Factorization by common factors

Multiplication and factorisation are two directions of the same road. Multiplication starts from factors and expands them into a longer expression; factorisation starts from the expanded expression and walks back to find the factors that produced it. The simplest and most useful way to make that return trip is to spot a factor that every single term shares, and pull it outside a bracket. This one skill — finding the highest common factor of a set of terms and taking it out — is the entry point for the whole of factorisation. Every other technique you meet later, including reversing the algebraic identities and spotting differences of squares, works better once a common factor has already been removed. This article builds the method carefully: how to find the HCF of both numbers and letters, how to extend it to grouping when a common factor is hidden across pairs of terms, and how to avoid the sign and power mistakes that cost the most marks.

The core insight

Factorising by common factors reverses the distributive law $a(x+y)=ax+ay$ by finding the highest common factor of every term and writing the expression as HCF times what remains, extending to grouping when different pairs of terms share different factors.

The core idea

Expansion and factorisation are opposite operations built on the same rule, the distributive law. Expanding takes $a(x+y)$ and turns it into $ax+ay$ ; factorising takes $ax+ay$ and asks what single factor is common to both terms, then writes it back as a product.

ax+ay=a(x+y)

where: $a$ is the common factor shared by both terms, and $x$ , $y$ are what remains once $a$ has been divided out of each term. To factorise any expression by this method, you need the highest common factor, or HCF, of all its terms. The HCF has two parts that must both be found separately. The numerical part is the largest whole number that divides every coefficient in the expression exactly. The algebraic part is, for each variable that appears in every term, the lowest power of that variable found in any single term — not the highest, because a power higher than the smallest available would leave a negative power behind after division, which is not allowed. Once the full HCF is identified, every term is divided by it, and the leftover pieces are collected inside a bracket, with the HCF written outside. Take $12x^2y+18xy^2$ as an illustration: the numerical HCF of $12$ and $18$ is $6$ , both terms contain $x$ and $y$ , and the lowest power of each present in both terms is $x^1$ and $y^1$ , so the overall common factor is $6xy$ , giving

12x^2y+18xy^2=6xy(2x+3y)

Sometimes an expression has four terms and no single factor is common to all of them, but the terms can be arranged into two pairs, each of which does have its own common factor. This extension is called factorisation by grouping:

ax+ay+bx+by=a(x+y)+b(x+y)=(x+y)(a+b)

where: $a$ , $b$ are the factors common within each pair, and the bracket $(x+y)$ that appears in both grouped terms becomes a second common factor, extracted once more to finish the job. The figure below shows both routes side by side: a single shared factor pulled straight out of every term, and two separate pairwise factors that reveal a shared bracket only after the first round of extraction.

Derivation AWhy grouping produces a valid factorisation

The goal is to show, using nothing beyond the distributive law applied twice in a row, that an expression with four terms and no single overall common factor can still be factorised completely by pairing terms cleverly, and that the result you land on is guaranteed to expand back to the exact original expression with every term and sign intact.

Step 1. Start from the four-term expression $ax+ay+bx+by$ and group the terms into two pairs that each share a factor: the first two terms share $a$ , and the last two share $b$ . Writing the grouping explicitly, $ax+ay+bx+by=(ax+ay)+(bx+by)$ .

Step 2. Apply the distributive law in reverse to each pair separately. In the first pair, $a$ is common to both terms, so $ax+ay=a(x+y)$ . In the second pair, $b$ is common to both terms, so $bx+by=b(x+y)$ . The expression is now $a(x+y)+b(x+y)$ .

Step 3. Notice that the bracket $(x+y)$ is now itself a common factor of the two remaining terms, exactly as a single variable would be. Applying the distributive law in reverse one more time, treating $(x+y)$ as the shared factor, gives $a(x+y)+b(x+y)=(x+y)(a+b)$ .

Step 4. Confirm the result is correct by expanding forward again: $(x+y)(a+b)=ax+bx+ay+by$ , which is exactly the original four terms reordered. Because expansion and factorisation are exact opposites, this forward check proves the grouping was valid, and the same two-step logic — factor each pair, then factor the shared bracket — works for any four-term expression where the terms can be paired this way. Verified: $(x+y)(a+b)=ax+ay+bx+by$ by direct expansion, matching the original expression term for term.

