Why does 5c(2d + 1) expand to 10cd + 5c, not just 10cd?

The outside factor 5c must multiply every term inside the bracket. There are two terms inside: 2d and 1. So 5c × 2d = 10cd and 5c × 1 = 5c, giving 10cd + 5c. Writing only 10cd means the second multiplication was never carried out.

What is the first-term-only expanding brackets mistake?

The mistake is multiplying the outside of a bracket by the first term inside, then either copying the remaining terms unchanged or dropping them altogether. For example, writing 4(2d − 5) as 8d − 5 (copying −5) or as 8d (dropping it). The correct expansion is 8d − 20, because 4 × 5 = 20.

How do you expand and simplify 5(3x + 4) − 2(x − 1)?

Expand each bracket in turn. 5(3x + 4) = 15x + 20. −2(x − 1) = −2x + 2, because −2 × −1 = +2. Collecting: 15x + 20 − 2x + 2 = 13x + 22. The trap is treating −2 × −1 as −2, giving 13x + 18.

How does the expanding brackets mistake affect solving equations?

In 4(2d − 5) = 28, the first step is to expand the bracket correctly: 8d − 20 = 28, so 8d = 48 and d = 6. If you copy −5 unchanged to get 8d − 5 = 28, you get 8d = 33, which gives a non-integer answer and loses marks.

GCSE Maths Foundation · AQA · Algebraic manipulation

Expanding brackets: why 5c(2d + 1) is 10cd + 5c, not just 10cd

Asked to multiply out $5c(2d + 1)$ , a common Foundation error is to write only $10cd$ — multiplying the outside by the first term inside and then stopping. The second term, $+1$ , never gets multiplied, so the $+5c$ is lost.

The thirty-second fix: the outside of a bracket multiplies every term inside, not just the first. Count the terms in the bracket, then write one product for each — you should finish with exactly as many products as there were terms.

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How to spot it in your own work

You wrote $10cd$ for $5c(2d + 1)$ , dropping the $+5c$ .
You wrote $8d - 5$ for $4(2d - 5)$ , copying the $-5$ unchanged instead of multiplying it.
You wrote $13x + 18$ for $5(3x + 4) - 2(x - 1)$ by treating $-2 \times -1$ as $-2$ .
You wrote $3w + 10$ for $\dfrac{6w + 10}{2}$ , dividing only the first term in the numerator.

An exam question that triggers it

Here is the canonical AQA Foundation trigger, the shape of JUN22 Paper 2 Q17b:

Multiply out $5c(2d + 1)$ .

The misconception answer is $10cd$ , found by multiplying the outside by the first term $2d$ and then stopping. But the bracket holds two terms, and the $5c$ must reach both.

The correct answer is $10cd + 5c$ : $5c \times 2d = 10cd$ and $5c \times 1 = 5c$ .

Why students fall for this

The distributive law asks a student to do the same multiplication several times — once per term inside the bracket. Under exam pressure the first product feels like the whole job, so the pen stops after $5c \times 2d = 10cd$ . The remaining term is small (a bare $+1$ ) and easy to overlook, which is exactly why AQA put it there.

The same partial move shows up with a subtraction inside the bracket. In $4(2d - 5)$ , students multiply $4 \times 2d = 8d$ and then copy the $-5$ across unchanged, writing $8d - 5$ . The fix is the same: the multiplier reaches every term, so $4 \times 5 = 20$ and the answer is $8d - 20$ .

The fix: One arrow per term — as many products as there are terms inside

Before expanding, count the terms inside the bracket. Draw one arrow from the outside to each term, and write a separate product for every arrow. For $5c(2d + 1)$ : two terms inside means two products — $5c \times 2d = 10cd$ and $5c \times 1 = 5c$ , so $5c(2d + 1) = 10cd + 5c$ .

Carry the sign of each inside term into its product. For $a(b - c)$ : two terms → $ab - ac$ . If you finish with fewer products than there were terms, you have missed a multiplication.

Worked example

Multiply out $5c(2d + 1)$ .

Count the terms inside the bracket. There are two: $2d$ and $1$ . So expect two products.
Multiply the outside by the first term. $5c \times 2d = 10cd$ .
Multiply the outside by the second term. $5c \times 1 = 5c$ .
Write both products.
$5c(2d + 1) = 10cd + 5c$

The trap answer $10cd$ has only one product where there should be two — the second multiplication was never carried out.

Find out if this is costing you marks

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Common questions

Why does $5c(2d + 1)$ expand to $10cd + 5c$ , not just $10cd$ ?: The outside factor $5c$ must multiply every term inside. There are two terms: $2d$ and $1$ . So $5c \times 2d = 10cd$ and $5c \times 1 = 5c$ , giving $10cd + 5c$ . Writing only $10cd$ means the second multiplication was never carried out.
Why is $4(2d - 5)$ equal to $8d - 20$ , not $8d - 5$ ?: The $4$ multiplies both terms inside, including the $-5$ . So $4 \times 2d = 8d$ and $4 \times 5 = 20$ , giving $8d - 20$ . Copying the $-5$ across unchanged (to get $8d - 5$ ) skips the second multiplication.
How do you expand and simplify $5(3x + 4) - 2(x - 1)$ ?: Expand each bracket in turn. $5(3x + 4) = 15x + 20$ . $-2(x - 1) = -2x + 2$ , because $-2 \times -1 = +2$ . Collecting like terms: $15x + 20 - 2x + 2 = 13x + 22$ . The trap is treating $-2 \times -1$ as $-2$ , which gives the wrong answer $13x + 18$ .
Does dividing a bracket-like expression follow the same rule?: Yes. A denominator divides into every term in the numerator. In $\dfrac{6w + 10}{2}$ : $\dfrac{6w}{2} + \dfrac{10}{2} = 3w + 5$ . Dividing only the first term — leaving the $10$ unchanged to get $3w + 10$ — is the same first-term-only error.

Related misconceptions

Combining unlike terms: why 9x + y − 6x + y is 3x + 2y, not 5xyThe companion simplifying error — treating separate terms carelessly, where here you must treat every term separately.
Coefficient vs index: why y × y × y is y³, not 3yAnother algebra-structure error: which operation makes a coefficient and which makes a power.
Function machines: why 12 → [−4] → [×5] is 40, not −8Another spot where reading the structure of an expression matters as much as the arithmetic.

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