Expanding Brackets

Brackets are a compact way to write a multiplication. Expanding means multiplying it out in full, turning a tidy but closed expression into an open one you can collect and work with.

MYP 4Standard MathsAlgebraCriteria A · C · D~10 min read

Single brackets

The distributive law says the term outside a bracket multiplies every term inside it. So \(a(b + c) = ab + ac\). Nothing inside is left out.

Expand
To remove brackets by multiplying the outside term into each term inside, writing the multiplication out in full.
Worked example

Expand \(3(2x + 5)\).

1
Multiply the outside \(3\) by the first inside term: \(3 \times 2x = 6x\).
2
Multiply the \(3\) by the second inside term: \(3 \times 5 = 15\).
3
Write both results with the sign between them.
\(6x + 15\)

Minus signs outside

A minus sign outside the bracket multiplies in as well, and it flips the sign of every term inside. This is where most errors appear.

Flip every sign, not just the first

In \(-2(3x - 4)\), the \(-2\) hits both terms: \(-2 \times 3x = -6x\) and \(-2 \times -4 = +8\). The result is \(-6x + 8\), not \(-6x - 8\).

Double brackets and FOIL

To expand two brackets multiplied together, every term in the first must meet every term in the second. FOIL is a memory aid for the four products: First, Outer, Inner, Last.

Worked example

Expand \((x + 3)(x + 4)\).

1
First: \(x \times x = x^2\). Outer: \(x \times 4 = 4x\).
2
Inner: \(3 \times x = 3x\). Last: \(3 \times 4 = 12\).
3
Collect the two like middle terms: \(4x + 3x = 7x\).
\(x^2 + 7x + 12\)

FOIL is just "each times each"

The four letters only remind you to make all four products. If a bracket has more terms, drop the mnemonic and keep the rule: multiply every term by every term.

Expand then simplify

Once brackets are gone, collect like terms to finish. In questions with two separate brackets added together, expand each first, then combine.

Worked example

Expand and simplify \(2(x + 3) + 3(x - 1)\).

1
Expand the first bracket: \(2(x + 3) = 2x + 6\).
2
Expand the second bracket: \(3(x - 1) = 3x - 3\).
3
Collect like terms: \(2x + 3x = 5x\) and \(6 - 3 = 3\).
\(5x + 3\)

Check yourself

1. Expand \(4(3x - 2)\). +

\(4 \times 3x = 12x\) and \(4 \times -2 = -8\). Answer: \(12x - 8\).

2. Expand \((x + 5)(x + 2)\). +

First \(x^2\), Outer \(2x\), Inner \(5x\), Last \(10\). The middle terms give \(2x + 5x = 7x\). Answer: \(x^2 + 7x + 10\).

3. Expand and simplify \(3(x + 2) - 2(x - 4)\). +

\(3(x+2) = 3x + 6\) and \(-2(x - 4) = -2x + 8\). Collecting: \(3x - 2x = x\) and \(6 + 8 = 14\). Answer: \(x + 14\).


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