Synthetic division: A faster way to factor polynomials

Factoring a cubic can involve several pages of long division. Synthetic division produces the same quotient and remainder in a much shorter calculation, particularly when the divisor is a linear factor such as x − 2. Once practised, it can save time without hiding the algebra.

Students encounter this kind of polynomial factoring in senior mathematics under the Australian Curriculum, although assessment names differ across the country. An HSC in New South Wales, a VCE in Victoria and SACE in South Australia all reward sound algebraic reasoning. Synthetic division supports those goals and prepares students for the mathematics encountered after school.

The method is most useful when students understand both its procedure and its purpose. CurryPlum's algebra guides, calculators and step-by-step solutions offer ways to check work, while focused practice turns the shortcut into a reliable exam skill.

The idea behind the shortcut

Synthetic division divides a polynomial by a term of the form x − c. It returns a quotient with a degree one lower than the original polynomial and a remainder. A remainder of zero means x − c is a factor, which is the factor theorem at work.

Only multiplication and addition are required. Write the coefficients in descending powers of x, including zeros for missing terms, then place c beside them. Bring down the first coefficient, multiply it by c, add the result to the next coefficient and continue. The last number is the remainder; the others form the quotient.

The shortcut works because the remainder after dividing by x − c equals the polynomial's value at c. It suits a monic linear divisor, with a coefficient of 1 on x. A divisor such as 2x − 3 needs adjustment, so students should not treat every linear divisor as x − c.

A worked example

Consider P(x) = x³ − 5x² + 8x − 4 divided by x − 2. Here c = 2, and the coefficient list is 1, −5, 8, −4. Bring down 1 and multiply by 2 to get 2; add 2 to −5 to get −3. Multiply −3 by 2 to get −6 and add it to 8 to get 2.

Multiply 2 by 2 to get 4, then add 4 to −4 to obtain 0. This zero remainder confirms that x − 2 is a factor. The quotient coefficients are 1, −3, 2, giving x² − 3x + 2.

Since x² − 3x + 2 = (x − 1)(x − 2), the complete factorisation is P(x) = (x − 1)(x − 2)². Substitution checks the work: P(2) = 8 − 20 + 16 − 4 = 0.

Signs, roots and common variations

A plus sign changes the number used in the calculation. To divide by x + 3, use c = −3. For example, x³ + 6x² + 11x + 6 divided by x + 3 gives quotient x² + 3x + 2 and remainder 0, revealing the factorisation (x + 1)(x + 2)(x + 3).

Missing powers of x must still be represented. Dividing x³ + 2x + 1 by x − 1 requires the list 1, 0, 2, 1; the zero belongs to the missing x² term. The calculation gives quotient x² + x + 3 and remainder 4, matching 1 + 2 + 1 = 4.

Repeated factors need no different procedure. Dividing x³ − 6x² + 11x − 6 by x − 1 gives x² − 5x + 6; dividing that by x − 2 gives x − 3. The original expression factors as (x − 1)(x − 2)(x − 3).

When the shortcut is worth using

The biggest saving occurs when a polynomial has several possible linear factors. Rather than writing a full long-division layout for each guess, a student can test likely roots quickly and then factor the resulting quadratic. For students aiming for an ATAR and university mathematics, those saved minutes can go towards checking work or completing another question.

The method is especially helpful with cubic expressions, where one factor reduces the workload immediately. A graphing calculator can confirm where a curve crosses the x-axis, but synthetic division provides the exact factor in symbolic form.

Students in Sydney, Melbourne or Adelaide who balance classes, part-time work and tutoring can fit ten minutes of focused practice into a busy day. Australia's online homework-assistance services can also complement teacher explanations, while worked resources from SLD Online can provide another way to revisit the process.

Habits that prevent errors

Most errors arise from missing coefficients, mishandled negative numbers or confusing the quotient with the original polynomial. A dependable routine is to write the divisor first, identify c and label the final number as the remainder. These habits protect accuracy under exam pressure.

A non-zero remainder remains useful information, not a failed attempt. Dividing x² + 3x + 2 by x − 1 gives quotient x + 4 and remainder 6, so the full result is (x − 1)(x + 4) + 6. This equals the original quadratic and illustrates the remainder theorem.

A practical revision routine

Begin with short, focused practice. Read one worked example, reproduce it without looking, then attempt two or three similar divisions with different signs and missing terms. Finish by checking each answer independently; a calculator can support the check, but it should not conceal a sign error.

Keep a one-page record of divisor forms, the meaning of each synthetic row and the errors you made. This turns scattered practice into a targeted revision plan and makes any tutoring session more productive.

On a clean sheet, divide x³ − 6x² + 11x − 6 by x − 2, verify the remainder and factor the quadratic. Your next concrete step is to repeat the polynomial with the divisor changed to x − 3 and confirm that the quotient factors as (x − 1)(x − 2).