Split the Middle Term and the Quadratic Falls Apart
Learn to factorise quadratics by splitting the middle term, handle a leading coefficient greater than one, and factorise sums and differences of cubes.
How do you factorise something with no common factor and no square pattern?
You split the middle term into two pieces chosen so that grouping becomes possible.
Take . There is no common factor, and it is not a perfect square. But look for two numbers whose product is and whose sum is — namely and . Then
The became purely so that grouping would work, and it did. Nothing was added or removed — one term was written as a sum of two.
This is the method for every quadratic in the chapter, and it is grouping from the first part applied to a three-term expression. This page covers the second part of the ICSE Class 8 Mathematics chapter on factorisation.
Take . There is no common factor, and it is not a perfect square. But look for two numbers whose product is and whose sum is — namely and . Then
The became purely so that grouping would work, and it did. Nothing was added or removed — one term was written as a sum of two.
This is the method for every quadratic in the chapter, and it is grouping from the first part applied to a three-term expression. This page covers the second part of the ICSE Class 8 Mathematics chapter on factorisation.
Formula
What are the cube factorisation identities?
Two new patterns join those of the first part:
The sign rules are worth stating precisely, because they are easy to muddle:
- The first bracket takes the same sign as the original expression
- The middle term of the second bracket takes the opposite sign
- The outer terms of the second bracket are always positive
So a sum of cubes gives , and a difference gives . There is only ever one negative sign in the whole factorisation, and it is never on or .
**The coefficient is , not .** The middle term is , not . Writing would make the bracket a perfect square, which it is not — and the whole factorisation would then be wrong.
Why these are true. Expand the difference form:
The four middle terms cancel in pairs, leaving . That cancellation is exactly what the coefficient of is there to produce.
Unlike squares, a sum of cubes does factorise. A sum of two squares has no factors, but a sum of two cubes always has as a factor — a genuine difference between the two cases, and one examiners like to test.
The sign rules are worth stating precisely, because they are easy to muddle:
- The first bracket takes the same sign as the original expression
- The middle term of the second bracket takes the opposite sign
- The outer terms of the second bracket are always positive
So a sum of cubes gives , and a difference gives . There is only ever one negative sign in the whole factorisation, and it is never on or .
**The coefficient is , not .** The middle term is , not . Writing would make the bracket a perfect square, which it is not — and the whole factorisation would then be wrong.
Why these are true. Expand the difference form:
The four middle terms cancel in pairs, leaving . That cancellation is exactly what the coefficient of is there to produce.
Unlike squares, a sum of cubes does factorise. A sum of two squares has no factors, but a sum of two cubes always has as a factor — a genuine difference between the two cases, and one examiners like to test.
How do you split the middle term when the leading coefficient is 1?
Find two numbers with the right product and the right sum. For the product is and the sum is .
Worked example 1 — all positive. . Product , sum , so and :
Worked example 2 — positive constant, negative middle. . Product , sum , so and :
Worked example 3 — negative constant. . Product , sum , so and :
Worked example 4 — negative constant, negative middle. . Product , sum , so and :
The sign rules that come out of these four. Read the constant first, then the middle term:
- Constant positive: both numbers have the same sign, and the middle term tells you which — positive middle means both positive, negative middle means both negative.
- Constant negative: the numbers have opposite signs, and the larger one carries the sign of the middle term.
Deciding the signs before hunting for the numbers halves the work, because you then only need pairs of the right kind.
How to search for the pair. List the factor pairs of the constant and test their sums. For a product of : and sum to ; and sum to ; and sum to . The list is short, and it is always finite.
When no pair works. Not every quadratic factorises over whole numbers. For the factor pairs of are only and , summing to , not . That expression simply does not factorise at this level, and exhausting the list is how you know — not a feeling that it looks hard.
Checking your answer. Expand it back. . One line, complete certainty.
Worked example 1 — all positive. . Product , sum , so and :
Worked example 2 — positive constant, negative middle. . Product , sum , so and :
Worked example 3 — negative constant. . Product , sum , so and :
Worked example 4 — negative constant, negative middle. . Product , sum , so and :
The sign rules that come out of these four. Read the constant first, then the middle term:
- Constant positive: both numbers have the same sign, and the middle term tells you which — positive middle means both positive, negative middle means both negative.
- Constant negative: the numbers have opposite signs, and the larger one carries the sign of the middle term.
