Factoring Trinomials (a = 1)

Fibo  /  Math  /  Algebra  /  Trinomials a=1
Skill: factoring-trinomials-a1

Factoring Trinomials (a = 1)

Every x² + bx + c hides two numbers: they multiply to c and add to b. Find them and the trinomial splits into (x + m)(x + n). Master the sign patterns, always check GCF first, and finish by solving with the zero-product property.

1 Understand

The core idea in plain language.

What it is. To factor x2 + bx + c, find two numbers m, n with m·n = c and m + n = b. Then x2 + bx + c = (x + m)(x + n). If no such integers exist, the trinomial is prime over the integers.

Why it matters. This is the most-factored shape in algebra — quadratics, rational expressions, and equation solving all lean on it. The sign pattern tells you the signs of m and n before you even list pairs.

Where it is used. Solving quadratic equations · simplifying rational expressions · finding x-intercepts of parabolas.

2 See It

Diagrams that make the idea visual.

Two numbers, two clues

x2 + 7x + 12: the numbers multiply to 12 and add to 7. Factor pairs of 12: (1,12), (2,6), (3,4). Only 3 + 4 = 7 — so x2 + 7x + 12 = (x + 3)(x + 4).

x² + 7x + 12 multiply to 12 · add to 7 (1, 12) → 13 (no) (2, 6) → 8 (no) (3, 4) → 7 (checks) (x + 3)(x + 4) check: 3·4 = 12 and 3 + 4 = 7
Factor pairs of 12: only (3, 4) adds to 7. So x² + 7x + 12 = (x + 3)(x + 4).

The sign pattern

The signs of b and c reveal the signs of m and n instantly: c positive means m, n share a sign (b decides which); c negative means m, n have opposite signs and the bigger one takes the sign of b.

c > 0 → same sign b > 0: (x + m)(x + n) · b < 0: (x − m)(x − n) c < 0 → opposite signs (x + m)(x − n) — bigger |·| takes b’s sign x² − 5x + 6: c > 0, b < 0 (x − 2)(x − 3)
Sign rules: c > 0 → same sign (b picks); c < 0 → opposite signs.

3 Worked Examples

Follow each step. The pattern is always the same.

Example 1 Factor: x² + 7x + 12
  1. Need m·n = 12, m + n = 7. Pairs of 12: (1,12), (2,6), (3,4).
  2. 3 + 4 = 7 — the pair is (3, 4).
x² + 7x + 12 = (x + 3)(x + 4)

Check: (x + 3)(x + 4) = x² + 7x + 12. Both clues satisfied.

Example 2 Factor: x² − 5x + 6
  1. c > 0 and b < 0 → both numbers negative.
  2. Need m·n = 6, m + n = −5: (−2) + (−3) = −5.
x² − 5x + 6 = (x − 2)(x − 3)

Check: (−2)(−3) = 6 (checks); (−2) + (−3) = −5 (checks).

Example 3 Factor: x² + 2x − 15
  1. c < 0 → opposite signs; the bigger absolute value takes the sign of b (+).
  2. Pairs of 15: (1,15), (3,5). 5 − 3 = 2 — the pair is (+5, −3).
x² + 2x − 15 = (x + 5)(x − 3)

Check: 5·(−3) = −15 (checks); 5 + (−3) = 2 (checks).

Example 4 GCF first: 2x² + 10x + 12
  1. GCF of the terms is 2 — pull it out first: 2(x² + 5x + 6).
  2. Now factor x² + 5x + 6: need product 6, sum 5 → (2, 3).
2x² + 10x + 12 = 2(x + 2)(x + 3)

Check: 2(x + 2)(x + 3) = 2(x² + 5x + 6) = 2x² + 10x + 12 (checks).

Example 5 Solve: x² − 4x + 3 = 0
  1. Factor: need product 3, sum −4 → (−1, −3): (x − 1)(x − 3) = 0.
  2. Zero product: x − 1 = 0 or x − 3 = 0.
x = 1 or x = 3

Check: 1 − 4 + 3 = 0 (checks); 9 − 12 + 3 = 0 (checks).

4 Common Mistakes

These errors show up on almost every quiz. Spot them now and they will never cost you points.

Mistake 1: right product, wrong sum
Wrong
x2 − 7x + 12 = (x + 3)(x + 4).
Right
(x − 3)(x − 4) — 3·4 = 12 (checks) but 3 + 4 = 7 ≠ −7. The sum clue rules.
Rule: both clues must hold — product AND sum.
Mistake 2: skipping the GCF
Wrong
Hunting two numbers for 2x2 + 10x + 12 directly.
Right
Pull the GCF first: 2(x + 2)(x + 3). The two-numbers method assumes a = 1.
Rule: GCF first, always — then the a = 1 method applies.
Mistake 3: calling it prime too soon
Wrong
x2 + 2x − 15 is prime (no factor pair of 15 adds to 2).
Right
(x + 5)(x − 3) — with c < 0 the numbers have opposite signs: 5 + (−3) = 2.
Rule: for c < 0, list pairs with opposite signs.

5 Quick Check

Try each one on paper first, then reveal the answer.

1. Factor x² − 9x + 20.
Answer
(x − 4)(x − 5).
2. Factor x² + 2x − 15.
Answer
(x + 5)(x − 3).
3. Solve x² − 4x + 3 = 0 by factoring.
Answer
x = 1 or x = 3.

Key Points to Remember

  • x² + bx + c = (x + m)(x + n) where m·n = c and m + n = b.
  • c > 0 → m, n share a sign; b picks which: b > 0 both +, b < 0 both −.
  • c < 0 → m, n have opposite signs; the bigger |·| takes the sign of b.
  • GCF first — the two-numbers method needs a = 1.
  • No integer pair? The trinomial is prime over the integers.
  • Always multiply back: both the product and the sum must check.
  • Factored form solves equations: (x + m)(x + n) = 0 gives x = −m, −n.
Practice problems Try the tool See it on FiboMap
Fibo · Learn, Practice & Explore Math