Four Terms and No Common Factor: Group Them in Pairs
Learn to take out the highest common factor, factorise four-term expressions by grouping, and recognise the difference of two squares and perfect square patterns.
What do you do when nothing is common to all four terms?
You take the terms two at a time, and the common factor appears where there was none before.
Consider . No single quantity divides all four terms. But group the first two and the last two:
Now is common to both parts, and taking it out gives
The common factor was never in the terms individually — it was created by the grouping. That is the whole technique, and it turns four-term expressions from impossible into routine.
This page covers the first part of the ICSE Class 8 Mathematics chapter on factorisation.
Consider . No single quantity divides all four terms. But group the first two and the last two:
Now is common to both parts, and taking it out gives
The common factor was never in the terms individually — it was created by the grouping. That is the whole technique, and it turns four-term expressions from impossible into routine.
This page covers the first part of the ICSE Class 8 Mathematics chapter on factorisation.
Formula
What are the standard factorisation patterns?
Factorising is expanding run backwards. Every identity from the special-products chapter becomes a factorisation pattern when read from right to left:
Alongside these sit two general methods:
- Highest common factor — take out everything that divides every term
- Grouping — pair the terms so that each pair yields a common bracket
The order to work in. Always look for a common factor first, before any pattern. A common factor left inside makes the remaining expression harder than it needs to be, and an answer that still contains one is not fully factorised.
How to find the highest common factor of algebraic terms. Take the HCF of the coefficients, and the lowest power of each variable present in every term.
**Worked example — the HCF of , and .**
- Coefficients , , have HCF
- appears as , , — lowest power is
- appears as , , — lowest power is
So the HCF is , and
Check by expanding: , , . All three match.
The mistake to avoid. Taking out and writing only two terms in the bracket. The bracket must have as many terms as the original expression — three in, three out. Counting them is a free check that catches a dropped term every time.
Alongside these sit two general methods:
- Highest common factor — take out everything that divides every term
- Grouping — pair the terms so that each pair yields a common bracket
The order to work in. Always look for a common factor first, before any pattern. A common factor left inside makes the remaining expression harder than it needs to be, and an answer that still contains one is not fully factorised.
How to find the highest common factor of algebraic terms. Take the HCF of the coefficients, and the lowest power of each variable present in every term.
**Worked example — the HCF of , and .**
- Coefficients , , have HCF
- appears as , , — lowest power is
- appears as , , — lowest power is
So the HCF is , and
Check by expanding: , , . All three match.
The mistake to avoid. Taking out and writing only two terms in the bracket. The bracket must have as many terms as the original expression — three in, three out. Counting them is a free check that catches a dropped term every time.
How do you factorise by grouping?
Pair the terms, take a factor out of each pair, and check that the two brackets match.
Worked example 1 — the straightforward case.
Worked example 2 — when signs must be managed. Factorise .
Grouping the first pair and the last pair:
The second pair needed care: , since and . Only by taking out a negative factor do the two brackets come out the same.
Check by expanding: , which is the original expression with its terms in a different order. Correct.
The signal that the grouping worked. The two brackets must be identical. If they differ — say you get and — then a sign is wrong, and taking out the negative of your chosen factor from the second pair will fix it, since .
Worked example 3 — when the first pairing fails. Factorise .
Pairing as written: has nothing common, and neither pairing helps. Rearrange first:
Grouping is not restricted to the order given. You may pair any two terms, so if the obvious pairing fails, look for two terms that share something and put them together.
Worked example 4 — a common factor hiding as well. Factorise .
Worked example 5 — grouping after a common factor. Factorise .
First take out the common :
Then group inside:
Had the been left in, the grouping would still have worked but every number would have been three times larger. Common factor first is not a preference — it is what keeps the arithmetic small.
Worked example 1 — the straightforward case.
Worked example 2 — when signs must be managed. Factorise .
Grouping the first pair and the last pair:
The second pair needed care: , since and . Only by taking out a negative factor do the two brackets come out the same.
Check by expanding: , which is the original expression with its terms in a different order. Correct.
