Algebraic Long Division: GCSE Higher Step-by-Step

When is algebraic long division used?

At GCSE Higher, you use algebraic long division to:

• Factorise a cubic expression when one linear factor is known or can be guessed.
• Find the quotient and remainder when one polynomial is divided by another.
• Verify that a given expression is a factor of a polynomial (remainder = 0 means exact factor).

If you know that (x − a) is a factor of a cubic f(x), you can also check by substituting x = a into f(x). If f(a) = 0, the factor is confirmed — then divide to find the remaining quadratic and factorise that.

How does algebraic long division work — step by step?

General method: Divide polynomial P(x) by divisor D(x).

1. Write the polynomial in descending powers. Insert 0 as a placeholder coefficient for any missing power.
2. Divide the leading term of P by the leading term of D — this gives the first term of the quotient.
3. Multiply the full divisor by that term and subtract it from P.
4. Bring down the next term and repeat.
5. Stop when the degree of the remainder is less than the degree of D.

Worked example 1 — exact division

Divide x³ + 2x² − 5x − 6 by (x − 2).

         x² + 4x + 3
       ________________
x − 2 | x³ + 2x² − 5x − 6
         x³ − 2x²
         ─────────────
               4x² − 5x
               4x² − 8x
               ─────────────
                     3x − 6
                     3x − 6
                     ─────────────
                           0

Step-by-step:

1. x³ ÷ x = x². Multiply: x²(x − 2) = x³ − 2x². Subtract: (x³ + 2x²) − (x³ − 2x²) = 4x².
2. Bring down: 4x² − 5x. Divide: 4x² ÷ x = 4x. Multiply: 4x(x − 2) = 4x² − 8x. Subtract: (4x² − 5x) − (4x² − 8x) = 3x.
3. Bring down: 3x − 6. Divide: 3x ÷ x = 3. Multiply: 3(x − 2) = 3x − 6. Subtract: 0.

Result: x³ + 2x² − 5x − 6 = (x − 2)(x² + 4x + 3)

Now factorise the quadratic: x² + 4x + 3 = (x + 1)(x + 3).

Full factorisation: (x − 2)(x + 1)(x + 3)

Verification: expand (x − 2)(x + 1) = x² − x − 2, then (x² − x − 2)(x + 3) = x³ + 3x² − x² − 3x − 2x − 6 = x³ + 2x² − 5x − 6 ✓

Worked example 2 — division with a remainder

Divide 2x³ − 3x + 1 by (x + 2).

Note: the x² term is missing, so insert 0x²:

2x³ + 0x² − 3x + 1 divided by (x + 2).

1. 2x³ ÷ x = 2x². Multiply: 2x²(x + 2) = 2x³ + 4x². Subtract: (2x³ + 0x²) − (2x³ + 4x²) = −4x².
2. Bring down: −4x² − 3x. Divide: −4x² ÷ x = −4x. Multiply: −4x(x + 2) = −4x² − 8x. Subtract: (−4x² − 3x) − (−4x² − 8x) = 5x.
3. Bring down: 5x + 1. Divide: 5x ÷ x = 5. Multiply: 5(x + 2) = 5x + 10. Subtract: (5x + 1) − (5x + 10) = −9.

Result: (2x³ − 3x + 1) ÷ (x + 2) = 2x² − 4x + 5 remainder −9.

Written fully: 2x³ − 3x + 1 = (x + 2)(2x² − 4x + 5) − 9.

Check by substituting x = −2: 2(−8) + 6 + 1 = −16 + 6 + 1 = −9 ✓ (remainder theorem).

How do you find a factor to use as the divisor?

When the question asks you to "fully factorise" a cubic f(x) = x³ + bx² + cx + d, try the integer factors of d as candidates for x = a (so that (x − a) is a factor):

For Example 1 above, d = −6. Test a = 2: f(2) = 8 + 8 − 10 − 6 = 0 ✓ → (x − 2) is a factor.

Frequently asked questions

Can I use inspection (matching coefficients) instead of long division?

Yes — for simple cubics, inspection (writing x³ + 2x² − 5x − 6 = (x − 2)(x² + px + q) and equating coefficients) is quicker. Long division is the reliable method for more complex polynomials or when inspection is unclear. Both are acceptable at GCSE.

What if there is a remainder in the division?

A non-zero remainder means the divisor is not a factor. The result is written as: P(x) = D(x) × Q(x) + R, where R is the remainder (a constant when dividing a cubic by a linear). The question will usually tell you to "show that" the remainder has a particular value, so compute it and compare.

Is there a quicker way to find the remainder without dividing?

Yes — the remainder theorem: the remainder when f(x) is divided by (x − a) equals f(a). Substitute x = a into the original polynomial. This gives the remainder instantly, but not the quotient. For the full quotient, long division or inspection is needed.

Does the leading coefficient have to be 1?

No. The method works the same when the leading coefficient is not 1 — for example dividing 3x³ + 5x² − x − 2 by (x + 1). At step 1, you divide 3x³ by x to get 3x² as the first quotient term. Just follow the same procedure; the numbers are a little messier but the steps are identical.

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Work through algebraic long division with Professor Pi at aitutors.me.