The factoring trinomials calculator above splits ax2 + bx + c into a product of two binomials, and — more usefully — shows you the search that got there. Most factoring tools print an answer that appears from nowhere. This one lists the factor pairs of a×c it tested, marks the pair that worked, and walks through the split-the-middle-term grouping so you can reproduce the method on the next question without the tool.
Arb Digital publishes this as part of a free algebra set built on one principle: for a technique that is assessed on the working, an answer without the working is close to worthless.
What This Factoring Trinomials Calculator Does
Enter integer values for a, b and c. The calculator first pulls out any common factor shared by all three terms, because factoring a GCF out at the start makes every later step smaller. It then computes the product a×c and searches for two integers that multiply to that product and add to b. Those two numbers let you split the middle term and factor by grouping.
If no such integer pair exists, the trinomial is prime over the integers, and the calculator says so explicitly rather than inventing a factorisation. It also gives the discriminant and the exact roots in that case, so you can still write the factored form using surds if the question allows it.
How to Use It
- Write the trinomial in descending order, ax2 + bx + c, with the squared term first.
- Enter the three integer coefficients with their signs. For 6x2 + 11x − 35 that is 6, 11 and −35.
- Press Factor. The binomial factors, the target product, the winning pair and the roots all update.
- Read the pair search line to see which candidate pairs were tried and rejected, and why one of them worked.
- Expand your answer to check. Multiplying the two binomials back out must return the trinomial you started with.
The AC Method, Step by Step
When a = 1, factoring is easy: find two numbers that multiply to c and add to b, and the factors are (x + m)(x + n). When a is anything else, that shortcut breaks, and the AC method is the reliable replacement.
Multiply a by c. Find two integers m and n with m × n = ac and m + n = b. Rewrite the middle term as mx + nx, which changes nothing because m + n = b. You now have four terms, and grouping the first two and the last two produces a shared binomial factor that you can pull out.
The reason it always works is worth understanding rather than memorising. Since (ax + m)(ax + n) expands to a2x2 + a(m + n)x + mn = a2x2 + abx + ac, dividing by a returns exactly ax2 + bx + c. So the factorisation is (ax + m)(ax + n) ÷ a, and the division always distributes cleanly across the two binomials whenever an integer pair exists. Wolfram MathWorld's overview of polynomial factorization covers the general theory behind this.
A Fully Worked Example
Factor 6x2 + 11x − 35.
Step 1 — common factor. The GCF of 6, 11 and 35 is 1, so nothing comes out.
Step 2 — the target product. ac = 6 × (−35) = −210. Because the product is negative, the two numbers must have opposite signs, and because their sum is +11, the positive one must be larger in magnitude.
Step 3 — the search. Work through the factor pairs of 210 and check each difference: 1 and 210 differ by 209; 2 and 105 by 103; 3 and 70 by 67; 5 and 42 by 37; 6 and 35 by 29; 7 and 30 by 23; 10 and 21 by 11. That is the pair. Assign signs to make the sum +11, giving m = 21 and n = −10. Check: 21 × (−10) = −210 and 21 + (−10) = 11.
Step 4 — split and group. 6x2 + 21x − 10x − 35. Group as (6x2 + 21x) + (−10x − 35). Factor each group: 3x(2x + 7) − 5(2x + 7). The bracket (2x + 7) is common to both.
Step 5 — write the answer. (2x + 7)(3x − 5). Expanding to check gives 6x2 − 10x + 21x − 35 = 6x2 + 11x − 35. The roots follow immediately: x = −7/2 and x = 5/3.
How to Know Instantly Whether a Trinomial Will Factor
You do not have to search blindly. Compute the discriminant b2 − 4ac first. If it is a perfect square, an integer factor pair exists and the search is guaranteed to succeed. If it is not a perfect square, no integer pair exists and no amount of trial and error will find one — the trinomial is prime over the integers.
For the worked example, b2 − 4ac = 121 + 840 = 961 = 312, a perfect square, which is why the search terminated. For x2 + x + 1 the discriminant is 1 − 4 = −3, which is not just non-square but negative, so the trinomial has no real roots and cannot factor over the reals at all. The square root calculator is the quickest way to check whether a discriminant is a perfect square before you begin.
The Sign Rules That Cut the Search in Half
Once you know the signs of the product and the sum, most candidate pairs eliminate themselves before you test them:
- ac positive, b positive: both numbers are positive.
- ac positive, b negative: both numbers are negative.
- ac negative: the numbers have opposite signs, and the one matching the sign of b has the larger magnitude.
That third rule is the one people skip, and it is the reason so many students find the right pair of magnitudes but attach the wrong signs. In the worked example, ac was negative and b was positive, so the larger number, 21, had to be the positive one. Getting that backwards gives −21 and 10, which sum to −11 and factor the wrong trinomial entirely.
Special Patterns Worth Recognising First
Three patterns short-circuit the whole method and are worth checking before you start any search.
A difference of squares, a2x2 − c2 with no middle term, factors instantly as (ax + c)(ax − c). A perfect square trinomial, where a and c are both perfect squares and b equals twice the product of their roots, factors as a single binomial squared — 9x2 + 30x + 25 is (3x + 5)2, and its discriminant is exactly zero. A trinomial with c = 0 is not really a trinomial at all: x factors straight out, leaving x(ax + b).
There is no sum-of-squares factorisation over the integers. a2x2 + c2 does not factor over the reals, and writing it as (ax + c)(ax − c) — a very common slip — is simply wrong, because that product has a minus sign in the middle. Wolfram MathWorld's note on the difference of squares sets out precisely which of these identities hold and which do not.
Factoring Trinomials in Disguise
Plenty of expressions that do not look quadratic factor with exactly this method after a substitution. x4 − 5x2 + 4 is a quadratic in x2: set u = x2, factor u2 − 5u + 4 as (u − 1)(u − 4), then substitute back to get (x2 − 1)(x2 − 4), which factors further into (x − 1)(x + 1)(x − 2)(x + 2). The same trick handles trinomials in x6, in sin x, or in any repeated expression. Enter the coefficients of the substituted form here, then reverse the substitution yourself.
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Browse Free Tools Talk To UsCommon Mistakes to Avoid
- Forgetting the GCF first. 12x2 + 30x − 18 is far easier once 6 comes out, and leaving it in makes the target product nine times larger than it needs to be.
- Multiplying only a by the constant's sign. The target is a×c including the sign of c, not the absolute value.
- Assigning the signs to the wrong number. When the product is negative, the number matching the sign of b must be the larger of the two.
- Dropping the GCF from the final answer. If you took a 6 out at the start, it belongs in front of the brackets at the end.
- Not expanding to check. Multiplying the binomials back out takes ten seconds and catches almost every sign error.
Related Free Tools From Arb Digital
Get exact roots for a prime trinomial with the quadratic equation solver, and test whether a discriminant is a perfect square with the square root calculator. The LCM and GCF calculator handles the common-factor step, the prime number checker speeds up listing factor pairs, and the fraction calculator tidies the rational roots that come out at the end. The full free online tools hub has the rest.
Frequently Asked Questions
Multiply the leading coefficient by the constant term, find two integers that multiply to that product and add to the middle coefficient, split the middle term using them, then factor the four terms by grouping.
Compute the discriminant, b squared minus four a c. If it is a perfect square, an integer factorisation exists. If it is not, the trinomial is prime over the integers.
It means no pair of binomials with integer coefficients multiplies to give the trinomial. It can still be factored using surds or complex numbers, but not with whole numbers.
Yes. Removing a common factor makes every subsequent number smaller and shortens the factor pair search considerably. Just remember to write it back in front of the final answer.
Find two numbers that multiply to the constant term and add to the middle coefficient. The factors are then x plus the first number, times x plus the second number.
Not over the real numbers. Only a difference of squares factors, into the sum and the difference of the two roots. Writing a sum of squares that way is a common and incorrect shortcut.
Multiply the binomials back out. If expanding does not return the original trinomial exactly, including every sign, the factorisation is wrong somewhere.
This calculator is provided for study and reference. Always expand your factored answer to confirm it reproduces the original trinomial before submitting it.
