Expanding Single Brackets

You can visualise this using an area model. Imagine a rectangle with height 3 and width \( (x + 4) \). The total area is the sum of the two parts: \( 3 \times x = 3x \) and \( 3 \times 4 = 12 \).

Alternatively, think of repeated addition: \( 3(x + 4) \) means \( (x + 4) + (x + 4) + (x + 4) = 3x + 12 \).

Expanding correctly: \( 2(x \; – \; 5) = 2x \; – \; 10 \), because the 2 multiplies both terms inside the bracket. Writing \( 2x \; – \; 5 \) means only the \( x \) was multiplied by 2 — the 5 was left unchanged.

A quick substitution confirms it: let \( x = 10 \). Then \( 2(10 \; – \; 5) = 2 \times 5 = 10 \), but \( 2 \times 10 \; – \; 5 = 20 \; – \; 5 = 15 \). Since \( 10 \neq 15 \), the two expressions are not equivalent. The multiplier must apply to every term inside the bracket.

Multiply each term inside the bracket by \( -2 \): \( -2 \times x = -2x \), and \( -2 \times (-3) = +6 \) because a negative times a negative gives a positive. So the result is \( -2x + 6 \).

Many students expect the answer to be \( -2x \; – \; 6 \) because they see the minus sign and the 3 and assume the result is “minus 6.” But the original expression has minus 3 inside the bracket, and \( -2 \times -3 = +6 \). Testing with \( x = 3 \): \( -2(3 \; – \; 3) = -2 \times 0 = 0 \), and \( -2(3) + 6 = -6 + 6 = 0 \). ✓

\( 3(x + 5) = 3x + 15 \) — this is a linear expression (the highest power of \( x \) is 1). But \( x(x + 5) = x^2 + 5x \) — this is a quadratic expression because multiplying \( x \times x \) creates an \( x^2 \) term.

When the term outside the bracket is a number, the degree of the expression stays the same. But when the term outside is a variable, it increases the power of every term by one. This is why expanding \( x(x + 5) \) produces something fundamentally different from expanding \( 3(x + 5) \).

Example: \( 6(x + 1) = 6x + 6 \)

Another: \( 3(2x + 7) = 6x + 21 \)

Creative: \( -2(-3x + 4) = 6x \; – \; 8 \) — using two negatives. Or a multi-variable example like \( 2(3x + y) = 6x + 2y \), which still satisfies the condition.

Trap: \( 2(x + 6) = 2x + 12 \) — a student might pick this because the number 6 appears in the bracket. But the coefficient of \( x \) here is 2, not 6. The 6 contributes to the constant term, not the \( x \) term.

Example: \( 5(2x + 3) \)

Another: \( 10(x + 1.5) \)

Creative: \( -5(-2x \; – \; 3) \) — using a negative multiplier with negative terms inside. Or \( 2.5(4x + 6) \) — using a decimal multiplier that students rarely consider.

Trap: \( 10(x + 15) = 10x + 150 \) — a student might see the 10 and 15 in the target expression and put them straight into the bracket. But the multiplier must act on the constant too: \( 10 \times 15 = 150 \), not 15.

Example: \( 2(3x + 6) \) and \( 6(x + 2) \) — both give \( 6x + 12 \).

Another: \( 4(x + 5) \) and \( 2(2x + 10) \) — both give \( 4x + 20 \).

Creative: \( -3(-2x \; – \; 4) \) and \( 6(x + 2) \) — both give \( 6x + 12 \). Or \( 0.5(8x + 2) \) and \( 2(2x + 0.5) \) — both give \( 4x + 1 \). Using negatives or decimals makes this feel very different even though the principle is the same.

Trap: \( 3(x + 2) \) and \( 2(x + 3) \) — they look like rearrangements of each other, but \( 3(x + 2) = 3x + 6 \) while \( 2(x + 3) = 2x + 6 \). The constant terms happen to be the same, but the \( x \) terms differ. Swapping the multiplier and one of the terms inside doesn’t preserve the expansion.

Example: \( x(x + 3) = x^2 + 3x \)

Another: \( 2x(x \; – \; 1) = 2x^2 \; – \; 2x \)

Creative: \( x(x + 0) = x^2 \) — technically a bracket expansion. Or expand to multi-variables: \( x(x + y) = x^2 + xy \), which is quadratic in terms of the variables’ degree.