Deciding the signs before hunting for the numbers halves the work, because you then only need pairs of the right kind.
How to search for the pair. List the factor pairs of the constant and test their sums. For a product of : and sum to ; and sum to ; and sum to . The list is short, and it is always finite.
When no pair works. Not every quadratic factorises over whole numbers. For the factor pairs of are only and , summing to , not . That expression simply does not factorise at this level, and exhausting the list is how you know — not a feeling that it looks hard.
Checking your answer. Expand it back. . One line, complete certainty.
How do you factorise when the leading coefficient is not 1?
Multiply the first and last coefficients, then split the middle term using that product.
For , find two numbers whose **product is and whose sum is . Then group.
Worked example 1.** Factorise .
Here , and the sum needed is . Factor pairs of : and give ; and give . That is the pair.
Split the middle term as :
Group and factor each pair:
Notice the second group gave a factor of , which must still be written. Leaving the bracket bare — writing and then losing the — is a common slip that produces a missing term.
Check: . Correct.
Worked example 2 — with a negative middle term. Factorise .
. The constant is positive and the middle term negative, so both numbers are negative. Pairs of : and sum to ; and sum to . That is the pair.
The second group needed a negative factor taken out, so that both brackets read .
Check: . Correct.
Worked example 3 — a common factor first. Factorise .
Every coefficient is even, so take out :
Now with sum , giving and :
Had the been left in, the product to split would have been and the search longer. Common factor first, always.
Worked example 4 — a sum of cubes. Factorise .
Recognise and , so and :
**Check at **: the original gives , and the factors give . Correct.
Worked example 5 — a difference of cubes with the constant first. Factorise .
Here and :
The order matters: since the expression is , the first bracket is , not . Identify which cube is being subtracted from which before writing anything.
Worked example 6 — combining methods. Factorise .
Common factor first:
Three steps, each one a pattern already met, and the answer is only complete after the third.
Worked example 7 — a harder pattern. Factorise .
This looks impossible until you add and subtract :
Now it is a difference of squares:
**Check at , **: the original gives , and the factors give . Correct.
The trick of completing a square by adding and subtracting the same quantity is worth noticing here, because it reappears throughout later algebra.
For , find two numbers whose **product is and whose sum is . Then group.
Worked example 1.** Factorise .
Here , and the sum needed is . Factor pairs of : and give ; and give . That is the pair.
Split the middle term as :
Group and factor each pair:
Notice the second group gave a factor of , which must still be written. Leaving the bracket bare — writing and then losing the — is a common slip that produces a missing term.
Check: . Correct.
Worked example 2 — with a negative middle term. Factorise .
. The constant is positive and the middle term negative, so both numbers are negative. Pairs of : and sum to ; and sum to . That is the pair.
The second group needed a negative factor taken out, so that both brackets read .
Check: . Correct.
Worked example 3 — a common factor first. Factorise .
Every coefficient is even, so take out :
Now with sum , giving and :
Had the been left in, the product to split would have been and the search longer. Common factor first, always.
Worked example 4 — a sum of cubes. Factorise .
Recognise and , so and :
**Check at **: the original gives , and the factors give . Correct.
Worked example 5 — a difference of cubes with the constant first. Factorise .
Here and :
The order matters: since the expression is , the first bracket is , not . Identify which cube is being subtracted from which before writing anything.
Worked example 6 — combining methods. Factorise .
Common factor first:
Three steps, each one a pattern already met, and the answer is only complete after the third.
Worked example 7 — a harder pattern. Factorise .
This looks impossible until you add and subtract :
Now it is a difference of squares:
**Check at , **: the original gives , and the factors give . Correct.
The trick of completing a square by adding and subtracting the same quantity is worth noticing here, because it reappears throughout later algebra.
Exam tip
Exam tip: decide the signs before hunting for numbers
For , find two numbers with **product and sum **. For , use **product ** and sum , then group.
Read the constant's sign first. Positive constant means the two numbers share a sign, decided by the middle term. Negative constant means opposite signs, with the larger number carrying the middle term's sign.
List the factor pairs and test their sums. For a product of : and give , and give . The list is short and finite — if none works, the expression does not factorise, as with .
Common factor first: , and taking the out keeps the product to split small.
**Write the ** when a group leaves a bare bracket: .
For cubes, the first bracket takes the same sign, the middle term of the second takes the opposite sign, and the outer terms are always positive. The middle coefficient is **, not .