The signal that the grouping worked. The two brackets must be identical. If they differ — say you get and — then a sign is wrong, and taking out the negative of your chosen factor from the second pair will fix it, since .
Worked example 3 — when the first pairing fails. Factorise .
Pairing as written: has nothing common, and neither pairing helps. Rearrange first:
Grouping is not restricted to the order given. You may pair any two terms, so if the obvious pairing fails, look for two terms that share something and put them together.
Worked example 4 — a common factor hiding as well. Factorise .
Worked example 5 — grouping after a common factor. Factorise .
First take out the common :
Then group inside:
Had the been left in, the grouping would still have worked but every number would have been three times larger. Common factor first is not a preference — it is what keeps the arithmetic small.
How do you use the difference of squares and perfect squares?
A difference of two squares always factorises. A sum of two squares does not.
Worked example 1. Factorise .
Recognise and :
Worked example 2 — factorising twice. Factorise .
Since and :
But is itself a difference of squares:
And is a sum of squares, which does not factorise, so the work stops there.
Always check whether your factors factorise again. Stopping at is the standard incomplete answer in this topic.
Worked example 3 — after a common factor. Factorise .
The difference of squares was invisible until the came out — another reason the common factor goes first.
Worked example 4 — a numerical use. Evaluate without squaring anything.
Two squares of five digits each, reduced to a single small multiplication.
Worked example 5 — a perfect square trinomial. Factorise .
Test the pattern: the outer terms are and , and twice their product is
which matches the middle term exactly. So
The test for a perfect square, in order. First and last terms must be perfect squares; then the middle term must equal twice the product of their roots. If the middle term does not match, the expression is not a perfect square and needs a different method.
Worked example 6 — a near miss. Is a perfect square?
The outer terms are still and , but twice the product of the roots is , not . So no — and writing here would be wrong. Checking the middle term is not optional.
Worked example 7 — with a negative middle term. Factorise .
Outer terms: and . Twice the product: , matching in size. The middle term is negative, so
Worked example 1. Factorise .
Recognise and :
Worked example 2 — factorising twice. Factorise .
Since and :
But is itself a difference of squares:
And is a sum of squares, which does not factorise, so the work stops there.
Always check whether your factors factorise again. Stopping at is the standard incomplete answer in this topic.
Worked example 3 — after a common factor. Factorise .
The difference of squares was invisible until the came out — another reason the common factor goes first.
Worked example 4 — a numerical use. Evaluate without squaring anything.
Two squares of five digits each, reduced to a single small multiplication.
Worked example 5 — a perfect square trinomial. Factorise .
Test the pattern: the outer terms are and , and twice their product is
which matches the middle term exactly. So
The test for a perfect square, in order. First and last terms must be perfect squares; then the middle term must equal twice the product of their roots. If the middle term does not match, the expression is not a perfect square and needs a different method.
Worked example 6 — a near miss. Is a perfect square?
The outer terms are still and , but twice the product of the roots is , not . So no — and writing here would be wrong. Checking the middle term is not optional.
Worked example 7 — with a negative middle term. Factorise .
Outer terms: and . Twice the product: , matching in size. The middle term is negative, so
Exam tip
Exam tip: common factor first, then check every bracket again
Always take out the highest common factor first. Take the HCF of the coefficients and the lowest power of each variable present in every term: the HCF of , and is .
Count the terms. Three terms in the original means three inside the bracket.
For grouping, pair the terms so each pair leaves the same bracket. If you get and , take out the negative instead — they differ only by a sign.
If the given order does not group, rearrange the terms. Any two may be paired.
A difference of two squares always factorises; a sum of two squares does not.
Check whether your factors factorise again: , not .
For a perfect square, test both conditions: the outer terms must be squares and the middle term must be twice the product of their roots. fails, because twice the product is .
And expand your answer back — it takes one line and proves the factorisation.
Count the terms. Three terms in the original means three inside the bracket.
For grouping, pair the terms so each pair leaves the same bracket. If you get and , take out the negative instead — they differ only by a sign.
