How to Factor a Quadratic Equation: 3 Methods With Worked Examples
Knowing how to factor a quadratic equation is one of the core skills in high school algebra — it shows up in tests, standardized exams, and every math course that follows. A quadratic equation in standard form looks like ax² + bx + c = 0, and factoring means rewriting that expression as a product of two simpler binomials so you can find the values of x that make the equation true. Students often ask how to factor a quadratic equation quickly on a timed test, and the answer depends on the type of quadratic — whether a equals 1, whether a special pattern applies, or whether the AC method is needed. This guide walks through all three approaches in order from simplest to most general, shows every step on real numerical examples, and ends with a set of practice problems so you can test yourself before a test.
Contents
- 01What Is Factoring a Quadratic Equation?
- 02Method 1: How to Factor a Quadratic Equation When a = 1
- 03Three Worked Examples Using the Factor-Pair Method
- 04Method 2: How to Factor a Quadratic Equation When a ≠ 1 (The AC Method)
- 05AC Method — Three More Worked Examples
- 06Method 3: Special Factoring Patterns
- 07Common Mistakes When Factoring Quadratic Equations
- 08Practice Problems: Factor These Quadratic Equations
- 09When Factoring Doesn't Work — and What to Do Instead
- 10FAQ — How to Factor a Quadratic Equation
What Is Factoring a Quadratic Equation?
A quadratic equation has the standard form ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. Factoring means rewriting the left side as a product of two binomials: (px + q)(rx + s) = 0. Once the equation is in factored form, you apply the zero-product property — if the product of two factors is zero, then at least one factor must equal zero. This turns one quadratic equation into two simple linear equations, each of which is trivial to solve. For example, (x + 3)(x + 4) = 0 immediately gives x = −3 or x = −4. The power of factoring is that it converts a potentially tricky quadratic into two one-step equations. However, factoring only gives neat, rational answers when the discriminant b² − 4ac is a perfect square (0, 1, 4, 9, 16, 25, …). When it is not, you need the quadratic formula — but for a large portion of textbook and test problems, factoring is the faster route. The three methods covered in this guide are: (1) the factor-pair method for monic quadratics where a = 1, (2) the AC method for non-monic quadratics where a ≠ 1, and (3) special patterns like perfect square trinomials and difference of squares. Each one is a distinct technique with its own decision criteria, but they all rest on the same logical foundation: zero-product property.
Zero-product property: if (x + p)(x + q) = 0, then x = −p or x = −q. This is the engine that makes factoring useful.
Method 1: How to Factor a Quadratic Equation When a = 1
When the leading coefficient a equals 1, the quadratic has the monic form x² + bx + c = 0. This is the most common form in introductory algebra, and the factor-pair method handles it in four steps. The key insight is that if the factored form is (x + p)(x + q), expanding it gives x² + (p + q)x + pq. That means p + q = b (the middle coefficient) and p × q = c (the constant). Your job is to find two numbers whose sum is b and whose product is c. With practice, this takes under a minute for small integers.
1. Step 1 — Write the equation in standard form
Make sure the equation is arranged as x² + bx + c = 0 with zero on the right side. If the equation is presented as x² − 3x = 10, subtract 10 from both sides first: x² − 3x − 10 = 0. Never try to identify b and c until the right side is zero.
2. Step 2 — Identify b and c
Read b and c directly from the standard form, including their signs. In x² − 3x − 10 = 0, we have b = −3 and c = −10. The sign is part of the coefficient — do not strip it away.
3. Step 3 — List factor pairs of c and find the right pair
Write out pairs of integers whose product equals c, then check which pair sums to b. For c = −10: the factor pairs are (1, −10), (−1, 10), (2, −5), (−2, 5). Check sums: 1 + (−10) = −9, no. (−1) + 10 = 9, no. 2 + (−5) = −3, yes! The pair is (2, −5).
4. Step 4 — Write the factored form and solve
Use the pair to write (x + 2)(x − 5) = 0. Apply the zero-product property: x + 2 = 0 gives x = −2, and x − 5 = 0 gives x = 5. Always check both answers by substituting back: for x = −2: (−2)² − 3(−2) − 10 = 4 + 6 − 10 = 0 ✓. For x = 5: 25 − 15 − 10 = 0 ✓.
