Factoring a quadratic trinomial is the single most-reused skill in the rest of algebra, and which technique to reach for depends on two quick checks: is the leading coefficient \(1\), and does the trinomial match one of two recognizable patterns? This lesson organizes those checks into one process and then drills each branch on its own.
The Factoring Decision Process
- Factor out the GCF first, always. See Greatest Common Factor — this step shrinks every number involved and should never be skipped.
- Check for a difference of squares or a perfect square trinomial. Recognizing either pattern on sight is faster than any other method.
- If the leading coefficient is \(1\), find two numbers that multiply to \(c\) and add to \(b\).
- If the leading coefficient isn’t \(1\), use the ac method from Factoring by Grouping.
- If no integer pair works anywhere above, the quadratic is prime over the integers, and factoring simply doesn’t apply.
Factoring a Trinomial with Leading Coefficient 1
When the leading coefficient is \(1\), the trinomial factors into two binomials whose constant terms multiply to \(c\) and add to \(b\).
$$ x^{2}+7x+12 $$
Two numbers that multiply to \(12\) and add to \(7\): \(3\) and \(4\).
$$ x^{2}+7x+12 = (x+3)(x+4) $$
| \(c\)‘s sign | \(b\)‘s sign | What that means for the two numbers |
|---|---|---|
| Positive | Positive | Both numbers positive |
| Positive | Negative | Both numbers negative |
| Negative | Either | One positive, one negative (their signs must differ) |
Factoring a Trinomial with Leading Coefficient Other Than 1: The AC Method
When the leading coefficient isn’t \(1\), the same “multiply and add” idea needs one adjustment: the two numbers multiply to \(a \cdot c\), not just \(c\), and the middle term gets split before grouping. This is covered in full in Factoring by Grouping; the short version:
$$ 2x^{2}+11x+5 \qquad ac = 2\cdot 5 = 10 $$
Two numbers that multiply to \(10\) and add to \(11\): \(10\) and \(1\).
$$ 2x^{2}+10x+x+5 = 2x(x+5)+1(x+5) = (x+5)(2x+1) $$
Recognizing the Difference of Squares and Perfect Square Trinomial
Two patterns from Multiplying Polynomials factor instantly once recognized, without needing any number-finding step at all.
| Pattern to spot | Factors as | Example |
|---|---|---|
| \(a^{2}-b^{2}\) (difference of squares) | \((a+b)(a-b)\) | \(x^{2}-49=(x+7)(x-7)\) |
| \(a^{2}+2ab+b^{2}\) (perfect square) | \((a+b)^{2}\) | \(x^{2}+14x+49=(x+7)^{2}\) |
| \(a^{2}-2ab+b^{2}\) (perfect square) | \((a-b)^{2}\) | \(x^{2}-14x+49=(x-7)^{2}\) |
Checking whether the first and last terms are perfect squares, and whether the middle term equals twice the product of their roots, takes only a few seconds and saves the entire multiply-and-add search when it succeeds.
When a Quadratic Can’t Be Factored
If no integer pair multiplies to \(ac\) and adds to \(b\) — checked systematically through every factor pair of \(ac\) — the quadratic is prime over the integers. It may still have real solutions (found with the quadratic formula), just not ones that correspond to integer-coefficient binomial factors.
$$ x^{2}+x+1 \qquad ac = 1 \qquad \text{factor pairs of } 1: \; (1,1) \text{ only} $$
\(1+1=2\), not \(1\) (the value of \(b\)) — no factor pair works, so this trinomial is prime.
Worked Example A: Leading Coefficient of 1
Factor \(x^{2}-2x-15\).
Step 1 — find two numbers that multiply to \(-15\) and add to \(-2\):
$$ -5 \text{ and } 3 $$
Step 2 — write the two binomials:
$$ (x-5)(x+3) $$
Answer: \((x-5)(x+3)\)
Worked Example B: The AC Method
Factor \(6x^{2}-7x-3\).
Step 1 — compute \(ac\):
$$ ac = 6\cdot(-3) = -18 $$
Step 2 — find two numbers that multiply to \(-18\) and add to \(-7\):
$$ -9 \text{ and } 2 $$
Step 3 — split the middle term and group:
$$ 6x^{2}-9x+2x-3 = 3x(2x-3)+1(2x-3) $$
Step 4 — factor out the matching binomial:
$$ (2x-3)(3x+1) $$
Answer: \((2x-3)(3x+1)\)
Worked Example C: Recognizing a Difference of Squares
Factor \(9x^{2}-25\).
Step 1 — recognize both terms as perfect squares:
$$ 9x^{2} = (3x)^{2} \qquad 25 = 5^{2} $$
Step 2 — apply the difference of squares pattern:
$$ (3x+5)(3x-5) $$
Answer: \((3x+5)(3x-5)\)
Worked Example D: Factoring Out a GCF First, Then a Trinomial
Factor \(3x^{2}+15x+18\).
Step 1 — factor out the GCF of \(3\) first:
$$ 3x^{2}+15x+18 = 3(x^{2}+5x+6) $$
Step 2 — factor the remaining trinomial (leading coefficient \(1\)):
$$ x^{2}+5x+6 = (x+2)(x+3) $$
Answer: \(3(x+2)(x+3)\). Removing the GCF first turned numbers that would have needed the ac method (\(3,15,18\)) into a simple leading-coefficient-of-\(1\) trinomial.
Common Mistakes to Avoid
- Skipping the GCF check. A trinomial like \(3x^{2}+15x+18\) is far easier to factor once the \(3\) is removed first — trying the ac method on the original numbers works but takes needless extra steps.
- Finding a pair that adds to \(b\) but doesn’t multiply to \(ac\) (or \(c\)). Both conditions must hold at the same time; a pair satisfying only one of them is not usable.
