Polynomials and Factoring
Factoring Polynomials: the Five Standard Methods
Common factors, grouping, reverse FOIL, the special forms and perfect squares — the five methods, the order to try them in, and what to do when none of them works.
What this page covers
- Extract the greatest common monomial or binomial factor
- Factor a four-term expression by grouping
- Factor a quadratic trinomial by reversing the FOIL expansion
- Recognise and apply the special forms, and identify a prime quadratic
Factoring is the distributive law run backwards
Expanding turns a(b + c) into ab + ac. Factoring turns ab + ac back into a(b + c). There is no new law involved, only a harder search: expanding is mechanical, while factoring requires spotting what the common part is.
The reason it is worth the effort is the zero-product property. A product equal to zero forces one of its factors to be zero, so a factored polynomial hands over its roots immediately. An unfactored one does not.
Method 1: common factors
The source begins with pure arithmetic, which makes the idea unmistakable before any algebra appears.
The greatest common factor is found by taking each prime that appears in every term, raised to the lowest power it carries anywhere. The same rule applies to variables.
Worked example — monomial factor
Factorise 3x5 - 24x4 + 12x3.
| Term | Numeric part | Power of x |
|---|---|---|
| 3x5 | 3 | x5 |
| -24x4 | -24 = -23 · 3 | x4 |
| 12x3 | 12 = 22 · 3 | x3 |
Every term carries a 3, and the lowest power of x is x3. The greatest common factor is 3x3.
Expanding gives 3x5 - 24x4 + 12x3. Always multiply back; it takes seconds and catches every arithmetic slip.
The quadratic left inside does not factor over the integers: b2 - 4ac = 64 - 16 = 48, which is positive but not a perfect square. It factors over the reals as (x - 4 - 2√3)(x - 4 + 2√3).
The common factor need not be a monomial
The source's second example takes out a product of two binomial powers, which is the same idea applied to larger building blocks.
Worked example — binomial factors
Factorise 12(x2 - 1)4(3x + 1)3 + 8x(x2 - 1)3(3x + 1)4.
The shared parts are 4, (x2 - 1)3 and (3x + 1)3 — in each case the lowest power present.
One further step is available that the source does not take: x2 - 1 is itself a difference of two squares, so (x2 - 1)3 = (x - 1)3(x + 1)3. Factoring is not finished until every factor is prime.
Method 2: grouping
With four terms and no factor common to all of them, split into two pairs, factor each pair, and hope the same bracket appears twice. If it does, that bracket is itself a common factor.
Worked example — the source's case
Factorise 3x2 + 4xy - 3xt - 4ty.
- Split into pairs. (3x2 + 4xy) - (3xt + 4ty). Note the sign: both of the last two terms were negative, so a minus comes out in front and both signs inside flip to positive.
- Factor each pair. The first gives x(3x + 4y); the second gives t(3x + 4y).
- Take out the shared bracket. x(3x + 4y) - t(3x + 4y) = (x - t)(3x + 4y).
(x - t)(3x + 4y) = 3x2 + 4xy - 3xt - 4ty. Correct.
Step 1 is where grouping usually fails. Bracketing -(3xt + 4ty) requires both inner signs to flip. Writing -(3xt - 4ty) instead changes the expression and the shared bracket will not appear.
If the shared bracket does not appear, try pairing the terms differently before concluding the method fails. Here (3x2 - 3xt) + (4xy - 4ty) gives 3x(x - t) + 4y(x - t) = (3x + 4y)(x - t) — the same answer by a different pairing.
Method 3: reverse FOIL
For a trinomial x2 + px + q with leading coefficient 1, expand the target form and compare coefficients.
Find two numbers whose product is the constant term and whose sum is the coefficient of x. Those two numbers are a and b.
The signs narrow the search considerably before any arithmetic is done.
| Constant term | Coefficient of x | The two numbers are |
|---|---|---|
| Positive | Positive | Both positive |
| Positive | Negative | Both negative |
| Negative | Either | One positive, one negative; the larger in size carries the sign of the x coefficient |
Worked examples from the source
| Trinomial | Need product | Need sum | Numbers | Factors |
|---|---|---|---|---|
| x2 + 5x + 6 | 6 | 5 | 3, 2 | (x + 3)(x + 2) |
| x2 - 5x - 6 | -6 | -5 | -6, 1 | (x - 6)(x + 1) |
| x2 - 13x - 30 | -30 | -13 | -15, 2 | (x - 15)(x + 2) |
| x2 - 13x + 36 | 36 | -13 | -9, -4 | (x - 9)(x - 4) |
The source lays out the candidate factor pairs for -6 explicitly — (-2) × 3, 2 × (-3), (-1) × 6, 1 × (-6) — and then picks the pair with the required sum. Writing out the candidates is a reliable habit when the constant has many factors.