For monic quadratics: find p and q where p × q = c and p + q = b. Then the factored form is (x + p)(x + q) = 0.
Three Worked Examples Using the Factor-Pair Method
Working through examples builds the pattern recognition needed to factor quickly. Each example below uses the same four-step process and highlights a slightly different sign situation. Cover the solutions and attempt each problem yourself before reading the answer.
1. Example 1 (Both factors positive) — x² + 8x + 15 = 0
b = 8, c = 15. Factor pairs of 15: (1, 15), (3, 5). Sums: 1 + 15 = 16, no. 3 + 5 = 8, yes. Factored form: (x + 3)(x + 5) = 0. Solutions: x = −3 or x = −5. Check x = −3: 9 − 24 + 15 = 0 ✓. Check x = −5: 25 − 40 + 15 = 0 ✓. When b and c are both positive, both numbers in the pair are positive.
2. Example 2 (Mixed signs) — x² − 2x − 24 = 0
b = −2, c = −24. Because c is negative, one number in the pair is positive and one is negative. Factor pairs of −24 where one has each sign: (4, −6), (−4, 6), (3, −8), (−3, 8), and others. Sums: 4 + (−6) = −2, yes! Factored form: (x + 4)(x − 6) = 0. Solutions: x = −4 or x = 6. Check x = 6: 36 − 12 − 24 = 0 ✓. Check x = −4: 16 + 8 − 24 = 0 ✓.
3. Example 3 (Both factors negative) — x² − 11x + 28 = 0
b = −11, c = 28. Because c is positive and b is negative, both numbers in the pair are negative. Factor pairs of 28 (both negative): (−1, −28), (−2, −14), (−4, −7). Sums: −1 + (−28) = −29, no. −2 + (−14) = −16, no. −4 + (−7) = −11, yes! Factored form: (x − 4)(x − 7) = 0. Solutions: x = 4 or x = 7. Check x = 4: 16 − 44 + 28 = 0 ✓. Check x = 7: 49 − 77 + 28 = 0 ✓.
Sign rule quick-check: c > 0 and b > 0 → both factors positive. c > 0 and b < 0 → both factors negative. c < 0 → factors have opposite signs.
Method 2: How to Factor a Quadratic Equation When a ≠ 1 (The AC Method)
When the leading coefficient a is not 1, the factor-pair method needs a modification called the AC method (also called the split-the-middle-term method or the grouping method). The idea is to multiply a × c, find two numbers multiplying to that product and adding to b, use them to rewrite the middle term as two separate terms, then factor by grouping. This method always works for any factorable quadratic, regardless of how large a is.
1. Step 1 — Compute the product a × c
Multiply the leading coefficient by the constant term. For 6x² + 11x + 4 = 0, compute 6 × 4 = 24. This product is the new target for your factor pair.
2. Step 2 — Find two numbers multiplying to a × c and adding to b
For 6x² + 11x + 4, you need two numbers multiplying to 24 and adding to 11. Factor pairs of 24: (1, 24), (2, 12), (3, 8), (4, 6). Sums: 3 + 8 = 11, yes. The pair is (3, 8).
3. Step 3 — Split the middle term using the pair
Replace the 11x term with 3x + 8x (using the pair in either order): 6x² + 3x + 8x + 4 = 0. The equation is algebraically identical — you have only rewritten the middle term.
4. Step 4 — Factor by grouping
Group the four terms in pairs: (6x² + 3x) + (8x + 4) = 0. Factor the GCF from each group: 3x(2x + 1) + 4(2x + 1) = 0. The binomial (2x + 1) appears in both groups, so factor it out: (2x + 1)(3x + 4) = 0.
5. Step 5 — Apply the zero-product property and solve
2x + 1 = 0 gives x = −1/2. 3x + 4 = 0 gives x = −4/3. Check x = −1/2: 6(1/4) + 11(−1/2) + 4 = 1.5 − 5.5 + 4 = 0 ✓. Check x = −4/3: 6(16/9) + 11(−4/3) + 4 = 32/3 − 44/3 + 12/3 = 0/3 = 0 ✓.
