FA4, HA4 - Collecting terms and expanding brackets | Eduqas GCSE Mathematics

Different expressions can describe exactly the same calculation. Learn to collect like terms, multiply a single term over brackets, and combine these methods while keeping every sign. These are the collecting and single-bracket skills in FA4 and HA4.

Collecting like terms

If one person writes $3(x+4)$ for a length and another writes $3x+12$ , can both be right? We will connect the two forms, starting with the terms we are allowed to combine.

Three bags containing $x$ counters each, together with two more identical bags, contain $3x+2x=5x$ counters. This is addition of five copies of the same quantity. It is different from $3x+2$ , where the extra two counters are loose: there are not five bags.

Like terms

Like terms have exactly the same variable part, including every power. Their coefficients can differ. Thus $3x$ , $-x$ and $\frac12x$ are like terms; $x$ , $x^2$ and $y$ are not like one another.

Collect like terms by adding their signed coefficients. Constants collect with constants. In $7x-2x$ , the second coefficient is $-2$ , so $7+(-2)=5$ . A missing coefficient is $1$ , and $-x$ has coefficient $-1$ .

Worked example 1: two groups of terms

Question. Simplify $7a+3b-2a+5b$ .

Plan. Group matching variable parts without losing their signs.

\begin{aligned} 7a+3b-2a+5b&=(7-2)a+(3+5)b &&\text{group like terms}\\ &=5a+8b &&\text{add coefficients} \end{aligned}

Answer. $5a+8b$ .

Check. At $a=2$ , $b=3$ , the original is $14+9-4+15=34$ , and the answer gives $10+24=34$ .

The rule is unchanged when the coefficients are fractions. Keep them exact and use a common denominator.

Worked example 2: fractional coefficients

Question. Simplify $\frac34a+3b-\frac14a+5b$ .

Plan. Collect the same two groups, adding fractional coefficients in the $a$ -group.

\begin{aligned} \frac34a+3b-\frac14a+5b &=\left(\frac34-\frac14\right)a+(3+5)b\\ &=\frac12a+8b \end{aligned}

Answer. $\frac12a+8b$ .

Check. At $a=4$ , $b=1$ , the original gives $3+3-1+5=10$ . The answer gives $2+8=10$ .

Powers remain part of the label you match. For instance, $4x^2+2x-x^2=3x^2+2x$ ; the $x^2$ -terms combine, but the $x$ -term stays separate.

Watch out

$3a+4b=7ab$ is wrong: addition of unlike terms does not create a product. At $a=b=1$ , both happen to give $7$ , so that single numerical check would miss the error. At $a=2$ , $b=1$ , the values are $10$ and $14$ .

Use a fresh pair of fractional coefficients.

Single brackets

Multiplying a bracket means multiplying every term inside it.

A rectangle with height $3$ and width $x+4$ has area $3(x+4)$ . Split its width into $x$ and $4$ : the two areas are $3x$ and $12$ . Adding those areas gives $3x+12$ , the same total.

[DIAGRAM: asset_name: Splitting a product into two areas; asset_slug: distributing-area; caption: Splitting the width shows why both terms are multiplied by 3.; description: A rectangle of height 3 and total width x plus 4 is split into a blue region of width x and area 3x and a sage region of width 4 and area 12. Length x is positive in this area model.]

Expanding a bracket

To expand a bracket, distribute the outside multiplier to every term inside. For any numbers for which the expressions are defined,

k(a+b)=ka+kb

The same distributive rule works with subtraction and negative numbers.

The rectangle explains the rule for positive lengths. The algebraic rule also works when a letter or multiplier is negative; negative numbers do not need to represent physical lengths.

Worked example 3: a positive multiplier

Question. Expand $4(2x-3)$ .

Plan. Multiply both the $2x$ -term and the $-3$ -term by $4$ .

\begin{aligned} 4(2x-3)&=4\times2x+4\times(-3)\\ &=8x-12 \end{aligned}

Answer. $8x-12$ .

Check. At $x=5$ , $4(10-3)=28$ , and $8(5)-12=28$ .

Now change only the outside sign. A negative times a negative is positive.

Worked example 4: a negative multiplier

Question. Expand $-4(2x-3)$ .

Plan. Include the minus sign in the multiplier for both terms.

\begin{aligned} -4(2x-3)&=(-4)\times2x+(-4)\times(-3)\\ &=-8x+12 \end{aligned}

Answer. $-8x+12$ .

Check. At $x=5$ , the original is $-4(7)=-28$ . The expanded form gives $-40+12=-28$ .

A lone minus sign before brackets means multiplication by $-1$ : $-(x-6)=-x+6$ .

Watch out

$-4(2x-3)=-8x-12$ misses the second sign change. Write $(-4)\times(-3)=12$ before collecting anything.

Apply a negative multiplier to a new bracket.

Combined expressions

When an expression contains brackets and several terms, expand first and collect afterwards.

Keep each term's sign attached while expanding. In particular, the expression $A-3(B)$ means $A+(-3)(B)$ : the multiplier of the whole bracket is $-3$ .

Worked example 5: two positive multipliers

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

Plan. Expand each bracket, then collect the $x$ -terms and constants separately.

\begin{aligned} 2(3x+4)+3(x-2)&=6x+8+3x-6 &&\text{four products}\\ &=9x+2 &&\text{collect like terms} \end{aligned}

Answer. $9x+2$ .

Check. At $x=1$ , the original gives $2(7)+3(-1)=11$ . The answer gives $9+2=11$ .

Change the join between brackets from addition to subtraction. This changes the multiplier of the second bracket, including its constant.

Worked example 6: subtracting a bracket

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

Plan. Treat the second multiplier as $-3$ , then collect.

\begin{aligned} 2(3x+4)-3(x-2)&=6x+8-3x+6 &&\text{distribute signed multipliers}\\ &=3x+14 &&\text{collect like terms} \end{aligned}

Answer. $3x+14$ .

Check. Use $x=1$ , so the second bracket is negative. The original gives $2(7)-3(-1)=17$ . The answer gives $3+14=17$ .

A numerical check can catch an error but cannot establish equality for every value. Expansion and collection explain why the forms agree for every allowed $x$ . Avoid choosing a check value that makes the term you are checking disappear.

Exam tip

For “Expand and simplify”, finish both jobs. A correct line of expanded terms makes your method visible; your final expression must also collect the like terms.

The opening expressions $3(x+4)$ and $3x+12$ agree because multiplying the bracket produces exactly those two terms.

Quick Recap

Collect terms only when their variable parts and powers match.
Add signed coefficients; fraction coefficients stay exact.
Multiply every term in a bracket by the signed outside multiplier.
Expand first, collect second, then check at a useful numerical value.

Both expressions in the opening question are correct: $3(x+4)=3x+12$ .

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FA4, HA4 - Collecting terms and expanding brackets

Collecting like terms

Single brackets

Combined expressions

Quick Recap

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