Trap: \( 3x(x^2 \; – \; 5) = 3x^3 \; – \; 15x \) — this does contain an \( x^2 \) inside the bracket, but expanding it produces an \( x^3 \) term, making it a cubic expression, not a quadratic.

It depends on what is inside the bracket. For a standard binomial like \( 3(x + 2) \), it is true: one term (the product) becomes two (\( 3x + 6 \)).

However, if the bracket contains a single term, like \( 3(4x) \), it expands to \( 12x \), which is still just one term. Or if the multiplier is zero, \( 0(x + 4) = 0 \), which is one term. It only produces two terms if the bracket contains two unlike terms.

This is the distributive property: \( a(x + b) = a \times x + a \times b = ax + ab \). It holds for all values of \( a \) and \( b \), not just whole numbers — the question restricts to whole numbers, but the rule itself is universal.

Students should recognise that the multiplier always acts on every term inside the bracket. This is the fundamental principle behind all bracket expansion.

Expanding a bracket does not change the value of an expression — it changes the form. \( 3(x + 2) \) and \( 3x + 6 \) are equal for every value of \( x \). Expanding is a rearrangement, not an operation that increases size.

This addresses a subtle misconception: some students feel that “spreading out” an expression somehow makes it bigger, perhaps influenced by the word “expanding.” In algebra, expanding means removing brackets — it’s a structural change, not a numerical one.

The constant term after expanding is \( ab \). This is larger than \( b \) when \( a > 1 \) (and \( b > 0 \)): for example, \( 3(x + 4) = 3x + 12 \), and 12 > 4. But if \( a = 1 \), we get \( 1(x + 4) = x + 4 \), and the constant stays the same.

If \( a \) is a fraction between 0 and 1, the constant actually shrinks: \( \frac{1}{2}(x + 4) = \frac{1}{2}x + 2 \), and 2 < 4. And if \( a \) is negative, the constant changes sign entirely: \( -3(x + 4) = -3x \; – \; 12 \), and \( -12 < 4 \).

The student has only multiplied the first term in the bracket by 5 and left the second term unchanged. This is the “partial distribution” misconception — the most common error in expanding brackets.

The 5 must multiply every term inside the bracket: \( 5 \times x = 5x \) and \( 5 \times 3 = 15 \). The correct answer is \( 5x + 15 \). A quick check: if \( x = 1 \), then \( 5(1 + 3) = 5 \times 4 = 20 \), but \( 5(1) + 3 = 8 \). Clearly wrong.

The answer is correct, but the reasoning is dangerously wrong. The student’s rule — “the sign inside the bracket is always what you keep” — only works when the multiplier is positive.

To prove this fails, ask the student to expand \( -4(x \; – \; 3) \). Using their rule, they would write \( -4x \; – \; 12 \) (keeping the minus). But the correct expansion is \( -4x + 12 \), because \( -4 \times -3 = +12 \). Getting the right answer with wrong reasoning gives students false confidence that breaks down in harder problems.

The student treats the minus sign as “make everything negative” rather than as multiplication. A helpful strategy is to use the “Invisible One”: rewrite the expression as \( -1(x \; – \; 7) \). Now it looks like a standard multiplication.

Multiplying by \( -1 \): \( -1 \times x = -x \) and \( -1 \times (-7) = +7 \). The correct answer is \( -x + 7 \). The term inside was already \( -7 \), and multiplying a negative by a negative gives a positive.

The student has combined the brackets before expanding, treating the two separate expressions as if they could be merged into one bracket. You cannot add the multipliers (2 + 3) and separately add the bracket contents — this violates how the distributive property works.

The correct approach is to expand each bracket separately first: \( 2(x + 3) = 2x + 6 \) and \( 3(x + 1) = 3x + 3 \). Then collect like terms: \( 2x + 6 + 3x + 3 = 5x + 9 \). The student’s answer of \( 10x + 20 \) is very different from the correct answer of \( 5x + 9 \). A quick check with \( x = 0 \) confirms: \( 2(3) + 3(1) = 6 + 3 = 9 \), not 20.