AC method in one sentence: find two numbers multiplying to a × c and adding to b, split the middle term with them, then factor by grouping.
AC Method — Three More Worked Examples
The AC method can feel abstract until you practice it several times. Each example below picks a different pair structure so you see how the method handles signs. The step that trips students up most often is the grouping — if both groups share a common binomial factor, the grouping is correct; if they do not, swap the order of the two middle terms and try again.
1. Example 4 — 2x² + 7x + 3 = 0
a × c = 2 × 3 = 6. Find two numbers multiplying to 6 and adding to 7: (1, 6) → 7, yes. Split: 2x² + x + 6x + 3 = 0. Group: x(2x + 1) + 3(2x + 1) = 0. Factor: (x + 3)(2x + 1) = 0. Solutions: x = −3 or x = −1/2. Check x = −3: 2(9) + 7(−3) + 3 = 18 − 21 + 3 = 0 ✓.
2. Example 5 (Negative middle) — 3x² − 10x + 8 = 0
a × c = 3 × 8 = 24. Need two numbers multiplying to 24 and adding to −10. Because both the product (24, positive) and the sum (−10, negative) have these sign conditions, both numbers must be negative. Factor pairs of 24 (both negative): (−4, −6) → sum = −10, yes. Split: 3x² − 4x − 6x + 8 = 0. Group: x(3x − 4) − 2(3x − 4) = 0. Factor: (x − 2)(3x − 4) = 0. Solutions: x = 2 or x = 4/3. Check x = 2: 12 − 20 + 8 = 0 ✓.
3. Example 6 (Negative constant) — 4x² + 4x − 15 = 0
a × c = 4 × (−15) = −60. Need two numbers multiplying to −60 and adding to 4. One number positive, one negative. Try pairs: (10, −6) → sum = 4, yes. Split: 4x² + 10x − 6x − 15 = 0. Group: 2x(2x + 5) − 3(2x + 5) = 0. Factor: (2x − 3)(2x + 5) = 0. Solutions: x = 3/2 or x = −5/2. Check x = 3/2: 4(9/4) + 4(3/2) − 15 = 9 + 6 − 15 = 0 ✓.
Method 3: Special Factoring Patterns
Some quadratics fit recognizable algebraic identities and can be factored in one line without any trial and error. Memorizing these patterns saves time on timed tests and helps you spot elegant solutions that the AC method would handle more slowly. There are three patterns worth knowing at the algebra level: perfect square trinomials, difference of two squares (which is technically a binomial, not a trinomial), and sum or difference of cubes (relevant if your course covers cubic expressions). For standard quadratics, the first two are the most important.
1. Pattern 1 — Perfect Square Trinomial
A perfect square trinomial has the form a²x² ± 2abx + b². It factors as (ax ± b)². Recognition cues: the first and last terms are perfect squares, and the middle term is exactly twice the product of their square roots. Example: x² + 10x + 25. First term: x² = (x)². Last term: 25 = (5)². Middle term: 10x = 2 × x × 5 ✓. Factored: (x + 5)². Solution: x = −5 (repeated root). Another example: 4x² − 12x + 9 = (2x − 3)², giving x = 3/2 as a repeated root.
2. Pattern 2 — Difference of Squares
An expression of the form a²x² − b² factors as (ax + b)(ax − b). The middle term is zero (b = 0 in standard form), so the sum-product requirement reduces to: find two numbers multiplying to −b² and summing to 0. Examples: x² − 49 = (x + 7)(x − 7), giving x = ±7. 9x² − 16 = (3x + 4)(3x − 4), giving x = 4/3 or x = −4/3. 25x² − 4 = (5x + 2)(5x − 2), giving x = ±2/5. Caution: a sum of squares such as x² + 49 does NOT factor over the real numbers.