Worked example

Factorise $15a^3b^2-20a^2b^3+25a^2b^2$ and check the result. Finding the correct HCF for a three-term algebraic expression is the single most-tested skill in this topic, since it requires handling a numerical HCF and two separate variable powers together, and any factorisation is worthless unless the result, when expanded back out, exactly reproduces every term of the original expression with the correct sign.

Step 1. Find the numerical HCF. The coefficients are $15$ , $20$ and $25$ . The largest number dividing all three exactly is $5$ .

Step 2. Find the HCF of each variable separately. Looking at the powers of $a$ across all three terms — $a^3$ , $a^2$ , $a^2$ — the lowest is $a^2$ . Looking at the powers of $b$ — $b^2$ , $b^3$ , $b^2$ — the lowest is $b^2$ . So the algebraic part of the HCF is $a^2b^2$ .

Step 3. Combine both parts. The overall common factor is $5a^2b^2$ .

Step 4. Divide every term by the HCF to find what belongs inside the bracket: $15a^3b^2\div5a^2b^2=3a$ , $-20a^2b^3\div5a^2b^2=-4b$ , and $25a^2b^2\div5a^2b^2=5$ .

Step 5. Write the HCF outside a bracket containing these three results, keeping the signs exactly as they divided out.

15a^3b^2-20a^2b^3+25a^2b^2=5a^2b^2(3a-4b+5)

Step 6. Verify by expanding the answer back out: $5a^2b^2(3a-4b+5)=15a^3b^2-20a^2b^3+25a^2b^2$ , which matches the original expression term for term.

Factorised form: $5a^2b^2(3a-4b+5)$ .

Competitive stretch

Grouping extends the same idea to four terms with no single shared factor. Take $ax-ay+bx-by$ : pairing gives $(ax-ay)+(bx-by)=a(x-y)+b(x-y)$ , and since both pieces now share the bracket $(x-y)$ , the final factorisation is $(x-y)(a+b)$ . Checking by expansion, $(x-y)(a+b)=ax+bx-ay-by=ax-ay+bx-by$ , confirming the grouping was done correctly.

Common traps

Extracting only the numerical part of the HCF and forgetting the shared variable powers, or the reverse — both parts must be checked separately every time.
Taking too high a power of a variable as part of the HCF; the correct choice is always the lowest power appearing in any term, never the highest.
Dropping or flipping a sign when the leading term is negative, for example writing $-4x-8$ as $-4(x-2)$ instead of the correct $-4(x+2)$ .
Stopping before the HCF is fully removed, such as writing $6x+9$ as $3(2x+3)$ correctly but then leaving a partly-factored form like $1(6x+9)$ that has not actually pulled anything out.
Skipping the final check: multiplying the factorised answer back out is the fastest way to catch a wrong HCF or a dropped sign before it costs marks.

Ask the tutor: Want a few more mixed common-factor and grouping problems, or a check on how to spot the right way to pair four terms?

How do I know when to stop dividing out common factors?

Keep dividing until the numbers left inside the bracket have no common factor greater than $1$ , and no variable is present in every remaining term. A quick test is to look at your final bracket: if every term inside it still shares a number or letter, the HCF you removed was not actually the highest one, and you need to go back and take out more. Practising the expand-back check after every problem trains you to spot this automatically.

What if the terms have no common factor at all except 1?

Then the expression genuinely cannot be factorised by this method, and taking out $1$ achieves nothing useful. In that case the expression is either already in its simplest form, or it needs a different technique entirely, such as recognising one of the algebraic identities in reverse (for example a difference of squares) rather than a common monomial factor.

Why does grouping only work for four-term expressions?

Grouping is not limited to exactly four terms in principle, but four terms arranged into two pairs is the pattern you will meet at this stage, because it is the simplest case where no single factor divides every term but a hidden common bracket still exists. The key requirement is not the term count but that the terms can be split into groups, each of which shares its own factor, and that those groups then produce a matching bracket that can be factored out a second time.

Factorization by common factors

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads