Mistakes · Algebra
5 Common Factoring Mistakes and How to Catch Them
Every one of these is caught by expanding the answer. Ten seconds, and it tests the result rather than the search.
1. Not taking out the common factor first
2x² + 10x + 12 factors as 2(x + 2)(x + 3). Going straight to the trinomial method means hunting for a pair that multiplies to 12 and adds to 10 — which does not exist in the form you need, because the greatest common factor is still in the way.
It is the first of the common factoring mistakes for a reason: it does not produce a wrong answer so much as an impossible search, and the time is lost before anything is written down.
2. Getting the signs of the pair wrong
For x² − 5x + 6 the pair is −2 and −3, not 2 and 3. The rule is decided by the constant first and the middle term second.
| Constant c | Middle term b | The pair |
|---|---|---|
| positive | positive | both positive |
| positive | negative | both negative |
| negative | either | one of each, and the larger takes b's sign |
A positive constant forces the two numbers to share a sign, because only two positives or two negatives multiply to a positive. Only then does the middle term decide which sign that is.
3. Trying to factor a sum of squares
x² − 9 factors as (x + 3)(x − 3). x² + 9 does not factor over the real numbers at all, and writing (x + 3)(x + 3) is wrong twice over — that expands to x² + 6x + 9.
The difference of squares needs the minus sign. There is no matching pattern for a sum, which is one of the genuine asymmetries in algebra rather than something you have forgotten.
4. Cancelling across a sum
In (x² + 5x)/x the x cancels, because x is a factor of the whole numerator once it is written as x(x + 5). In (x² + 5)/x nothing cancels at all, because 5 has no x in it.
- (x² + 5x) ÷ xfactor the top first
- x(x + 5) ÷ x = x + 5now the x is a factor and cancels
- (x² + 5) ÷ xnothing to cancel — leave it
Cancelling removes a factor of the whole top and the whole bottom. If either is a sum, it has to be factored before anything can go.
5. Assuming everything factors
x² + x + 1 is irreducible, and no amount of searching will produce integer factors. Recognising that is a correct answer rather than an incomplete one.
The discriminant settles it before the search begins: compute b² − 4ac. A perfect square means integer factors exist; anything else means they do not. For x² + x + 1 it is −3, so there are no real roots either.
The check that catches all five
Expand your factors and compare with what you started from.
It works because expanding is a different operation from factoring, not a repetition of it. A missed common factor, a sign error, a forced factorisation and a bad cancellation each produce something that expands to the wrong expression.
For partial credit it also matters that the check is visible. Writing the expansion out shows the answer was verified rather than guessed.
Questions about common factoring mistakes
Why should I take out the common factor first?
Because it makes the remaining numbers smaller and often turns an impossible search into an easy one. 2x² + 10x + 12 has no usable pair until the 2 comes out.
Does x² + 9 factor?
Not over the real numbers. Only a difference of squares factors; a sum of squares has no real factorisation.
When can I cancel in an algebraic fraction?
Only when the same factor divides the entire numerator and the entire denominator. If either is a sum, factor it first — nothing cancels across a plus sign.
How do I know an expression will not factor?
Check the discriminant b² − 4ac. If it is not a perfect square, no integer factorisation exists and you can stop looking.