Algebra 1 Intermediate

Factoring Trinomials

Factor any trinomial ax² + bx + c into (px + q)(rx + s) step by step — handles the simple x² + bx + c case (a = 1) and the harder ax² + bx + c case (a ≠ 1) using the AC method with factor-pair grouping. Also finds roots and verifies with FOIL.

Simple method when a = 1
AC method when a ≠ 1
Factor pair table shown
FOIL verification included
Live
Enter Trinomial ax² + bx + c
Coefficients: + x +
x² + 5x + 6
Enter integers for a, b, and c. Use negative values by typing a minus sign.
Examples
Result
Enter values for a, b, and c above and press Factor Trinomial to see the factored form and step-by-step solution.
Factored Form
Roots (zeros)
a
a·c (target)
discriminant
Step-by-Step Solution
Factoring When a = 1: x² + bx + c
Find p, q such that p · q = c and p + q = b

When the leading coefficient is 1, factoring is a hunt for two numbers that multiply to c and add to b. Once found, the answer is simply (x + p)(x + q).

Example: x² + 5x + 6 — we need p·q = 6 and p+q = 5. Testing pairs: 1×6 = 6 but 1+6 = 7 ✗; 2×3 = 6 and 2+3 = 5 ✓. So the answer is (x + 2)(x + 3).

Negative c: one factor is positive and one is negative. Positive c, negative b: both factors are negative.

Always check your answer by FOIL: multiply back and confirm you get the original trinomial.
AC Method When a ≠ 1: ax² + bx + c
AC = a·c → split bx → factor by grouping

When a ≠ 1, find two numbers p and q such that p·q = a·c and p+q = b. Then rewrite the middle term as px + qx and factor by grouping in two pairs.

Example: 2x² + 7x + 3 — AC = 2·3 = 6, and we need p+q = 7. Since 1+6 = 7 and 1·6 = 6 ✓, rewrite as 2x² + x + 6x + 3, group as x(2x+1) + 3(2x+1), then factor out (2x+1) to get (x+3)(2x+1).

  • Compute AC = a × c first.
  • List factor pairs of AC; find the pair that sums to b.
  • Split the middle term, then factor by grouping.
  • If no integer pair works, the trinomial is prime over ℤ.

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