1. Algebraic Long Division

Video coming soon2 worked examples

Theory

Before partial fractions can be used, the fraction must be proper — the degree of the numerator lower than the degree of the denominator. When it is not, we divide first, and the tool for that is algebraic long division.

The process is identical to long division with numbers. Dividing $745$ by $6$ gives $124$ remainder $1$ , which we can write as:

\frac{745}{6} = 124 + \frac{1}{6}

The same layout works with polynomials. Dividing a polynomial $f(x)$ by $g(x)$ produces a quotient $q(x)$ and a remainder $r(x)$ :

\frac{f(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)}

At each stage, divide the leading term of what is left by the leading term of the divisor, multiply the whole divisor by that result, and subtract. Stop when the degree of what remains is lower than the degree of the divisor — that leftover is the remainder, and the fraction $\frac{r(x)}{g(x)}$ is guaranteed to be proper.

The Golden Rule: compare degrees before you do anything else. Only divide the leading terms to get each new term of the quotient, and keep going only while the degree of the remainder is at least the degree of the divisor.

⚠️ Common Examiner Traps

Missing terms: write in placeholders such as $0x^2$ so that like terms stay in line — this is the single most common source of error.
Sign errors on subtraction: you subtract the whole product, so every sign in it changes. Bracket it before subtracting.
Stopping at the wrong point: continue only while the degree of the remaining expression is greater than or equal to the degree of the divisor.
Checking your answer: multiply back — $g(x)\,q(x) + r(x)$ must return the original numerator exactly.

Worked examples

Example 1

Divide $x^3 + 2x^2 - 5x + 1$ by $x - 2$ , stating the quotient and remainder.

Step 1: Divide the leading terms: $x^3 \div x = x^2$ . Multiply the divisor by $x^2$ and subtract:

x^2(x-2) = x^3 - 2x^2 \quad\implies\quad (x^3 + 2x^2) - (x^3 - 2x^2) = 4x^2

Step 2: Bring down the next term to give $4x^2 - 5x$ . Now $4x^2 \div x = 4x$ :

4x(x-2) = 4x^2 - 8x \quad\implies\quad (4x^2 - 5x) - (4x^2 - 8x) = 3x

Step 3: Bring down the $+1$ to give $3x + 1$ . Now $3x \div x = 3$ :

3(x-2) = 3x - 6 \quad\implies\quad (3x + 1) - (3x - 6) = 7

Step 4: The remainder $7$ has lower degree than $x-2$ , so we stop. The quotient is $x^2 + 4x + 3$ and the remainder is $7$ :

\frac{x^3 + 2x^2 - 5x + 1}{x-2} = x^2 + 4x + 3 + \frac{7}{x-2}

Check: $(x-2)(x^2+4x+3) + 7 = x^3 + 2x^2 - 5x - 6 + 7 = x^3 + 2x^2 - 5x + 1$ ✓

Example 2

Express $\dfrac{2x^3 - 3x + 5}{x^2 + 1}$ as a quotient plus a proper fraction.

Step 1: The numerator has no $x^2$ term, so insert a placeholder before dividing:

2x^3 + 0x^2 - 3x + 5 \ \div \ (x^2 + 1)

Step 2: Divide the leading terms: $2x^3 \div x^2 = 2x$ . Multiply and subtract:

2x(x^2+1) = 2x^3 + 2x

(2x^3 + 0x^2 - 3x + 5) - (2x^3 + 2x) = 0x^2 - 5x + 5

Step 3: What remains, $-5x + 5$ , has degree $1$ , which is lower than the degree of $x^2+1$ . So the division stops here:

\frac{2x^3 - 3x + 5}{x^2+1} = 2x + \frac{5 - 5x}{x^2+1}

Check: $(x^2+1)(2x) + (5-5x) = 2x^3 + 2x + 5 - 5x = 2x^3 - 3x + 5$ ✓