Common factor extraction

Look at $12x^2y + 18xy^2$ and it seems like a finished expression — two terms, added together, nothing more to do. But every term here is secretly built from the same block, $6xy$ , multiplied by something else. Spot that shared block, pull it to the front, and the expression collapses to $6xy(2x+3y)$ — a product instead of a sum. That single move, pulling out a common factor, is the doorway into the whole Factorisation chapter. It reverses multiplication: instead of expanding brackets you are handed an expanded expression and asked to rebuild the brackets. Every other factorising technique in Class 8 — grouping, identities, splitting the middle term — assumes you can already spot and extract a common factor cleanly, so it is worth making automatic. This article walks through finding the highest common factor (HCF) of an expression's terms, extracting it correctly (including tricky signs and leftover $1$ s), and checking the result by expanding back.

The core insight

Every term of an expression shares a common factor when their coefficients share a numerical HCF and their variables share common letters, so writing $k\cdot p + k\cdot q = k(p+q)$ turns the sum into a product — the reverse of the distributive law.

The core idea

Factorisation means writing an expression as a product of factors instead of a sum of terms. The easiest kind of factorisation happens when every single term shares a common factor — a number, a variable, or both. That shared piece is lifted out in front of a bracket, and what is left of each term after dividing by it goes inside the bracket.

ax + ay = a(x+y)

where: $a$ is the common factor present in both terms, and $x$ , $y$ are what remain once each term is divided by $a$ . Checking this is easy: expand $a(x+y)$ using the distributive law and you land back on $ax+ay$ , so factorising and expanding undo each other. For expressions with several terms and more than one variable, the common factor itself has two pieces that must each be hunted down separately.

\text{common factor} = (\text{HCF of the coefficients}) \times (\text{lowest power of each variable common to every term})

where: the HCF of the coefficients is the ordinary highest common factor of the plain numbers in front of each term, and the lowest power of each variable means that if a letter such as $x$ appears in every term, you take whichever power of $x$ is smallest across all the terms — never the largest, and never a variable missing from even one term. A few points make this systematic rather than a guessing game.

Scan every term's numerical coefficient first and find their HCF using the usual method (listing factors, or repeated division).
For each variable that appears in every term, compare the powers it carries in each term and keep only the lowest one; a variable absent from any single term cannot be part of the common factor at all.
Multiply the numerical HCF and the variable part together — that product is what gets written outside the bracket.
Divide each original term by this common factor to find what stays inside the bracket, keeping every sign intact.

Figure

A bar diagram showing three terms each split into a shared common factor block and a leftover block, with the shared blocks aligned to show they are identical across all terms

Derivation AThe distributive law in reverse

The goal is to show, algebraically, that pulling out the HCF of the terms of an expression and dividing each term by it always produces a bracket that, when multiplied back out by the distributive law, exactly reproduces the original expression — so the two directions of factorisation and expansion are true inverses of each other, not just a lucky coincidence for small examples.

Step 1. Start from an expression with terms $T_1, T_2, \ldots, T_n$ , and suppose $k$ is a factor common to every one of them. That means each term can be written as $T_i = k \cdot q_i$ for some quotient $q_i$ — this is exactly what " $k$ is a factor of $T_i$ " means.

Step 2. Add the terms in this form: $T_1 + T_2 + \cdots + T_n = k q_1 + k q_2 + \cdots + k q_n$ . Nothing has changed numerically; every term has simply been rewritten to expose the shared block $k$ .

Step 3. Apply the distributive law in reverse: $ka+kb = k(a+b)$ for any two terms, and extending this to $n$ terms gives $kq_1 + kq_2 + \cdots + kq_n = k(q_1+q_2+\cdots+q_n)$ . This is the factorised form — $k$ outside, the quotients added inside the bracket.

Step 4. Verify the reverse direction. Multiply $k(q_1+q_2+\cdots+q_n)$ back out using the ordinary (forward) distributive law: $k$ multiplies each term inside the bracket in turn, giving $kq_1+kq_2+\cdots+kq_n$ , which is precisely the expression you started with. So factorising by common factor extraction and then expanding the bracket always returns the original expression — the two processes are exact inverses, which is why "expand the bracket back" is always a valid check.

