The AC method — for when the number in front of x² is not 1. Two brackets, and there is a reliable way to find them.
Factor completely
3x2 − 10x + 3
Find two numbers multiplying to 3 × 3 = 9 and adding to −10: that is −1 and −9. Split and group: x(3x − 1) − 3(3x − 1) = (3x − 1)(x − 3). Neither A nor B multiplies back. A puts the 3 and the 1 in the wrong brackets: (3x − 3)(x − 1) = 3x² − 6x + 3. B drops the split pair −9 and −1 straight into the brackets, as if the x² had no 3 in front: (3x − 9)(x − 1) = 3x² − 12x + 9.
Keep the sign that sits in front of each number.
3x2 − 10x + 3
Multiply the two ends.
When a is not 1 you cannot just look at c. The pair you need lives in a × c, because the 3 in front will be split between the brackets.
Test every factor pair of 9. Only one pair can do both jobs.
| Pair | Product | Sum | Works? |
|---|---|---|---|
| 1 and 9 | 9 | 10 | no |
| 3 and 3 | 9 | 6 | no |
| −3 and −3 | 9 | −6 | no |
| −1 and −9 | 9 | −10 | ✓ yes |
The product is positive, so the two numbers share a sign. The sum is negative, so that shared sign is minus. You only had to test negative pairs.
3x2 − 10x + 3
−x − 9x = −10x
3x2 − x − 9x + 3
Nothing changed in value — −x − 9x is still −10x. Four terms is what makes grouping possible.
(3x2 − x) + (−9x + 3)
GCF is x · GCF is −3
x(3x − 1) − 3(3x − 1)
3x2 ÷ x = 3x
−x ÷ x = −1
−9x ÷ −3 = 3x
3 ÷ −3 = −1
Pull out −3, not 3. The minus is what makes the second bracket come out as (3x − 1) and match the first.
x(3x − 1) − 3(3x − 1)
The ringed bracket appears twice. The gold pieces are what is left outside it.
(3x − 1)(x − 3)
If the two brackets do not match, the GCF step went wrong — go back and check the sign you pulled out.
Run it backwards to find a missing coefficient. If 3x + 1 is a factor of 6x² + kx − 4, the other bracket must give 6x² and −4 at the ends: (3x + 1)(2x − 4). Multiply out: 6x² − 12x + 2x − 4 = 6x² − 10x − 4, so k = −10.
Factor −2x² + x + 6.
−(2x² − x − 6)
take out −1 first: every sign flips
ac = −12: pair −4 and 3, sum −1
product negative: opposite signs
2x(x − 2) + 3(x − 2) = (2x + 3)(x − 2)
split −x as −4x + 3x, group
−(2x + 3)(x − 2) = (2x + 3)(2 − x)
x = 0: 6 = −(3)(−2) ✓
Different: with a negative a, factor out −1 so the bracket starts positive — and keep the minus in the answer.
Factor 3x² − 7xy + 2y².
ac = 6: pair −6 and −1, sum −7
same steps; y rides along
3x² − 6xy − xy + 2y²
each piece keeps its xy
3x(x − 2y) − y(x − 2y) = (3x − y)(x − 2y)
x = y = 1: −2 = (2)(−1) ✓
Different: the last term of each bracket carries a y, so the y² at the end comes out right.
Factor 2x² + 5x + 4.
ac = 8: 1 + 8 = 9, 2 + 4 = 6
sum 5 is needed; positives only
no pair works
does not factor (it is prime)
Different: when every factor pair of ac fails, the trinomial is prime over the integers — b² − 4ac = 25 − 32 is negative, which confirms it.
Writing (2x + 1)(x + 4) because the ends match gives 2x² + 9x + 4 — the middle is wrong. If the table has no ✓, the honest answer is “not factorable”.
Factor completely: 4x4 − 35x² − 9. The powers are 4, 2, 0 — a trinomial in x², so let u = x² and use the same five moves.
4u² − 35u − 9, ac = −36
u = x²
pair −36 and 1: sum −35 ✓
product −36
4u² − 36u + u − 9 = 4u(u − 9) + 1(u − 9)
split and group
(4u + 1)(u − 9) = (4x² + 1)(x² − 9)
put x² back
(4x² + 1)(x − 3)(x + 3)
x² − 9 is a difference of squares
The block can be a bracket too. For (x − 2)² + 3(x − 2) − 10, let u = x − 2: u² + 3u − 10 = (u + 5)(u − 2), and putting x − 2 back gives (x + 3)(x − 4).
| Before you stop, ask | Example |
|---|---|
| a common factor first? | 6x² − 15x − 9 = 3(2x² − 5x − 3) = 3(2x + 1)(x − 3) |
| a bracket that is a difference of squares? | x² − 9 = (x − 3)(x + 3) |
| a sum of squares? | 4x² + 1 and x² + 9 do not factor — leave them |
(4x² + 1)(x² − 9) multiplies back correctly, so it looks finished, and it will be one of the choices. “Completely” means every bracket has been checked — a leftover x² − 9, or a 2 still inside a bracket, is not complete.
Factoring items say “factor completely” or “which is a factor of”. Take out the common factor first, then factor the trinomial, then look at every bracket again. For “which is a factor” items, multiplying two choices back together is often quicker than factoring from scratch.
−9x − x = −10x
3x2 − 10x + 3 ✓ back to the start
Factor completely
6x2 + 2x − 20
Take the common 2 out first: 2(3x² + x − 10), which factors to 2(3x − 5)(x + 2). A and C multiply out to the same expression but are not fully factored — each still carries a 2 inside a bracket. "Completely" means the common factor sits out front.
Hint: every term shares a common factor. Pull it out first, then run the AC method on what is left inside.
