Multiplying Out Brackets

In algebra, "expanding" or "multiplying out" means removing the brackets by multiplying every term inside the bracket by the term outside (or by the other bracket).

1. Single Brackets

The Rule: Multiply the term on the outside by everything inside. $$ a(b + c) = ab + ac $$
Example: \(3(2x + 5)\)
1. Multiply \(3\) by \(2x\) → \(6x\)
2. Multiply \(3\) by \(+5\) → \(+15\)
Answer: \(6x + 15\)
Watch Out for Negatives!
If the outside number is negative, it changes the signs of the terms inside.
\(-2(x - 4) = -2x + 8\) (Because \(-2 \times -4 = +8\)).

2. Double Brackets (FOIL)

To expand \((x + a)(x + b)\), calculate 4 products. Many people use FOIL:
Example: \((x + 3)(x - 2)\)
1. First: \(x \times x = x^2\)
2. Outside: \(x \times -2 = -2x\)
3. Inside: \(3 \times x = +3x\)
4. Last: \(3 \times -2 = -6\)

Combine: \(x^2 - 2x + 3x - 6\)
Simplify: \(x^2 + x - 6\)

3. Squared Brackets

Common Mistake:
\((x + 3)^2\) is NOT \(x^2 + 9\).
You must write the bracket twice and use FOIL!
Example: \((x + 5)^2\)
Rewrite it: \((x + 5)(x + 5)\)
Expand: \(x^2 + 5x + 5x + 25\)
Simplify: \(x^2 + 10x + 25\)

4. Triple Brackets & Cubed Brackets

To solve \((x+1)(x+2)(x+3)\), take it step-by-step.

Strategy: 1. Ignore the 3rd bracket for a moment.
2. Expand the first two brackets and simplify.
3. Multiply that result by the 3rd bracket.
Example: \((x+2)(x+3)(x+4)\)
Step 1: Expand \((x+2)(x+3)\)
Result: \(x^2 + 5x + 6\) Step 2: Multiply by \((x+4)\)
\((x^2 + 5x + 6)(x + 4)\)

Multiply everything in the left bracket by \(x\): \(x^3 + 5x^2 + 6x\)
Multiply everything in the left bracket by \(4\): \(+4x^2 + 20x + 24\)

Step 3: Add them up
\(x^3 + 9x^2 + 26x + 24\)

Practice Questions & Solutions

Attempt these on paper first!

Q1: Expand \(4(2x - 3)\)
\(4 \times 2x = 8x\)
\(4 \times -3 = -12\)
Answer: \(8x - 12\)
Q2: Expand \(x(x^2 + 5)\)
\(x \times x^2 = x^3\) (Add powers)
\(x \times 5 = 5x\)
Answer: \(x^3 + 5x\)
Q3: Expand and Simplify \(-2(3x - 4) - 3(x + 1)\)
First bracket: \(-6x + 8\)
Second bracket: \(-3x - 3\)
Combine: \(-6x - 3x + 8 - 3\)
Answer: \(-9x + 5\)
Q4: Expand \((x + 4)(x + 7)\)
\(x^2 + 7x + 4x + 28\)
Simplify middle terms: \(7x + 4x = 11x\)
Answer: \(x^2 + 11x + 28\)
Q5: Expand \((2x - 3)(x + 5)\)
First: \(2x \times x = 2x^2\)
Outside: \(2x \times 5 = 10x\)
Inside: \(-3 \times x = -3x\)
Last: \(-3 \times 5 = -15\)
Simplify: \(10x - 3x = 7x\)
Answer: \(2x^2 + 7x - 15\)
Q6: Expand \((x - 6)^2\)
Write it twice: \((x - 6)(x - 6)\)
Expand: \(x^2 - 6x - 6x + 36\)
Note: \(-6 \times -6 = +36\)
Answer: \(x^2 - 12x + 36\)
Q7: Expand \((3x + 2)^2\)
Write it twice: \((3x + 2)(3x + 2)\)
First: \(3x \times 3x = 9x^2\)
Outside: \(3x \times 2 = 6x\)
Inside: \(2 \times 3x = 6x\)
Last: \(2 \times 2 = 4\)
Answer: \(9x^2 + 12x + 4\)
Q8: Expand \((x + 1)(x + 2)(x + 3)\)
1. \((x+1)(x+2) = x^2 + 3x + 2\)
2. Multiply by \((x+3)\):
\((x^2 + 3x + 2)(x + 3)\)
\(x(x^2 + 3x + 2) = x^3 + 3x^2 + 2x\)
\(3(x^2 + 3x + 2) = + 3x^2 + 9x + 6\)
Add columns:
Answer: \(x^3 + 6x^2 + 11x + 6\)
Q9: Expand \((x - 2)^3\)
1. Write as \((x-2)(x-2)(x-2)\)
2. First two: \(x^2 - 4x + 4\)
3. Multiply by last \((x-2)\):
\((x^2 - 4x + 4)(x - 2)\)
Multiply by \(x\): \(x^3 - 4x^2 + 4x\)
Multiply by \(-2\): \(-2x^2 + 8x - 8\)
Answer: \(x^3 - 6x^2 + 12x - 8\)
Q10: Challenge - Expand \((x + 1)(x - 1)(x^2 + 1)\)
Spot the shortcut! \((x+1)(x-1)\) is the Difference of Two Squares.
Result: \(x^2 - 1\)

Now multiply that by the last bracket:
\((x^2 - 1)(x^2 + 1)\)
This is another Difference of Two Squares!
First: \(x^2 \times x^2 = x^4\)
Outside/Inside cancel out (\(+x^2 - x^2\)).
Last: \(-1 \times 1 = -1\)
Answer: \(x^4 - 1\)