Worked example

Factorise $12x^2y + 18xy^2$ , then extend the same method to a three-term expression, $15a^3b^2 - 20a^2b^3 + 5a^2b^2$ .

Step 1. HCF of the coefficients $12$ and $18$ . Listing factors, $12 = 2^2\times3$ and $18 = 2\times3^2$ ; the common factors are $2$ and $3$ , so the HCF is $2\times3 = 6$ .

Step 2. HCF of the variable parts. The first term carries $x^2y^1$ , the second carries $x^1y^2$ . For $x$ , the lowest power present in both terms is $x^1$ ; for $y$ , the lowest power present in both is $y^1$ . So the shared variable factor is $xy$ .

Step 3. Combine: the overall common factor is $6 \times xy = 6xy$ .

Step 4. Divide each term by $6xy$ : $12x^2y \div 6xy = 2x$ , and $18xy^2 \div 6xy = 3y$ . So $12x^2y + 18xy^2 = 6xy(2x+3y)$ .

Check: expanding $6xy(2x+3y)$ gives $12x^2y + 18xy^2$ back exactly, so the factorisation is confirmed.

$\textbf{12}x^2y + 18xy^2 = \textbf{6xy(2x+3y)}$

Competitive stretch

Now the three-term case, $15a^3b^2 - 20a^2b^3 + 5a^2b^2$ . The coefficients $15, 20, 5$ have HCF $5$ . The powers of $a$ present are $a^3, a^2, a^2$ , so the lowest is $a^2$ ; the powers of $b$ present are $b^2, b^3, b^2$ , so the lowest is $b^2$ . The common factor is $5a^2b^2$ . Dividing each term gives $15a^3b^2 \div 5a^2b^2 = 3a$ , $-20a^2b^3 \div 5a^2b^2 = -4b$ (the sign travels with the term), and $5a^2b^2 \div 5a^2b^2 = 1$ (never drop this trailing $1$ ).

$15a^3b^2 - 20a^2b^3 + 5a^2b^2 = \textbf{5}a^2b^2\textbf{(3a}-\textbf{4b}+\textbf{1)}$

Common traps

Forgetting the trailing "1." When a term divides evenly by the common factor to give exactly $1$ (not $0$ ), that $1$ must still appear inside the bracket — dropping it, as could happen with the $5a^2b^2$ term above, is a very common slip.
Taking out only part of the HCF. Pulling out $2x$ from $6x^2+8x$ instead of noticing the full common factor is available leaves the bracket only partly simplified; always check both the numerical HCF and every variable's lowest shared power.
Sign errors on negative terms. When a term inside the original expression is negative, its sign must be carried correctly into the bracket after division — $-20a^2b^3 \div 5a^2b^2$ must stay $-4b$ , not $4b$ .
Skipping the expand-back check. Not verifying the factorised form by multiplying the bracket out again is how small arithmetic mistakes go unnoticed, especially once there are three or more terms.

Ask the tutor: Want to practise finding the common factor in a four-term expression, or check whether you have pulled out the full HCF on a tricky example? Ask the AI tutor to work through more cases with you step by step.

How do I know when I have taken out the highest common factor and not just a common factor?

Check both parts separately. For the numbers, find the actual HCF of all the coefficients using factor trees or repeated division rather than stopping at the first common number you notice. For the letters, compare the power of each shared variable across every term and take the smallest one — that smallest shared power is as far as you can go. If, after dividing, the terms inside the bracket still share another common factor, you stopped too early.

What happens if there is no variable common to all the terms, only numbers?

That is perfectly fine — the common factor is then just a number. For example, $10p + 15q$ has no shared letter since $p$ and $q$ are different variables, but $10$ and $15$ share the HCF $5$ , so it factorises as $5(2p+3q)$ . A common factor extraction never requires both a numerical and a variable part; either one alone is enough to factorise.

Why does the bracket sometimes end up with a plain 1 or a negative number inside it?

Because dividing a term by the common factor does not always leave a variable behind. If a term equals the common factor exactly, dividing gives $1$ , and if the original term was negative, dividing preserves that negative sign. Both are ordinary results of division, not exceptions, so they must be written inside the bracket exactly as they come out — leaving out a $1$ or dropping a minus sign changes the value of the expression.

Common factor extraction

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads