Multiplying out brackets is easy — you just push through. The real magic is running that process backwards: starting from a messy expression and discovering the tidy product it came from. That's factoring, and here is the picture that makes it obvious.
What a polynomial is
A polynomial is nothing more than a sum of terms, where each term is a number times a power of \( x \). Here is one:
\[ 3x^2 + 2x - 5 \]The numbers out front — \( 3 \), \( 2 \) and \( -5 \) — are the coefficients. The highest power of \( x \) that appears is called the degree; this one has degree \( 2 \), so we call it a quadratic. Degree \( 1 \) is linear, degree \( 3 \) is cubic, and so on. The degree tells you, at a glance, the most important thing about a polynomial's shape and behaviour.
Terms that share the same power of \( x \) are called like terms — and like terms are the only ones you're ever allowed to combine.
The quick tour: adding and multiplying
Before we factor, let's run the machine forwards so you know exactly what we're undoing.
Adding and subtracting is just collecting like terms — add the coefficients of matching powers and leave everything else alone:
\[ (3x^2 + 2x - 5) + (x^2 - 4x + 1) = 4x^2 - 2x - 4 \]The \( x^2 \) terms join together, the \( x \) terms join together, the constants join together. Powers that don't match simply ride along untouched — they're different-sized pieces and can't be merged, just like you can't add apples to oranges.
Multiplying follows one rule: every term in the first bracket multiplies every term in the second. For two binomials, the popular nickname is FOIL — First, Outer, Inner, Last — which is just a way of remembering all four products:
\[ (x + 2)(x + 3) = \underbrace{x^2}_{\text{First}} + \underbrace{3x}_{\text{Outer}} + \underbrace{2x}_{\text{Inner}} + \underbrace{6}_{\text{Last}} = x^2 + 5x + 6 \]The two middle pieces, \( 3x \) and \( 2x \), are like terms, so they collapse into \( 5x \). Keep that picture in mind — in a moment we're going to walk it backwards.
FOIL is not a new rule. It's just "each term times each term" applied to the special case of two binomials. The same distributive law multiplies a trinomial by a binomial, or anything by anything — there are simply more products to keep track of. Never let the acronym make you think binomials are special.
The heart of it: factoring
Here is the secret of the whole lesson. Factoring is multiplication in reverse. You're handed an expanded polynomial like \( x^2 + 5x + 6 \) and asked to find the brackets it came from — \( (x + 2)(x + 3) \). Why bother? Because a product tells you things a sum never could: where the expression equals zero, how it simplifies, how to solve the equation built from it.
To see why the brackets multiply out the way they do, picture a rectangle. Make its width \( x + p \) and its height \( x + q \). Its area is the product \( (x + p)(x + q) \) — but we can also find that area by chopping the rectangle into four natural pieces and adding them up:
- one big \( x \times x \) square, contributing \( x^2 \);
- a strip \( x \) tall and \( p \) wide, and another \( x \) tall and \( q \) wide — together \( (p + q)x \);
- a little \( p \times q \) rectangle of unit squares in the corner, contributing \( pq \).
Add the four pieces and you have proved the identity at the centre of this entire topic:
\[ (x + p)(x + q) = x^2 + (p + q)x + pq \]Look hard at the right-hand side. The middle coefficient is the sum \( p + q \), and the constant is the product \( pq \). That single observation is the key to factoring every quadratic of this form.
Drag the two handles in the widget below to change \( p \) and \( q \). Watch the four coloured regions resize, and keep an eye on the running total: the middle coefficient is always \( p + q \) and the constant is always \( pq \). Try \( p = 2,\ q = 3 \) and confirm you get \( x^2 + 5x + 6 \).
Tip The whole rectangle picture is just area = length × width, read in two directions. Multiplying brackets builds the four pieces from the sides; factoring recovers the sides from the pieces. Same rectangle, opposite directions.
The factoring recipe for \( x^2 + bx + c \)
The area model hands you a recipe for free. To factor \( x^2 + bx + c \), you need two numbers \( p \) and \( q \) that
\[ p + q = b \qquad\text{and}\qquad p \cdot q = c. \]That's it — two numbers that multiply to \( c \) and add to \( b \). Once you have them, the factors are \( (x + p)(x + q) \).
How do you find them? Hunt through the factor pairs of \( c \) and check which pair adds to \( b \). Let's do the search out loud for \( x^2 + 5x + 6 \), where \( b = 5 \) and \( c = 6 \). The factor pairs of \( 6 \) are:
\[ 1 \times 6\ \ (\text{sum } 7), \qquad 2 \times 3\ \ (\text{sum } 5). \]The pair \( 2 \) and \( 3 \) multiplies to \( 6 \) and adds to \( 5 \) — a perfect match. So \( x^2 + 5x + 6 = (x + 2)(x + 3) \), exactly the brackets we started from. The search is short because you only ever test the factor pairs of the constant.
