C2.2; E2.2 - Expanding and taking out common factors

The distributive law

Can $3(x+4)$ and $3x+12$ really describe the same quantity for every $x$? Three bags, each containing $x$ counters and four extra counters, contain three lots of $x$ and twelve extras. You can count by bags or by types of counter.

This is the distributive law: a multiplier outside brackets multiplies every term inside. In symbols,

It follows from how multiplication distributes over addition. It works for real numbers, including zero, negative numbers and fractions; the bags are just a starting picture for positive whole numbers.

A split rectangle shows the same reasoning. For positive $x$, its total area can be calculated using the whole width or by adding the two smaller areas.

[DIAGRAM: asset_name: One rectangle counted two ways; asset_slug: distributive_area; caption: A rectangle of height three and width x plus four has area three x plus twelve.; description: A rectangle of height three is split into widths x and four. The left pale blue region is labelled three x and the right pale sage region twelve. Height three, widths x and four, and total area three times bracket x plus four equals three x plus twelve are labelled. The picture represents positive x; the algebraic distributive law also holds for other real values.]

Now multiply the outside factor by $x$. The same two products are required, but one contains two factors of $x$.

The factor can be a number, a letter or a product of both. When multiplying letter powers, count the factors: $x^2\times x=x\times x\times x=x^3$. When different letters occur, retain them both: $x\times y=xy$.

Use an intermediate line showing every product whenever there is a risk of missing one.

Expanding with negative terms

Subtraction inside a bracket does not change the rule. Read $b-c$ as $b+(-c)$, so $a(b-c)=ab-ac$. A negative outside factor changes the signs of both products, using the signed multiplication rules you already know.

For example, $-2(u-3)=(-2)u+(-2)(-3)=-2u+6$. The positive $6$ comes from multiplying two negative numbers; it is not a separate “bracket sign rule”.

Change only the sign of the outside factor. The whole result must now be the negative of the previous result.

A minus sign immediately before brackets means multiplication by $-1$. Therefore $-(a+b)=-a-b$ and $-(a-b)=-a+b$. For instance, $5-(2x-3)=5-2x+3=8-2x$. Both terms in the subtracted bracket change sign.

The number of different letters does not change a single-bracket expansion. Each product simply retains all its letter factors.

Taking out the highest common factor

You can run expansion backwards. Since $3x+12=3(x+4)$, the sum can be written as a product. This is factorisation. In a common-factor question, “factorise” means take out the highest common factor so that no further common factor remains inside.

For $12x^2$ and $18xy$, the coefficient HCF is $6$. Both contain an $x$, but only the second contains $y$, so the complete HCF is $6x$. Build the remaining bracket by dividing each term by the factor you have removed.

This division is a way to find the factors symbolically; the finished product identity is still valid at $x=0$. You are not imposing a new condition that $x\ne0$.

Now multiply both original terms by $y$. That new shared factor must also appear outside.

For a subtraction, keep the subtraction in the bracket. For example, $8t^2-12t=4t(2t-3)$. If a term equals the factor removed, its remainder is $1$: $5u+5=5(u+1)$, not $5u$.

Expanding your answer checks every original term and sign. This check establishes an identity for all values, including zero.

Quick Recap

• Expanding multiplies every term in a bracket by the complete outside factor.
• Negative signs follow the ordinary rules for multiplying signed numbers.
• Factorising reverses expansion; include all shared numerical and letter factors.
• Expand a factorised answer to check its terms and signs.

The opening expressions $3(x+4)$ and $3x+12$ agree for every real $x$: they are two forms of the same distributive-law calculation.