Examples · Algebra
Factoring Examples: 10 Worked Problems
Ten expressions, ordered so each one needs exactly one method more than the last.
Common factors first
The first three factoring examples need nothing but the greatest common factor.
- 3(2x + 3)3 divides both terms
- x(x + 7)every term has an x
- 4x²(x − 3)the shared factor is 4x², not just 4
In example 3 the greatest common factor includes the variable to the lowest power that appears. Taking out only the 4 leaves work undone.
Differences of squares
No middle term, and a minus sign: a² − b² factors as (a + b)(a − b).
- (x + 3)(x − 3)both terms are perfect squares
- 2(x² − 4)common factor of 2 first
- 2(x + 2)(x − 2)then the difference of squares
Trinomials with a leading coefficient of 1
Find two numbers that multiply to the constant and add to the middle coefficient.
- 2 × 3 = 6, 2 + 3 = 5the pair
- (x + 2)(x + 3)the factors
- −2 × −3 = 6, −2 + −3 = −5both negative
- (x − 2)(x − 3)the factors
- 3 × 3 = 9, 3 + 3 = 6the same number twice
- (x + 3)²so it is a perfect square
Six and seven differ only in the middle sign, and that flips both numbers negative. A positive constant forces the pair to share a sign; the middle term decides which.
A leading coefficient that is not 1, and grouping
Here the AC method splits the middle term so the four terms factor in pairs.
- ac = 12multiply the outer coefficients
- 6x² + 3x + 4x + 23 and 4 multiply to 12 and add to 7
- 3x(2x + 1) + 2(2x + 1)factor each pair
- (2x + 1)(3x + 2)the bracket is now the common factor
- x²(x + 2) + 3(x + 2)group the four terms in pairs
- (x + 2)(x² + 3)take out the shared bracket
Grouping is the same move as the last step of the AC method, applied to four terms that already arrive in that shape. The x² + 3 does not factor further over the reals.
The one that does not factor
Not every expression in a worked examples set has an answer, and recognising that is part of the skill.
- b² − 4ac = 1 − 4 = −3the discriminant
- irreduciblenot a perfect square, so no integer binomial factors exist
A quadratic factors into two binomials over the integers exactly when its discriminant is a perfect square. Checking that first turns a long fruitless search into one line of arithmetic, and 'irreducible' is the complete answer.
Checking every one by expanding
Check by expanding each answer and comparing. This is not optional politeness — it is the only step that tests the result rather than repeating the search that produced it.
| Example | Factored | Expands back to |
|---|---|---|
| 6 | (x + 2)(x + 3) | x² + 5x + 6 ✓ |
| 9 | (2x + 1)(3x + 2) | 6x² + 7x + 2 ✓ |
| 10 | (x + 2)(x² + 3) | x³ + 2x² + 3x + 6 ✓ |
The Factoring Calculator on this site runs exactly this check internally, and refuses to show a factorisation that does not expand back correctly.
Questions about factoring examples
What is the first step in any factoring problem?
Take out the greatest common factor, including the variable to the lowest power that appears. Everything after that works with smaller numbers.
How do I recognise a difference of squares?
Two terms, both perfect squares, with a minus sign between them. It factors as (a + b)(a − b). A sum of squares does not factor at all.
When do I need the AC method?
When the leading coefficient is not 1 and no simple pair works. Multiply a by c, find a pair for that product, split the middle term and factor in pairs.
What is factoring by grouping?
Splitting four terms into two pairs, factoring each pair, and taking out the bracket they then share. It is the last step of the AC method used on its own.