Algebra · step 3 of 13

Factoring & polynomials

Factoring converts a polynomial into a product, which is what makes equations solvable — a product is zero exactly when a factor is. The standard toolkit is: take out the common factor, recognise the special forms (difference of squares, difference and sum of cubes), split the middle term for quadratics, and use the factor theorem for cubics and beyond.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Take out the highest common factor firstAlways. 2x³ − 8x = 2x(x² − 4) = 2x(x−2)(x+2), and skipping this step hides the later factorisations.
Recognise the special formsa² − b² = (a−b)(a+b); a³ − b³ = (a−b)(a²+ab+b²); a³ + b³ = (a+b)(a²−ab+b²). Note a² + b² does not factor over ℝ.
Split the middle term for quadraticsFor ax² + bx + c find two numbers multiplying to ac and adding to b, then group.
Use the factor theorem for higher degreeIf P(r) = 0 then (x − r) is a factor. Test the divisors of the constant term, then divide out.

Worked example

Factor x³ − 4x² + x + 6 completely.

Candidate roots divide 6: ±1, ±2, ±3, ±6. Test x = −1: −1 − 4 − 1 + 6 = 0 ✓.
So (x + 1) is a factor. Divide: x³ − 4x² + x + 6 = (x + 1)(x² − 5x + 6).
Factor the quadratic: two numbers multiplying to 6 and adding to −5 are −2 and −3.
x² − 5x + 6 = (x − 2)(x − 3).

Answer. (x + 1)(x − 2)(x − 3).

Where marks get dropped

These are the specific errors that cost credit on factoring & polynomials questions — QED's rubric penalises each of them separately.

Trying to factor a² + b² over the reals. It is irreducible; only over ℂ does it split as (a + bi)(a − bi).
Stopping after one factorisation. x⁴ − 16 = (x²−4)(x²+4) still has (x²−4) = (x−2)(x+2) to extract.
Testing only positive candidate roots. Negative divisors of the constant are just as likely.

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Factoring & polynomials — frequently asked questions

What is the factor theorem?

P(r) = 0 if and only if (x − r) divides P(x). It converts root-finding into factoring and is the standard entry point for cubics.

Which candidate roots should I test?

By the rational root theorem, any rational root p/q has p dividing the constant term and q dividing the leading coefficient. For monic polynomials that means the divisors of the constant.

How do I divide out a factor?

Polynomial long division or synthetic division. Synthetic division is faster for linear factors and is worth practising until it is automatic.

The rest of Algebra

Foundational algebra to close prerequisite gaps. Each subtopic below has its own method, worked example and mark-losing traps.

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