Factoring Polynomials: Grouping and Trinomials
Factoring a polynomial means writing it as a product of simpler polynomials, for example x2 + 5x + 6 = (x + 2)(x + 3). In brief: first take out the greatest common factor, then look for a special pattern (difference of squares, perfect square), then use grouping for four terms or the product-sum method for trinomials. Always check by multiplying back.
Factoring is the reverse of multiplying. It is the main tool for solving polynomial equations, simplifying fractions and finding zeros of functions. This guide gives a reliable order of methods, explains factoring by grouping and factoring trinomials with leading coefficient 1 and not 1, and shows how to check your result.
How do you factor out the greatest common factor?
Find the largest number and the highest power of the variable that divide every term. Write it in front of a bracket containing the quotients. This is always the first step.
Example: 12x3 − 18x2 + 6x. The greatest common factor is 6x. So the polynomial is 6x(2x2 − 3x + 1). The bracket can be factored further: 2x2 − 3x + 1 = (2x − 1)(x − 1). The full factorization is 6x(2x − 1)(x − 1).
How does factoring by grouping work?
For four terms, split the polynomial into two pairs, factor out a common factor from each pair and hope that a common bracket appears.
- Group the terms in pairs.
- Factor out the greatest common factor in each pair.
- If the brackets match, factor out the common bracket.
Example 1
x3 + 3x2 + 2x + 6 = x2(x + 3) + 2(x + 3) = (x + 3)(x2 + 2).
Example 2: with a minus sign
2x3 − 6x2 − 5x + 15 = 2x2(x − 3) − 5(x − 3) = (x − 3)(2x2 − 5). When factoring a negative number from the second pair, change the signs inside: −5x + 15 = −5(x − 3).
How do you factor trinomials x² + bx + c?
Find two numbers p and q such that p · q = c and p + q = b. Then x2 + bx + c = (x + p)(x + q).
Example 3
x2 + 7x + 12. Numbers with product 12 and sum 7: 3 and 4. So (x + 3)(x + 4).
Example 4: negative terms
x2 − 2x − 15: product −15 and sum −2 give −5 and 3. So (x − 5)(x + 3). If c is positive, the two numbers have the same sign as b. If c is negative, they have opposite signs.
How do you factor trinomials ax² + bx + c?
Use the product-sum method (also called the AC method): multiply a · c, find two numbers with that product and sum b, split the middle term and factor by grouping.
Example 5: 6x² + 11x + 3
a · c = 18. The numbers with product 18 and sum 11 are 2 and 9. Split: 6x2 + 2x + 9x + 3 = 2x(3x + 1) + 3(3x + 1) = (3x + 1)(2x + 3). Check: 6x2 + 9x + 2x + 3 ✓.
Example 6: 4x² − 4x − 15
a · c = −60. The numbers are −10 and 6: sum −4. Split: 4x2 − 10x + 6x − 15 = 2x(2x − 5) + 3(2x − 5) = (2x − 5)(2x + 3).
Which special patterns should you recognise?
- Difference of squares: a2 − b2 = (a + b)(a − b).
- Perfect square trinomial: a2 ± 2ab + b2 = (a ± b)2.
- Sum and difference of cubes: a3 ± b3 = (a ± b)(a2 ∓ ab + b2).
Example: 9x2 − 30x + 25 = (3x − 5)2, since 2 · 3x · 5 = 30x. And 8x3 + 1 = (2x + 1)(4x2 − 2x + 1).
Factoring methods table
| Form | Method | Example |
|---|---|---|
| Any polynomial | Common factor first | 4x2 + 8x = 4x(x + 2) |
| Two terms | Difference of squares or cubes | x2 − 16 = (x + 4)(x − 4) |
| Three terms, a = 1 | Product and sum | x2 + 5x + 6 = (x + 2)(x + 3) |
| Three terms, a ≠ 1 | AC method | 2x2 + 7x + 3 = (2x + 1)(x + 3) |
| Four terms | Grouping | x3 + x2 + x + 1 = (x + 1)(x2 + 1) |
Where does factoring come from?
Babylonian scribes solved quadratic problems that are equivalent to factoring, and Greek geometric algebra used rectangles to represent products. The idea that polynomials can be broken into simple factors is the heart of the fundamental theorem of algebra, proved by Gauss in the early nineteenth century, which says that every polynomial factors completely over the complex numbers.
Where is it used today?
Factoring gives the zeros of a function: from (x − 5)(x + 3) = 0 you read x = 5 and x = −3. It simplifies rational expressions, as in (x2 − 9)/(x − 3) = x + 3, and it helps to find areas and dimensions: a rectangle of area x2 + 7x + 12 has sides x + 3 and x + 4. Cryptography relies on the difficulty of factoring large integers, an analogous problem.
Which mistakes are common?
Forgetting the common factor
2x2 + 8x + 6 should become 2(x + 1)(x + 3). Not removing 2 first makes the work harder.
Stopping too early
x4 − 16 = (x2 + 4)(x2 − 4) is not finished. Continue: (x2 + 4)(x + 2)(x − 2).
Sign errors in grouping
When you factor out a negative number from the second pair, change the signs of both terms.
Skipping the check
Multiply the factors back. This takes seconds and catches nearly all errors.