Algebra

Factoring Quadratics: The Four Patterns

Almost every factorable quadratic is one of four shapes. The skill is identifying which, not grinding through trial and error.

By the Gauth editorial teamHow this content is produced and checkedUpdated 10 September 2026

The idea

Factoring reverses multiplication. When you expand $(x + 3) (x + 5)$ you get $x^{2} + 8 x + 15$ , and every piece of that result has a source: the $x^{2}$ comes from the two leading terms, the $15$ from the two constants, and the $8 x$ from the two cross terms added together. Factoring asks you to run that machine backwards — to look at $x^{2} + 8 x + 15$ and recover the two brackets that produced it.

This matters because a product tells you something a sum never will. If $(x + 3) (x + 5) = 0$ , then one of the two brackets must be zero, because the only way real numbers multiply to zero is if one of them already is. That single observation, the zero-product property, is why factored form hands you the roots for free while the expanded form hides them.

The four patterns below are not four unrelated recipes. They are the four distinct shapes that expansion can produce, sorted by how many terms survive the collection step. Learning to recognise the shape is most of the work; the algebra afterwards is mechanical.

When you need it

The obvious trigger is an instruction that says “factor”. The more valuable skill is noticing the unmarked cases, where factoring is the move even though nobody asked for it:

Solving. Any polynomial equation set to zero. Factored form plus the zero-product property beats every other method when it applies.
Simplifying a fraction. $\frac{x ^{2} - 9}{x ^{2} + 7 x + 12}$ looks stuck until both parts are factored, at which point $(x + 3)$ cancels and it reduces to $\frac{x - 3}{x + 4}$ .
Inequalities. To decide where $x^{2} - x - 6 > 0$ you need the sign of each factor, which means you need the factors.
A limit that gives 0/0. The indeterminate form nearly always means a common factor is waiting to be cancelled from top and bottom.
Sketching. The factored form names every x-intercept without any further work.

Before committing time, spend five seconds on the discriminant. For $a x^{2} + b x + c$ , if $b^{2} - 4 a c$ is a perfect square, integer factors exist and are worth finding. If it is not, no integer pair will ever work and you should go straight to the quadratic formula rather than burning two minutes on a search that cannot succeed.

The method

Three of the four patterns are just expansions read right to left, so they are worth deriving once rather than memorising.

Difference of squares. Expand the candidate answer and watch the middle terms destroy each other:

(a - b) (a + b) = a^{2} + ab - ab - b^{2} = a^{2} - b^{2}

The $+ ab$ and $- ab$ cancel exactly, which is why the result has no middle term at all. That absence is the recognition cue: two terms, a subtraction, both perfect squares.

Perfect square trinomial. Same approach, different signs:

(a + b)^{2} = a^{2} + 2 ab + b^{2}, (a - b)^{2} = a^{2} - 2 ab + b^{2}

The middle term is twice the product of the two roots. That factor of two is the test. In $x^{2} + 8 x + 16$ , the outer roots are $x$ and $4$ , and $2 \cdot x \cdot 4 = 8 x$ matches, so it is $(x + 4)^{2}$ . In $x^{2} + 9 x + 16$ it does not match, and that quadratic does not factor over the integers at all.

Why the AC method works. Suppose the trinomial does factor, into $(p x + q) (r x + s)$ . Expanding gives:

(p x + q) (r x + s) = p r x^{2} + (p s + q r) x + q s

Matching coefficients against $a x^{2} + b x + c$ gives $a = p r$ , $b = p s + q r$ , and $c = q s$ . Now multiply the two pieces of the middle coefficient together:

(p s) (q r) = (p r) (q s) = a c

So $p s$ and $q r$ are two numbers whose product is $a c$ and whose sum is $b$ . Finding that pair and splitting the middle term into those two pieces recovers the four-term expression that grouping can undo. Nothing is being guessed.

The working procedure, in order:

Greatest common factor first. Take out the largest shared number and the lowest shared power of each variable. This shrinks every coefficient and frequently exposes a pattern that was invisible before.
Count the remaining terms. Two, three, or four. This single number narrows you to at most two candidate patterns.
Test the special patterns first. A difference of squares or a perfect square trinomial takes one line. Only fall through to AC when neither fits.
Run AC, then group. Split $b$ into the two numbers you found, group the four terms into pairs, and factor each pair. The bracket left over must be identical in both pairs; if it is not, you have a sign error.
Factor the results again.“Completely” means every remaining bracket is prime. Squares in disguise, like $x^{2} - 4$ appearing inside your answer, are easy to walk past.
Expand to check. Thirty seconds, and it catches every sign error you would otherwise carry into the next question.