3. Pattern 3 — Perfect Square Combined With a Constant Shift
Sometimes completing-the-square thinking helps factor expressions that aren't obviously recognizable. For x² + 6x + 8, you might notice x² + 6x = (x + 3)² − 9, so x² + 6x + 8 = (x + 3)² − 1 = (x + 3 + 1)(x + 3 − 1) = (x + 4)(x + 2). This approach reframes the factor-pair method geometrically and can speed up mental factoring for moderately large coefficients.
Quick pattern check before using the AC method: is the first term a perfect square? Is the last term a perfect square? Is the middle term twice their product? If yes to all three, it's a perfect square trinomial.
Common Mistakes When Factoring Quadratic Equations
Most errors in factoring quadratic equations come from a handful of recurring habits. Each one below is paired with a concrete prevention strategy. If you recognize your own errors in this list, those are the ones to practice most before a test.
1. Mistake 1 — Not rearranging to standard form first
If the equation is 2x² = 5x − 3, you cannot factor it as-is. Subtract 5x and add 3 to get 2x² − 5x + 3 = 0 before identifying a, b, and c. This mistake changes the coefficients and gives completely wrong factor pairs. Fix: before doing anything else, write 'Standard form: ___ = 0' and fill it in.
2. Mistake 2 — Forgetting the GCF before factoring
If all terms share a common factor, pull it out first. For 2x² + 10x + 12 = 0, the GCF is 2. Factor it out: 2(x² + 5x + 6) = 0, which simplifies to x² + 5x + 6 = 0. Then factor the monic trinomial: (x + 2)(x + 3) = 0. If you skip this step, you end up running the AC method on harder numbers unnecessarily.
3. Mistake 3 — Using the wrong sign in the factored form
The factored form (x + p)(x + q) uses + signs, and the solutions are x = −p and x = −q. If you find the pair (−3, 5) for a monic quadratic, the factored form is (x − 3)(x + 5) = 0, not (x + 3)(x − 5) = 0. The pair values go directly into the binomials with the opposite sign when solving. Writing the pair and the factored form side by side on paper reduces this error.
4. Mistake 4 — Stopping at the factored form without solving
Writing (x − 4)(x + 2) = 0 is not the final answer — you must apply the zero-product property and state x = 4 or x = −2. Many students lose a full mark by treating the factored form as the solution. Always complete the problem by writing x = ___.
5. Mistake 5 — Forcing factoring when it doesn't work
Not every quadratic factors over the integers. If you have tried all factor pairs of c and none sum to b, the equation either does not factor or requires the quadratic formula. A quick check: compute b² − 4ac. If the result is a perfect square, factoring will work. If not, move directly to x = (−b ± √(b² − 4ac)) / 2a. Spending five minutes searching for factor pairs that don't exist wastes time on a timed test.
6. Mistake 6 — Grouping error in the AC method
In the AC method, after splitting the middle term, the two groups must share a common binomial factor. If they don't, you either split incorrectly or made an arithmetic error. Double-check that your two numbers genuinely multiply to a × c and add to b, then try swapping the order of the split terms. For 6x² + 11x + 4 split as 6x² + 8x + 3x + 4: group as 2x(3x + 4) + 1(3x + 4) = 0 → (2x + 1)(3x + 4) = 0. Swapping the split terms sometimes makes the grouping easier to see.
If you can't find factor pairs after checking all options, compute b² − 4ac. A non-perfect-square result means the equation cannot be factored over the integers — use the quadratic formula instead.
Practice Problems: Factor These Quadratic Equations
The problems below are arranged in increasing difficulty. Attempt each one before reading the solution. For problems 1–4 the leading coefficient is 1. Problems 5–7 have a ≠ 1 and use the AC method. Problem 8 uses a special pattern. Problem 9 requires you to first extract the GCF, and problem 10 is a word problem where you must build the equation before factoring.
1. Problem 1 — x² + 9x + 18 = 0
Need p × q = 18 and p + q = 9. Pairs of 18: (1,18), (2,9), (3,6). Sum 3 + 6 = 9 ✓. Factored: (x + 3)(x + 6) = 0. Solutions: x = −3 or x = −6. Check x = −3: 9 − 27 + 18 = 0 ✓.