Watch the signs. If \( c \) is positive, \( p \) and \( q \) share a sign (both match the sign of \( b \)). If \( c \) is negative, they have opposite signs. So \( x^2 - 5x + 6 \) factors with two negatives, \( (x-2)(x-3) \), while \( x^2 + x - 6 \) factors with opposite signs, \( (x+3)(x-2) \).
Two shortcuts worth memorising
Always pull out a common factor first. Before anything else, check whether every term shares a factor — the greatest common factor, or GCF. Pulling it out shrinks the numbers and often reveals an easier problem underneath:
\[ 2x^2 + 10x + 12 = 2\,(x^2 + 5x + 6) = 2\,(x + 2)(x + 3). \]Difference of squares. When you see one square subtracted from another, you never need to search at all. The pattern is
\[ a^2 - b^2 = (a - b)(a + b). \]It falls straight out of the area model: take \( p = -b \) and \( q = +b \), so the sum \( p + q = 0 \) kills the middle term and the product \( pq = -b^2 \) is what's left. Whenever the middle term is missing and both ends are perfect squares with a minus sign between them, reach for this shortcut on sight.
- There's no common factor, so go straight to the recipe with \( b = 7 \) and \( c = 12 \): find two numbers that multiply to \( 12 \) and add to \( 7 \).
- List the factor pairs of \( 12 \) and their sums: \( 1 \times 12 \) (sum \( 13 \)), \( 2 \times 6 \) (sum \( 8 \)), \( 3 \times 4 \) (sum \( 7 \)). The last pair wins.
- So \( p = 3 \) and \( q = 4 \), giving \( x^2 + 7x + 12 = (x + 3)(x + 4) \).
- Check by expanding: \( (x+3)(x+4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12 \). ✓
The factorisation is \( \mathbf{(x + 3)(x + 4)} \), and the expansion check confirms it term by term.
- Notice there's no middle \( x \) term and both pieces are perfect squares: \( x^2 = (x)^2 \) and \( 9 = 3^2 \). That's the difference-of-squares signature.
- Apply \( a^2 - b^2 = (a - b)(a + b) \) with \( a = x \) and \( b = 3 \).
- So \( x^2 - 9 = (x - 3)(x + 3) \).
- Check: \( (x-3)(x+3) = x^2 + 3x - 3x - 9 = x^2 - 9 \). The middle terms cancel, exactly as the pattern promises. ✓
The factorisation is \( \mathbf{(x - 3)(x + 3)} \). Spotting the pattern turned a search into a single step.
Practice
Try each one yourself, then reveal the full solution.
1. Factor \( x^2 + 8x + 15 \).
There's no common factor, so use the recipe with \( b = 8 \) and \( c = 15 \): two numbers that multiply to \( 15 \) and add to \( 8 \).
The factor pairs of \( 15 \) are \( 1 \times 15 \) (sum \( 16 \)) and \( 3 \times 5 \) (sum \( 8 \)). The pair \( 3 \) and \( 5 \) fits.
So \( x^2 + 8x + 15 = \mathbf{(x + 3)(x + 5)} \). Check: \( (x+3)(x+5) = x^2 + 5x + 3x + 15 = x^2 + 8x + 15 \). ✓
2. Factor \( x^2 - 5x + 6 \).
Here \( b = -5 \) and \( c = 6 \). Since \( c \) is positive but \( b \) is negative, both numbers must be negative.
Test the negative factor pairs of \( 6 \): \( (-1)(-6) \) adds to \( -7 \); \( (-2)(-3) \) adds to \( -5 \). The pair \( -2 \) and \( -3 \) multiplies to \( 6 \) and adds to \( -5 \).
So \( x^2 - 5x + 6 = \mathbf{(x - 2)(x - 3)} \). Check: \( (x-2)(x-3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6 \). ✓
3. Factor \( x^2 - 16 \).
There's no middle term, and both pieces are perfect squares: \( x^2 = (x)^2 \) and \( 16 = 4^2 \). That's a difference of squares.
Apply \( a^2 - b^2 = (a - b)(a + b) \) with \( a = x \) and \( b = 4 \).
So \( x^2 - 16 = \mathbf{(x - 4)(x + 4)} \). Check: \( (x-4)(x+4) = x^2 + 4x - 4x - 16 = x^2 - 16 \). The middle terms cancel. ✓