If the given order does not group, rearrange the terms. Any two may be paired.
A difference of two squares always factorises; a sum of two squares does not.
Check whether your factors factorise again: , not .
For a perfect square, test both conditions: the outer terms must be squares and the middle term must be twice the product of their roots. fails, because twice the product is .
And expand your answer back — it takes one line and proves the factorisation.
Did you know
Why factorising is the harder direction
Expanding brackets is a procedure: multiply everything by everything, collect, done. Given any expression with brackets, you can grind out the answer without insight.
Factorising has no such guarantee. You are handed the finished product and asked which multiplication produced it — and that demands recognition rather than routine. It is the difference between following a recipe and tasting a dish to work out the ingredients.
This asymmetry is not a quirk of school algebra. It runs right through mathematics, and with numbers it becomes genuinely profound. Multiplying two large prime numbers together takes a moment. Taking the product back apart into those two primes, knowing only the product, is so laborious that the difficulty itself is put to work: the security of ordinary online banking and messaging rests on exactly that gap between how easy multiplication is and how hard factorisation is.
So when factorising feels harder than expanding, that is not a failing on your part. You are running a calculation backwards, and backwards is where the real mathematics usually lives.
The practical consequence for this chapter is simple. Because factorising relies on recognition, the patterns must be familiar enough to spot on sight — which is why expanding them repeatedly, until and jump out at you, is time well spent rather than busywork.
Factorising has no such guarantee. You are handed the finished product and asked which multiplication produced it — and that demands recognition rather than routine. It is the difference between following a recipe and tasting a dish to work out the ingredients.
This asymmetry is not a quirk of school algebra. It runs right through mathematics, and with numbers it becomes genuinely profound. Multiplying two large prime numbers together takes a moment. Taking the product back apart into those two primes, knowing only the product, is so laborious that the difficulty itself is put to work: the security of ordinary online banking and messaging rests on exactly that gap between how easy multiplication is and how hard factorisation is.
So when factorising feels harder than expanding, that is not a failing on your part. You are running a calculation backwards, and backwards is where the real mathematics usually lives.
The practical consequence for this chapter is simple. Because factorising relies on recognition, the patterns must be familiar enough to spot on sight — which is why expanding them repeatedly, until and jump out at you, is time well spent rather than busywork.
Key takeaways
Common factors, grouping and square patterns: quick revision
- Factorising is expanding backwards. Read the identities right to left: , , .
- Common factor first, always. HCF of the coefficients, lowest power of each variable in every term.
- — three terms in, three terms out.
- Grouping creates a common factor where none existed: .
- — the second pair needed a negative factor.
- The two brackets must be identical. , so a sign mismatch is fixable.
- Rearrange if needed: regroups as .
- Combine both methods: .
- Difference of squares: . A sum of squares does not factorise.
- Factorise twice where possible: ; and .
- Numerically: .
- Perfect square test: outer terms are squares and the middle term is twice the product of their roots. , but is not a perfect square since twice the product is .
- With a negative middle term: .
- Expand your answer back to check — one line, complete certainty.
Factorise ten mixed expressions in one sitting, deciding the method before you write anything. Choosing between common factor, grouping and a square pattern is the skill being tested; the algebra afterwards is the easy part.
- Common factor first, always. HCF of the coefficients, lowest power of each variable in every term.
- — three terms in, three terms out.
- Grouping creates a common factor where none existed: .
- — the second pair needed a negative factor.
- The two brackets must be identical. , so a sign mismatch is fixable.
- Rearrange if needed: regroups as .
- Combine both methods: .
- Difference of squares: . A sum of squares does not factorise.
- Factorise twice where possible: ; and .
- Numerically: .
- Perfect square test: outer terms are squares and the middle term is twice the product of their roots. , but is not a perfect square since twice the product is .
- With a negative middle term: .
- Expand your answer back to check — one line, complete certainty.
Factorise ten mixed expressions in one sitting, deciding the method before you write anything. Choosing between common factor, grouping and a square pattern is the skill being tested; the algebra afterwards is the easy part.