Algebraic expansions

Two-term squares. These are the two most-used identities in the whole chapter:

where: $a$ and $b$ are any two algebraic terms (numbers, letters, or whole expressions), and the middle term $2ab$ carries the sign of the original bracket.

Two-term cubes. The cubic versions extend the same idea one power further:

where: the coefficients $1,3,3,1$ are fixed, the power of $a$ falls from $3$ to $0$ while the power of $b$ rises from $0$ to $3$, and in the minus version the signs simply alternate $+,-,+,-$.

The linear-factor product. This one governs any product of two binomials that share the same first term $x$:

where: $a$ and $b$ are the two constants (with their own signs), the coefficient of $x$ is their sum, and the constant term is their product. This sum-and-product pattern is exactly what you will later reverse to factorise quadratics by splitting the middle term.

The three-term square. With three terms inside the bracket instead of two, every pair now contributes a cross term:

where: $a$, $b$, $c$ are the three terms with their signs attached, and there are exactly three cross terms — one for each of the three possible pairs — so none can be skipped. A useful habit: $a$, $b$, or $c$ in any identity can stand for a whole expression, not just a single letter. In $(2x+1-3y)^2$ you can read this as $(a+b+c)^2$ with $a=2x$, $b=1$, $c=-3y$, and the identity applies exactly as written provided every substitution — including its sign — is carried through carefully. It also helps to see the five identities as one family rather than five isolated formulae. The squared identities and the cubed identities describe the same underlying operation — multiplying a bracket by itself — carried one power further; the coefficients $1,2,1$ in a square and $1,3,3,1$ in a cube are simply how many times each combination of $a$'s and $b$'s appears when every term is multiplied against every term. The linear-factor product $(x+a)(x+b)$ is the odd one out structurally, since it multiplies two different binomials rather than squaring one, which is exactly why its middle term is a plain sum $(a+b)x$ rather than a doubled product $2ab$. Keeping that structural distinction clear is what prevents the traps described later in this article. Finally, the three-term square is not a new idea at all — it is the two-term square applied to a bracket that itself contains three pieces, so every one of the three possible pairs among $a$, $b$, $c$ contributes its own cross term, in the same way a single pair contributed one cross term in the two-term case.

Expand $(2x-3y)^2$, then use the result to expand $(2x-3y+1)^2$.

Step 1. Match the first bracket to a known pattern: $(2x-3y)^2$ has the form $(a-b)^2$ with $a=2x$ and $b=3y$.

Step 2. Apply the identity $(a-b)^2 = a^2-2ab+b^2$ and substitute:

Step 3. Simplify each term: $(2x)^2 = 4x^2$, $2(2x)(3y) = 12xy$, $(3y)^2 = 9y^2$, so

Step 4. Check with $x=1,y=1$: left side $(2-3)^2=1$; right side $4-12+9=1$. They match.

Step 5. Now extend to the three-term bracket $(2x-3y+1)^2$ by matching it to $(a+b+c)^2$ with $a=2x$, $b=-3y$, $c=1$:

Step 6. Simplify each of the six terms and collect:

So $(2x-3y+1)^2 = \mathbf{4x^2+9y^2+1-12xy+4x-6y}$.

Competitive stretch

Notice the first three terms, $4x^2-12xy+9y^2$, are exactly the two-term expansion from Step 3 — reusing an already-expanded pair inside a bigger bracket, instead of restarting from scratch, is a genuine time-saver on longer exam problems. The same shortcut works in reverse: whenever you spot a three-term square, try grouping two of the terms first and checking whether their combination matches a two-term identity you already know, before falling back on the full six-term $(a+b+c)^2$ expansion.

• Writing $(a+b)^2 = a^2+b^2$, dropping the middle term $2ab$ entirely — the single most common error at this level.
• Losing a minus sign in $(a-b)^3 = a^3-3a^2b+3ab^2-b^3$, especially on the $3a^2b$ or $b^3$ term.
• In $(x+a)(x+b)$, computing the sum for the coefficient correctly but the product wrongly (or vice versa) — work out sum and product separately and check both.
• In $(a+b+c)^2$, forgetting one of the three cross terms — there are three pairs, not two, and each contributes $2\times$ its product.
• Confusing $(x+a)(x+b)$ with a squared bracket and inserting a $2ab$-style middle term that does not belong there — one pattern squares a single bracket, the other multiplies two distinct binomials.
• Forgetting that the sum-and-product pattern in $(x+a)(x+b)=x^2+(a+b)x+ab$ is exactly the pattern reused, in reverse, to factorise a quadratic like $x^2+7x+12$ by finding two numbers with sum $7$ and product $12$ — so a shaky grip on the expansion direction shows up later as a shaky grip on factorisation too.

Ask the tutor: Stuck telling $(a-b)^2$ apart from $(x+a)(x+b)$ at a glance, or want more mixed practice on three-term squares? Ask your AI tutor to run you through a quick set until the pattern recognition is automatic.