A sum of squares does not factorise, but a sum of cubes** does.
Mind the order: .
And expand every answer back — or check numerically at .
Read the constant's sign first. Positive constant means the two numbers share a sign, decided by the middle term. Negative constant means opposite signs, with the larger number carrying the middle term's sign.
List the factor pairs and test their sums. For a product of : and give , and give . The list is short and finite — if none works, the expression does not factorise, as with .
Common factor first: , and taking the out keeps the product to split small.
**Write the ** when a group leaves a bare bracket: .
For cubes, the first bracket takes the same sign, the middle term of the second takes the opposite sign, and the outer terms are always positive. The middle coefficient is **, not .
A sum of squares does not factorise, but a sum of cubes** does.
Mind the order: .
And expand every answer back — or check numerically at .
Did you know
Why splitting the middle term works at all
The method can feel like a conjuring trick. You need two numbers, you find them by trial, and suddenly the expression factorises. Where does the requirement come from?
Work forwards instead. Multiply two general brackets:
The middle coefficient is and the constant is . So if a quadratic came from two such brackets, then must be the sum of the two numbers and their product. Hunting for a pair with that sum and product is not a trick — it is asking directly which brackets could have produced this expression.
The same reasoning explains the rule. Multiplying out gives a middle coefficient of , while the product of the outer coefficients is . The two pieces of the split, and , multiply to — exactly . So the product to aim for is rather than , and the reason is bookkeeping, not convention.
This is why the method sometimes finds nothing. If no pair of whole numbers has the required sum and product, then no pair of whole-number brackets multiplies to give the expression. The failure of the search is itself a proof — which is a more satisfying answer than I could not see it.
Work forwards instead. Multiply two general brackets:
The middle coefficient is and the constant is . So if a quadratic came from two such brackets, then must be the sum of the two numbers and their product. Hunting for a pair with that sum and product is not a trick — it is asking directly which brackets could have produced this expression.
The same reasoning explains the rule. Multiplying out gives a middle coefficient of , while the product of the outer coefficients is . The two pieces of the split, and , multiply to — exactly . So the product to aim for is rather than , and the reason is bookkeeping, not convention.
This is why the method sometimes finds nothing. If no pair of whole numbers has the required sum and product, then no pair of whole-number brackets multiplies to give the expression. The failure of the search is itself a proof — which is a more satisfying answer than I could not see it.
Key takeaways
Splitting the middle term and cubes: quick revision
- Splitting the middle term turns a three-term expression into four so that grouping works: .
- For , find two numbers with **product , sum **.
- The four sign cases: ; ; .
- Constant positive means the numbers share a sign (set by the middle term); constant negative means opposite signs, with the larger carrying the middle term's sign.
- Search the factor pairs: for product , sums are , , . If no pair works the expression does not factorise — is one such.
- For , use **product **: needs product , sum , so and , giving .
- needs product , sum , so and , giving .
- Common factor first: .
- **Write the ** when a group leaves a bare bracket: .
- Cubes: and — one negative sign only, and the middle coefficient is , not .
- ; at both sides give .
- — order matters.
- A sum of squares does not factorise; a sum of cubes does.
- Combining methods: .
- Adding and subtracting a term: ; at both sides give .
- The method works because — you are asking which brackets produced the expression.
Factorise twenty quadratics in one sitting, saying the sign pattern aloud before searching. Once the signs are automatic, the search shrinks to a handful of pairs and the whole topic becomes quick.
- For , find two numbers with **product , sum **.
- The four sign cases: ; ; .
- Constant positive means the numbers share a sign (set by the middle term); constant negative means opposite signs, with the larger carrying the middle term's sign.
- Search the factor pairs: for product , sums are , , . If no pair works the expression does not factorise — is one such.
- For , use **product **: needs product , sum , so and , giving .
- needs product , sum , so and , giving .
- Common factor first: .
- **Write the ** when a group leaves a bare bracket: .
- Cubes: and — one negative sign only, and the middle coefficient is , not .
- ; at both sides give .
- — order matters.
- A sum of squares does not factorise; a sum of cubes does.
- Combining methods: .
- Adding and subtracting a term: ; at both sides give .
- The method works because — you are asking which brackets produced the expression.
Factorise twenty quadratics in one sitting, saying the sign pattern aloud before searching. Once the signs are automatic, the search shrinks to a handful of pairs and the whole topic becomes quick.