Three worked examples

The first uses AC on a leading coefficient other than one. The second stacks two patterns. The third is the awkward one: it is a quadratic only if you squint, and finishing it takes four factors and three separate applications of the patterns.

Worked example

2 x^{2} + 7 x + 3

1
No common factor, three terms, and the first and last are not both perfect squares with a matching middle term. That leaves the AC method.
$a = 2, b = 7, c = 3 ⟹ a c = 6$
2
Find two integers multiplying to 6 and adding to 7. Both must be positive, so the candidates are 1 and 6, or 2 and 3. Only the first pair sums to 7.
$6 \times 1 = 6, 6 + 1 = 7$
3
Split the middle term into those two pieces. The order does not matter; either choice works.
$2 x^{2} + 6 x + x + 3$
4
Group in pairs and factor each pair. Write the 1 explicitly in the second group so the shared bracket is visible.
$2 x (x + 3) + 1 (x + 3)$
5
The bracket (x + 3) is common to both, so factor it out.
$(x + 3) (2 x + 1)$
6
Check by expanding: 2x² + x + 6x + 3 collects back to the original.
$(x + 3) (2 x + 1) = 2 x^{2} + x + 6 x + 3 = 2 x^{2} + 7 x + 3$

Answer

(x + 3) (2 x + 1)

Worked example

18 x^{3} - 50 x

1
Both terms share a factor of 2 and a factor of x. Take out 2x before looking for any pattern.
$18 x^{3} - 50 x = 2 x (9 x^{2} - 25)$
2
Two terms remain, subtracted. Check whether both are perfect squares: 9x² is (3x)² and 25 is 5². They are.
$9 x^{2} = (3 x)^{2}, 25 = 5^{2}$
3
Apply the difference of squares with a = 3x and b = 5.
$9 x^{2} - 25 = (3 x - 5) (3 x + 5)$
4
Reassemble with the GCF. Neither remaining bracket factors further, so this is complete.
$18 x^{3} - 50 x = 2 x (3 x - 5) (3 x + 5)$
5
Check the outer product and the inner product: 2x times 9x² is 18x³, and 2x times −25 is −50x.
$2 x (9 x^{2} - 25) = 18 x^{3} - 50 x$

Answer

2 x (3 x - 5) (3 x + 5)

Worked example

4 x^{4} - 37 x^{2} + 9

1
No common factor, and the powers are 4, 2 and 0. That is a quadratic in x², so treat x² as the variable and the usual patterns apply.
$4 (x^{2})^{2} - 37 (x^{2}) + 9$
2
Run AC on the disguised quadratic. The product is 4 × 9 = 36 and the sum must be −37, so both numbers are negative.
$(- 1) (- 36) = 36, (- 1) + (- 36) = - 37$
3
Split the middle term and group. Watch the sign on the second group: pulling out −9 rather than +9 is what leaves 4x² − 1, matching the first bracket exactly.
$4 x^{4} - x^{2} - 36 x^{2} + 9 = x^{2} (4 x^{2} - 1) - 9 (4 x^{2} - 1)$
4
Pull out the shared bracket.
$(x^{2} - 9) (4 x^{2} - 1)$
5
Stop and look again. Both brackets are themselves differences of squares, so this is not finished.
$x^{2} - 9 = (x - 3) (x + 3), 4 x^{2} - 1 = (2 x - 1) (2 x + 1)$
6
Assemble all four factors. A quartic can have up to four linear factors, so this count is the expected one.
$(x - 3) (x + 3) (2 x - 1) (2 x + 1)$
7
Check by expanding in pairs: (x² − 9)(4x² − 1) gives 4x⁴ − x² − 36x² + 9.
$(x^{2} - 9) (4 x^{2} - 1) = 4 x^{4} - 37 x^{2} + 9$

Answer

(x - 3) (x + 3) (2 x - 1) (2 x + 1)

Identify the pattern first

Work down this tree with the expression in front of you rather than guessing. Each question eliminates whole categories, and the answer at the bottom comes with the reason it beat the alternatives.

Which factoring pattern is this?

Does every term share a common factor?

Solve a factoring problem now

First solution free

Where people go wrong

Treating a sum of squares as factorable

$x^{2} + 16$ is not $(x + 4)^{2}$ , and it is not $(x + 4) (x - 4)$ either. Expanding $(x + 4)^{2}$ gives an $8 x$ that is not there, and $(x + 4) (x - 4)$ gives $x^{2} - 16$ . Over the reals a sum of squares is prime, and the reason is short: factors would mean roots, and $x^{2} = - 16$ has no real solution.