- Forgetting the sign rules for choosing the two numbers. A negative \(c\) always means the two numbers have opposite signs; a positive \(c\) means they share the same sign as \(b\).
- Misidentifying a perfect square trinomial. The middle term must equal exactly twice the product of the square roots of the first and last terms — \(x^{2}+10x+25\) is a perfect square (since \(2\cdot x\cdot5=10x\)), but \(x^{2}+8x+25\) is not, even though \(25\) is still a perfect square.
- Giving up too early and calling a quadratic prime. Check every factor pair of \(ac\), not just the first one or two that come to mind, before concluding no pair works.
- Applying the difference-of-squares pattern to a sum. \(x^{2}+49\) does not factor over the real numbers at all; the pattern only applies to a difference.
Where This Shows Up Later
- Solving quadratic equations. Once a quadratic is factored, the zero product property turns \((x-5)(x+3)=0\) directly into two simple linear equations.
- Factoring Higher-Degree Polynomials. Quartic and higher polynomials frequently reduce to this exact quadratic-factoring skill after a substitution or a grouping step.
- Simplifying Rational Expressions. Both the numerator and denominator of a rational expression usually need to be factored with these same techniques before any cancelling can happen.
- Graphing parabolas. A quadratic’s factored form directly reveals its x-intercepts, which is often the fastest way to sketch its graph.
Practice Problems
Work each problem yourself before opening the answer.
Problem 1. Factor \(x^{2}+9x+20\).
Show answer
Two numbers multiplying to \(20\) and adding to \(9\): \(4\) and \(5\).
$$ (x+4)(x+5) $$
Answer: \((x+4)(x+5)\)
Problem 2. Factor \(x^{2}-3x-40\).
Show answer
Two numbers multiplying to \(-40\) and adding to \(-3\): \(-8\) and \(5\).
$$ (x-8)(x+5) $$
Answer: \((x-8)(x+5)\)
Problem 3. Factor \(x^{2}-64\).
Show answer
Difference of squares: \(x^{2}-8^{2}\).
$$ (x+8)(x-8) $$
Answer: \((x+8)(x-8)\)
Problem 4. Factor \(x^{2}-18x+81\).
Show answer
Check: \(81=9^{2}\), and \(2\cdot x\cdot 9=18x\), so this is a perfect square trinomial.
$$ (x-9)^{2} $$
Answer: \((x-9)^{2}\)
Problem 5. Factor \(2x^{2}+9x+4\).
Show answer
Step 1 — compute \(ac\): \(2\cdot 4=8\).
Step 2 — find two numbers multiplying to \(8\) and adding to \(9\): \(8\) and \(1\).
Step 3 — split and group:
$$ 2x^{2}+8x+x+4 = 2x(x+4)+1(x+4) = (x+4)(2x+1) $$
Answer: \((x+4)(2x+1)\)
Problem 6. Factor \(4x^{2}+4x-3\).
Show answer
Step 1 — compute \(ac\): \(4\cdot(-3)=-12\).
Step 2 — find two numbers multiplying to \(-12\) and adding to \(4\): \(6\) and \(-2\).
Step 3 — split and group:
$$ 4x^{2}+6x-2x-3 = 2x(2x+3)-1(2x+3) = (2x+3)(2x-1) $$
Answer: \((2x+3)(2x-1)\)
Problem 7. Factor \(2x^{2}+12x+18\).
Show answer
Step 1 — factor out the GCF of \(2\):
$$ 2(x^{2}+6x+9) $$
Step 2 — recognize a perfect square trinomial (\(9=3^{2}\), \(2\cdot x\cdot 3=6x\)):
$$ 2(x+3)^{2} $$
Answer: \(2(x+3)^{2}\)
Problem 8. Factor \(5x^{2}-45\).
Show answer
Step 1 — factor out the GCF of \(5\):
$$ 5(x^{2}-9) $$
Step 2 — recognize a difference of squares:
$$ 5(x+3)(x-3) $$
Answer: \(5(x+3)(x-3)\)
Problem 9. Factor \(x^{2}+6x+10\), or determine that it is prime.
Show answer
Check every factor pair of \(10\): \((1,10)\) sums to \(11\); \((2,5)\) sums to \(7\). No pair sums to \(6\).
Answer: prime over the integers — it does not factor.
Problem 10. Factor \(3x^{2}-x-14\).
Show answer
Step 1 — compute \(ac\): \(3\cdot(-14)=-42\).
Step 2 — find two numbers multiplying to \(-42\) and adding to \(-1\): \(-7\) and \(6\).
Step 3 — split and group:
$$ 3x^{2}-7x+6x-14 = x(3x-7)+2(3x-7) = (3x-7)(x+2) $$
Answer: \((3x-7)(x+2)\)
Quick Reference
| Step | Move |
|---|---|
| 1. Always first | Factor out the GCF, if one exists |
| 2. Check the pattern | Difference of squares (\(a^{2}-b^{2}\)) or perfect square trinomial |
| 3. Leading coefficient \(1\) | Find two numbers multiplying to \(c\), adding to \(b\) |
| 4. Leading coefficient \(\neq 1\) | Use the ac method: multiply to \(ac\), add to \(b\), split and group |
| 5. No pair works anywhere | The quadratic is prime over the integers |
| Always check | Multiply the factored answer back out to confirm it reproduces the original trinomial |
Every branch of this lesson leans on two earlier ones: Greatest Common Factor for step 1 and Factoring by Grouping for the ac method in step 4. For factoring beyond degree \(2\), continue to Factoring Higher-Degree Polynomials. Build speed with the Factoring Quadratics Generator, or browse the rest of the Algebra lessons as new ones publish.