac{x^2+2x-1}{x+1} = x+1 - \dfrac{2}{x+1}$.
Write the quotient plus the remainder over the original divisor.
▪ solved

Worked example 4

Combining with partial fractions · Challenge
Express $\dfrac{2x^2+3x-1}{(x+2)(x-1)}$ in the form $C + \dfrac{A}{x+2} + \dfrac{B}{x-1}$.
Bus-stop long division of 2x squared plus 3x minus 1 by x squared plus x minus 2, ending with remainder x plus 3.
Dividing by the whole quadratic denominator $(x+2)(x-1)=x^2+x-2$ in one bus-stop step: quotient $2$, remainder $x+3$.
MethodRecognise it's improper, divide, then split what's left
1
$(x+2)(x-1) = x^2+x-2$. The numerator's degree (2) equals this — improper, so divide first.
Same-degree top and bottom always signals division is needed.
2
Leading terms match with quotient $2$: $2(x^2+x-2)=2x^2+2x-4$.
Only one division step is needed, since both are degree 2.
3
Subtract: $(2x^2+3x-1)-(2x^2+2x-4) = x+3$.
Remainder, degree 1 — lower than the denominator's degree 2.
4
So the expression equals $2 + \dfrac{x+3}{(x+2)(x-1)}$.
Quotient plus the remainder over the original denominator.
5
Split the remaining proper fraction: $x+3 = A(x-1)+B(x+2)$.
Now a standard partial fractions problem.
6
Let $x=-2$: $1 = -3A \Rightarrow A=-\dfrac{1}{3}$. Let $x=1$: $4=3B \Rightarrow B=\dfrac{4}{3}$.
Substitute each root in turn.
7
Result: $2 - \dfrac{1/3}{x+2} + \dfrac{4/3}{x-1}$.
Combine the polynomial part with the split fraction.
▪ solved
Look deeper — the full picture

This worked example closes the loop the whole topic has been building towards: an improper fraction (this lesson) whose proper remainder happens to have a repeated or distinct factorised denominator (the last two lessons) can always be fully decomposed. Check the degrees first, divide if needed, then split whatever proper fraction remains.

Common mistake

Trying to split an improper fraction straight into partial fractions without dividing first. Setting $\dfrac{2x^2+3x-1}{(x+2)(x-1)} = \dfrac{A}{x+2}+\dfrac{B}{x-1}$ directly cannot work: the right-hand side is always a proper fraction (degree of the top less than the bottom once combined), so it can never equal an improper one. Always compare the degrees of numerator and denominator before choosing a method.

Check yourself

Divide $x^3-7x+6$ by $(x-2)$, and use the result to fully factorise the cubic.
(Hint: there's no $x^2$ term, so treat its coefficient as 0.)

Bus-stop long division of x cubed plus 0x squared minus 7x plus 6 by x minus 2, ending with remainder 0.
$x^3-7x+6$ has no $x^2$ term, so a $0x^2$ placeholder keeps every column lined up: quotient $x^2+2x-3$, remainder $0$.
Method
1
$x^3+0x^2-7x+6 \div (x-2)$: quotient $x^2+2x-3$, remainder $0$.
2
Factorise: $x^2+2x-3 = (x+3)(x-1)$.
Two numbers multiplying to −3, summing to 2: 3 and −1.
3
So $x^3-7x+6 = (x-2)(x+3)(x-1)$.
▪ solved
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