Week 1, page 10 gives x2 - 13x + 36 = (x - 10)(x - 3). That is not correct: (x - 10)(x - 3) = x2 - 13x + 30. Either the constant or the factors were mistranscribed. The correct factorisation of x2 - 13x + 36 is (x - 9)(x - 4), since (-9)(-4) = 36 and -9 - 4 = -13; alternatively, x2 - 13x + 30 does factor as (x - 10)(x - 3). This page uses the corrected pairing. Multiplying back would have caught it, which is the argument for always doing so.
Method 4: the special forms
These are the expansions of the previous page, read right to left. Recognising them saves the whole search.
| Recognise | Factors as |
|---|---|
| a2 - b2 | (a - b)(a + b) |
| a2 + 2ab + b2 | (a + b)2 |
| a2 - 2ab + b2 | (a - b)2 |
| a3 - b3 | (a - b)(a2 + ab + b2) |
| a3 + b3 | (a + b)(a2 - ab + b2) |
Worked example — applying it twice
The first factor was itself a difference of two squares and split again. The second, x2 + y2, is a sum of squares and stops there over the reals.
Perfect squares
The source lists four, which are worth being able to write down without working.
The last is the general statement, and it is the pivot on which completing the square turns: given x2 + 2ax, the term that completes the square is a2.
When nothing works: prime polynomials
Not every polynomial factors, and recognising a dead end quickly is as useful as recognising a pattern. The source gives three examples in the margin.
| Polynomial | Discriminant b2 - 4ac | Verdict |
|---|---|---|
| x2 + 4x + 5 | 16 - 20 = -4 | Negative: no real roots, prime over the reals |
| x2 + 4x + 1 | 16 - 4 = 12 | Positive but not a perfect square: real factors exist but involve surds, so no integer factors |
| x2 + 9 | 0 - 36 = -36 | Negative: a sum of two squares, prime over the reals |
The discriminant settles the question mechanically, which is far quicker than exhausting the candidate pairs. It is developed in The Quadratic Formula and the Discriminant.
'Prime' is relative to the number system. x2 + 9 is prime over the reals but factors as (x + 3i)(x - 3i) over the complex numbers, and x2 + 4x + 1 is prime over the integers but factors as (x + 2 - √3)(x + 2 + √3) over the reals. Always know which system the question means.
A decision procedure
- Take out the greatest common factor — always, before anything else. This step is skipped more often than any other and makes every later step harder when it is.
- Count the terms. Two: try a special form. Three: try reverse FOIL, checking first whether it is a perfect square. Four: try grouping.
- Check the discriminant if the trinomial resists. A negative value means stop; a non-square positive value means the factors involve surds.
- Factor each factor again. A quartic often needs three passes. Nothing is finished until every bracket is prime.
- Multiply back. Expanding the answer must reproduce the original exactly. This single check catches almost every error.
Common mistakes
| Mistake | Correct | Remedy |
|---|---|---|
| Skipping the common factor | 3x5 - 24x4 + 12x3 = 3x3(x2 - 8x + 4) | Make it step 1 unconditionally |
| -(3xt + 4ty) written with one sign flipped | Both inner signs flip | Expand the bracket back immediately |
| Stopping at (x2 - y2)(x2 + y2) | (x - y)(x + y)(x2 + y2) | Ask of every factor whether it factors further |
| Trying to factor x2 + 9 | Prime over the reals | Check the discriminant first |
| x2 - 13x + 36 = (x - 10)(x - 3) | (x - 9)(x - 4) | Multiply back; the constant comes out 30 |
| Taking the highest rather than lowest power as common | The GCF uses the lowest power present in every term | Tabulate the powers before choosing |
Frequently asked questions
Which method should I try first?
Always the common factor. Taking out 3x3 from 3x5 - 24x4 + 12x3 leaves a quadratic that any other method can handle, whereas attacking the quintic directly is far harder. Skipping this step is the usual reason a factorisation stalls.
How do I know when a quadratic will not factor?
Compute the discriminant b2 - 4ac. If it is negative there are no real roots and the quadratic is prime over the reals. If it is a perfect square the factors have integer coefficients; if it is positive but not a perfect square, real factors exist but involve surds.
Is a factorisation with surds still a factorisation?
Yes. x2 + 2x - 2 has no integer factors but does factor as (x + 1 - √3)(x + 1 + √3). Whether that counts depends on the question: 'factor over the integers' and 'factor over the reals' are different requests, and it is worth being clear which is being asked.
Why does grouping need exactly four terms?
It does not, strictly, but four is the case where it works cleanly: two pairs, each with its own common factor, leaving a shared bracket. Six terms can sometimes be grouped in three pairs or two triples, but the shared-bracket step has to appear or the method gives nothing.
Source. Handwritten teaching notes, Week 1, pages 9-11.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.