2. Problem 2 — x² − 5x − 14 = 0
Need p × q = −14 and p + q = −5. Pair (−7, 2): −7 × 2 = −14 ✓ and −7 + 2 = −5 ✓. Factored: (x − 7)(x + 2) = 0. Solutions: x = 7 or x = −2. Check x = 7: 49 − 35 − 14 = 0 ✓.
3. Problem 3 — x² − 16x + 63 = 0
Need p × q = 63 and p + q = −16. Both negative since c > 0 and b < 0. Pairs (both negative): (−7, −9) → sum = −16 ✓. Factored: (x − 7)(x − 9) = 0. Solutions: x = 7 or x = 9. Check x = 9: 81 − 144 + 63 = 0 ✓.
4. Problem 4 — x² + x − 42 = 0
Need p × q = −42 and p + q = 1 (note b = 1, the coefficient of x). Opposite signs since c < 0. Pair (7, −6): 7 × (−6) = −42 ✓ and 7 + (−6) = 1 ✓. Factored: (x + 7)(x − 6) = 0. Solutions: x = −7 or x = 6. Check x = 6: 36 + 6 − 42 = 0 ✓.
5. Problem 5 — 3x² + 14x + 8 = 0
AC method: a × c = 3 × 8 = 24. Find pair multiplying to 24 adding to 14: (2, 12) → 14 ✓. Split: 3x² + 2x + 12x + 8 = 0. Group: x(3x + 2) + 4(3x + 2) = 0. Factor: (x + 4)(3x + 2) = 0. Solutions: x = −4 or x = −2/3. Check x = −4: 3(16) + 14(−4) + 8 = 48 − 56 + 8 = 0 ✓.
6. Problem 6 — 5x² − 13x + 6 = 0
AC method: a × c = 5 × 6 = 30. Find pair multiplying to 30 adding to −13: both negative since product positive and sum negative. (−3, −10) → product = 30 ✓ and sum = −13 ✓. Split: 5x² − 3x − 10x + 6 = 0. Group: x(5x − 3) − 2(5x − 3) = 0. Factor: (x − 2)(5x − 3) = 0. Solutions: x = 2 or x = 3/5. Check x = 2: 20 − 26 + 6 = 0 ✓.
7. Problem 7 — 6x² − x − 12 = 0
AC method: a × c = 6 × (−12) = −72. Opposite sign pair summing to −1: (8, −9) → 8 × (−9) = −72 ✓ and 8 + (−9) = −1 ✓. Split: 6x² + 8x − 9x − 12 = 0. Group: 2x(3x + 4) − 3(3x + 4) = 0. Factor: (2x − 3)(3x + 4) = 0. Solutions: x = 3/2 or x = −4/3. Check x = 3/2: 6(9/4) − (3/2) − 12 = 13.5 − 1.5 − 12 = 0 ✓.
8. Problem 8 (Special pattern) — 16x² − 25 = 0
Recognize the difference of squares: 16x² − 25 = (4x)² − 5² = (4x + 5)(4x − 5) = 0. Solutions: x = −5/4 or x = 5/4. Check x = 5/4: 16(25/16) − 25 = 25 − 25 = 0 ✓. No trial and error needed once the pattern is recognized.
9. Problem 9 (GCF first) — 4x² − 8x − 60 = 0
GCF of 4, 8, and 60 is 4. Factor out: 4(x² − 2x − 15) = 0. Since 4 ≠ 0, solve x² − 2x − 15 = 0. Need p × q = −15 and p + q = −2. Pair (−5, 3): −5 × 3 = −15 ✓ and −5 + 3 = −2 ✓. Factored: 4(x − 5)(x + 3) = 0. Solutions: x = 5 or x = −3. Check x = 5: 4(25) − 8(5) − 60 = 100 − 40 − 60 = 0 ✓.