Stopping at the first factorisation

$x^{4} - 16$ becomes $(x^{2} - 4) (x^{2} + 4)$ , which looks like an answer and is not one. The first bracket is still a difference of squares, so the complete form is $(x - 2) (x + 2) (x^{2} + 4)$ . The habit that fixes this: after every factorisation, look at each bracket individually and ask the pattern question again.

Losing the GCF on the way out

Having taken $2 x$ out of $18 x^{3} - 50 x$ , it is easy to factor $9 x^{2} - 25$ beautifully and then write $(3 x - 5) (3 x + 5)$ as the final answer. The $2 x$ is part of the expression. Expanding your answer catches this instantly, because the leading coefficient will be wrong by a factor of two.

Sign errors in the second group

In $4 x^{4} - x^{2} - 36 x^{2} + 9$ , the second pair is $- 36 x^{2} + 9$ . Factoring out $- 9$ leaves $4 x^{2} - 1$ , matching the first bracket. Factoring out $+ 9$ would leave $- 4 x^{2} + 1$ , which does not match, and students often conclude the method failed. It did not; the sign chosen was wrong. Always pull out the sign that makes the two brackets agree.

Practice

Factor each one completely. Where a difference of squares appears, write the bracket containing the minus sign first so the checker can match your answer.

One from each pattern

Answers are checked here — nothing is sent anywhere.

1
$x^{2} - 9 (minus-sign bracket first)$
2
$3 x^{2} - 27 x$
3
$x^{2} + 10 x + 25 (write as a square)$
4
$2 x^{2} - 5 x - 3 (the bracket with 2 x first)$

Where this goes next

Factoring and the formula are two routes to the same place, and the discriminant test above is really a question about which is cheaper — the full comparison sits in the guide to choosing a quadratic method. Every pattern here depends on recognising perfect powers, so if $9 x^{2} = (3 x)^{2}$ or $4 x^{4} = (2 x^{2})^{2}$ is not automatic yet, the rules for exponents and radicals derive exactly why those regroupings are legal. For a factorisation you cannot get to come out, the factoring calculator names the pattern it used at each stage, and the algebra solver will carry the factored form on into whatever the question actually asked for.

Frequently asked questions

How do I know a quadratic will factor over the integers before I try?

Compute the discriminant b^2 - 4ac. If it is a perfect square, the roots are rational and integer factors exist. If it is not, no amount of hunting will find them and you should switch to the quadratic formula immediately. For 6x^2 - x - 12 the discriminant is 1 + 288 = 289 = 17^2, so it factors; for 6x^2 - x - 11 it is 1 + 264 = 265, which is not a square, so it does not.

Why does the AC method work?

If ax^2 + bx + c equals (px + q)(rx + s), then expanding gives a = pr, b = ps + qr, and c = qs. Multiply the two middle pieces: (ps)(qr) = (pr)(qs) = ac. So the two numbers you are hunting for, ps and qr, must multiply to ac and add to b. The method is not a trick; it is a direct consequence of what expanding a pair of brackets does.

Can a sum of squares be factored?

Not over the real numbers. If x^2 + k with k positive had real factors, it would have real roots, but x^2 = -k has no real solution. Over the complex numbers it splits as (x + i*sqrt(k))(x - i*sqrt(k)). A difference of squares is the one that factors, and confusing the two is the most common false pattern match.

What does 'factor completely' actually require?

Keep factoring until no factor can be broken down further over the integers. That usually means pulling out the greatest common factor first, then applying a pattern, then checking whether any resulting factor is itself a difference of squares or another trinomial. x^4 - 16 is not finished at (x^2 - 4)(x^2 + 4), because the first bracket is still a difference of squares.

Should I bother factoring if I already know the quadratic formula?

Yes, for two reasons. Factored form gives the roots and the y-intercept at a glance, which matters when sketching or solving inequalities. And factoring is the operation you will use constantly in rational expressions, limits, and integration, where the point is cancellation rather than finding a numerical root.

Keep going

Factoring calculatorFactors completely and names the pattern applied at every stage.The quadratic formulaWhat to do when the discriminant is not a perfect square.Exponent and radical rulesWhy 9x² is a perfect square and 8x³ is a perfect cube, derived from one idea.Algebra solverFull working for equations, inequalities, and simplification.