1
The identity tells you exactly what to look for
Expanding (x + p)(x + q) gives x² + (p + q)x + pq. Matching coefficients turns factoring into two simultaneous conditions on two numbers — product c, sum b — and nothing else.
Lesson · Polynomials and Factoring · Factoring
Factoring Simple Trinomials
For x² + bx + c the whole problem reduces to finding two integers whose product is c and whose sum is b. The candidate list is finite, so the search either succeeds or proves that no integer factorisation exists.
(x + p)(x + q) = x² + (p + q)x + pq
so for x² + bx + c find p, q with p·q = c and p + q = b
x² − 7x + 12 → (−3)(−4) = 12, (−3) + (−4) = −7 → (x − 3)(x − 4)
One identity, read backwards
Core ideas
What this lesson establishes
This is the first genuine search on the course: a finite list of candidates, tested one at a time, with a definite answer either way.
2
The candidate list is finite, so the search is complete
p must divide c, and a nonzero integer has finitely many divisors. List the divisor pairs, check the sums, and you have either found the answer or shown there is none.
3
"It does not factor over the rationals" is a result, not a surrender
x² + x + 1 has divisor pairs (1, 1) and (−1, −1), with sums 2 and −2. Neither is 1, so no rational factorisation exists. That is a proved statement, with its number system named, and it is the correct final answer.
The material
Product and sum
Where the two conditions come from, how the signs narrow the search, and when to stop.
Multiply out (x + p)(x + q) and compare with x² + bx + c:
(x + p)(x + q) = x² + qx + px + pq = x² + (p + q)x + pq
x² + b x + c
Theorem · The factoring condition
A monic trinomial x² + bx + c factors as (x + p)(x + q) with p and q integers exactly when there are integers p, q with pq = c and p + q = b.
Two coefficients, two conditions. Because p must divide c, only finitely many p need testing.
This applies when the leading coefficient is 1. When it is not, either take it out as a common factor if you can, or use the AC method of “The AC Method”.
Reading the signs before you search
The two conditions constrain the signs, so you never need to test the whole list blindly:
- c > 0. The product is positive, so p and q have the same sign — and that sign is the sign of b, since they must sum to it. In x² − 7x + 12, both are negative.
- c < 0. The product is negative, so p and q have opposite signs. The one with the larger absolute value carries the sign of b.
- c = 0. There is no trinomial to search: x² + bx has a common factor x, and “Factoring Out the Greatest Common Factor” already handled it.
These are not extra rules to memorise. Each one is read straight off pq = c and p + q = b, and deriving them on the spot is safer than recalling them.
Why a finite search settles the question
Suppose no pair on the list works. Could there still be a factorisation using fractions — something like (x + 1/2)(x + 8)? For a monic trinomial the answer is no. If x² + bx + c with integer coefficients factors into linear factors with rational numbers, those numbers are necessarily integers, so they are already on the list.
“The Factor Theorem and Rational Roots” proves this as part of the rational root theorem. For now, take it as the reason the search is trustworthy: an empty result means the trinomial does not factor over the rationals, not merely that you did not find how.
Example · A complete search that fails
x² + x + 1. The divisor pairs of 1 are (1, 1) and (−1, −1), giving sums 2 and −2. Neither equals 1, and the list is exhausted. The trinomial is irreducible over the integers, and by the paragraph above, over the rationals too.
Interactive
x^2 + bx + c
Every pair that multiplies to c, with its sum
Method
Factoring x² + bx + c
The search is short if you use the signs to prune it first.
1
Check the leading coefficient is 1, and take out any GCF
If the polynomial is 2x² + 10x + 12, factor out the 2 first: 2(x² + 5x + 6). If the leading coefficient survives that, this lesson does not apply and “The AC Method” does.
2
Use the sign of c, then the sign of b
c positive means both numbers share the sign of b; c negative means opposite signs. This halves the list before you test anything.
3
List the divisor pairs of c and add each one
Work systematically from 1 upward so nothing is skipped. Stop at the first pair whose sum is b — there cannot be a second.
4
Write the factors and expand to check; if the list is empty, say so
The pair p, q gives (x + p)(x + q), and expanding must return the original. If no pair summed to b, the answer is that it does not factor over the rationals.
Product and sum
Score 0 / 0Choose the answer you can justify.
Two searches that succeed, one that does not
Every divisor pair listed, so you can see the searches finish.
A. x² − 7x + 12 c > 0, b < 0 → both numbers negative
(−1)(−12) sum −13
(−2)(−6) sum −8
(−3)(−4) sum −7 <-- match
answer (x − 3)(x − 4)
check x² − 4x − 3x + 12 = x² − 7x + 12
B. x² + 2x − 15 c < 0 → opposite signs
(1)(−15) sum −14
(−1)(15) sum 14
(3)(−5) sum −2
(−3)(5) sum 2 <-- match
answer (x − 3)(x + 5)
C. x² + x + 1 c > 0, b > 0 → both positive
(1)(1) sum 2 only pair; 2 ≠ 1
answer does not factor over the rationals
In B the pair (−3, 5) has the larger number positive, matching b = +2, exactly as the sign reading predicted. Listing all four pairs takes seconds and removes any guesswork about which arrangement is right.
C is the important one. The search ended not because it got difficult but because the list ran out, and an exhausted list is a proof. Compare x² + 2x + 1, where the pair (1, 1) sums to 2 and the answer is (x + 1)².
For a faded search, take x² − 2x − 24. The supplied sign decision is that the pair has opposite signs and the larger absolute value is negative. List the positive divisor pairs of 24, assign signs, and expand your result. The pair is 4 and −6, so the answer is (x + 4)(x − 6); choosing −4 and 6 gives the right product but the wrong middle coefficient.
Common mistakes
1
Matching the product but not the sumBoth conditions must hold. (x + 2)(x + 10) has product 20 but sum 12, so it is not a factorisation of x² + 9x + 20 — and expanding it says so immediately.
2
Guessing signs instead of reading them offThe sign of c decides whether the numbers agree in sign and the sign of b decides which way. Derived in two seconds, this removes the most common source of near-miss answers.
3
Treating an unsuccessful search as an unfinished oneWhen every divisor pair has been tested and none works, the trinomial does not factor over the rationals. Writing that is the complete answer; hunting for a cleverer pair is not.
Check your understanding
✓
Finish when you can factor by search and defend a negative result.
Given x² + bx + c you should list the divisor pairs of c in order, use the signs to skip half of them, and either produce the factors or state that none exist — naming the pairs you tested as your evidence. That habit is exactly what “The Factor Theorem and Rational Roots” formalises for polynomials of any degree.
Note: The lab shows the whole candidate list, including the sums that miss. That display is the point of it: watching a complete search come back empty on x² + x + 1 is what turns "I cannot factor this" into "this does not factor", which are very different statements.