10. Problem 10 (Word problem) — Rectangular Garden
A rectangular patio has a length 4 m longer than its width. The area is 45 m². Find the dimensions. Let width = x m, so length = (x + 4) m. Area equation: x(x + 4) = 45. Rearrange to standard form: x² + 4x − 45 = 0. Need p × q = −45 and p + q = 4. Pair (9, −5): 9 × (−5) = −45 ✓ and 9 + (−5) = 4 ✓. Factored: (x + 9)(x − 5) = 0. Solutions: x = −9 (discard — length cannot be negative) or x = 5. Width = 5 m, length = 9 m. Check: 5 × 9 = 45 m² ✓.
When Factoring Doesn't Work — and What to Do Instead
Factoring is not always possible, and knowing when to stop trying saves significant time on timed assessments. A quadratic factors over the integers if and only if the discriminant b² − 4ac is a perfect square (0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, …). If b² − 4ac equals any other non-negative number, the roots exist but are irrational, and the quadratic formula is the right tool. If b² − 4ac is negative, the roots are complex (non-real), and neither factoring nor the standard quadratic formula gives real solutions. Consider the equation x² + x + 1 = 0: b² − 4ac = 1 − 4 = −3. This is negative, so there are no real solutions and you cannot factor a quadratic equation of this type over the real numbers. Compare that with x² + x − 6 = 0: b² − 4ac = 1 + 24 = 25, which is 5², so the equation factors as (x + 3)(x − 2) = 0, giving x = −3 or x = 2. The decision tree is simple: compute the discriminant first. Perfect square → factor. Non-perfect-square positive → quadratic formula for irrational roots. Negative → no real solutions. Building this habit means you will never spend more than 30 seconds deciding which method to use. For a full walkthrough of the quadratic formula including worked examples with irrational roots, see the related article on how to use the quadratic equation linked below.
Before spending more than 30 seconds hunting for factor pairs, compute b² − 4ac. If it's not a perfect square, stop factoring and use the quadratic formula.
FAQ — How to Factor a Quadratic Equation
These are the questions students ask most often when learning how to factor a quadratic equation. The answers focus on practical mechanics — what to actually write and decide during a problem rather than abstract theory.
1. What is the fastest way to check if a quadratic can be factored?
Compute the discriminant: b² − 4ac. If the result is a perfect square (0, 1, 4, 9, 16, 25, etc.), the quadratic can be factored over the integers. If not, use the quadratic formula. This check takes about 10 seconds and tells you immediately which approach to use.
2. Does the AC method work when a = 1?
Yes, the AC method works for any quadratic — when a = 1, a × c = c, so you are just finding two numbers multiplying to c and adding to b, which is exactly the factor-pair method. The two methods are identical in the monic case. For non-monic quadratics, the AC method is the reliable general approach.
3. Do I have to factor, or can I always just use the quadratic formula?
You can always use the quadratic formula — it works for every quadratic equation without exception. Factoring is a faster option for problems with rational roots, but it is never required. Many teachers expect you to show factoring when the roots are integers or simple fractions, because it demonstrates conceptual understanding. If the test or homework does not specify a method, you may use whichever approach you prefer.
4. What if I can't find factor pairs after trying all combinations?
First double-check your arithmetic by multiplying out a couple of candidate pairs. Then compute b² − 4ac. If it is not a perfect square, the equation genuinely cannot be factored over the integers and you should switch to the quadratic formula. You have not made an error — not every quadratic has integer roots.
5. Is there a shortcut for quadratics with large coefficients?
For large coefficients, the AC method combined with systematic listing is the most reliable approach. However, a shortcut worth knowing: after computing a × c, focus only on factor pairs near the square root of |a × c|. If a × c = 120, the square root is about 10.9, so pairs near (10, 12) or (8, 15) are likely candidates. This narrows the search from checking every pair to checking 3–4 near the middle.
6. Can I factor a quadratic that has a common factor but a ≠ 1 after factoring?
Yes — and you must. For 6x² + 18x + 12 = 0, the GCF is 6: factor it out to get 6(x² + 3x + 2) = 0. Now factor the monic trinomial inside the parentheses: 6(x + 1)(x + 2) = 0. The solutions are x = −1 or x = −2. Always factor out the GCF first before deciding whether the remaining trinomial has a = 1 or a ≠ 1.
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