The Sign in the Quadratic Factor Is Always the Opposite One
Split the middle term of a quadratic whether the leading coefficient is one or not, factorise a sum or difference of cubes with the right signs, and take a composite expression apart completely.
Which sign goes in the quadratic factor when you factorise a sum of cubes?
The two cube identities look almost identical, and there is one place where they differ in a way that is very easy to get backwards.
Look at the middle sign of the long bracket in each. For the sum of cubes it is minus. For the difference of cubes it is plus.
The sign in the quadratic factor is always the opposite of the sign in the linear factor.
That is worth fixing firmly, because it is the single most common error in this chapter — and it is easy to check. Multiply out and the middle terms must cancel:
Four terms cancel in pairs, leaving exactly the two cubes. **If you had used instead, nothing would have cancelled** and you would have been left with extra terms — which is the fastest way to remember which sign belongs where.
This page covers the second part of the ICSE Class 9 Mathematics chapter on factorisation: splitting the middle term of a quadratic trinomial when the leading coefficient is and when it is not, the sum and difference of cubes, and factorising a composite expression completely.
Look at the middle sign of the long bracket in each. For the sum of cubes it is minus. For the difference of cubes it is plus.
The sign in the quadratic factor is always the opposite of the sign in the linear factor.
That is worth fixing firmly, because it is the single most common error in this chapter — and it is easy to check. Multiply out and the middle terms must cancel:
Four terms cancel in pairs, leaving exactly the two cubes. **If you had used instead, nothing would have cancelled** and you would have been left with extra terms — which is the fastest way to remember which sign belongs where.
This page covers the second part of the ICSE Class 9 Mathematics chapter on factorisation: splitting the middle term of a quadratic trinomial when the leading coefficient is and when it is not, the sum and difference of cubes, and factorising a composite expression completely.
How do you split the middle term of a simple quadratic?
**Find two numbers whose product is the constant term and whose sum is the coefficient of .**
For you need two numbers and with
Then split the middle term into and factorise by grouping.
Worked example 1. Factorise .
Product , sum . The pairs multiplying to are , and — and . So the numbers are and :
Worked example 2 — both signs negative in the middle. Factorise .
Product and sum . Since the product is positive and the sum negative, both numbers are negative: and .
Worked example 3 — a negative constant. Factorise .
Product and sum . A negative product means the numbers have opposite signs, and the larger in size must be positive to give a positive sum: and .
Worked example 4. Factorise .
Product and sum , so opposite signs with the larger one negative: and .
Check: . Agreed.
Now the sign reasoning as a rule, because it halves the search every time.
- If the constant is positive, the two numbers have the same sign — both positive if the middle coefficient is positive, both negative if it is negative
- If the constant is negative, the two numbers have opposite signs, and the one with the larger size carries the sign of the middle coefficient
So look at the constant's sign first and the middle term's sign second. That tells you what kind of pair you are hunting before you list any factors, and it means you never test a pair with the wrong signs.
And notice that this method is the identity of the previous chapter read backwards. — so factorising is the same statement used from right to left. Expansion and factorisation are one identity in two directions, which is why every factorisation can be checked by expanding.
For you need two numbers and with
Then split the middle term into and factorise by grouping.
Worked example 1. Factorise .
Product , sum . The pairs multiplying to are , and — and . So the numbers are and :
Worked example 2 — both signs negative in the middle. Factorise .
Product and sum . Since the product is positive and the sum negative, both numbers are negative: and .
Worked example 3 — a negative constant. Factorise .
Product and sum . A negative product means the numbers have opposite signs, and the larger in size must be positive to give a positive sum: and .
Worked example 4. Factorise .
Product and sum , so opposite signs with the larger one negative: and .
Check: . Agreed.
Now the sign reasoning as a rule, because it halves the search every time.
- If the constant is positive, the two numbers have the same sign — both positive if the middle coefficient is positive, both negative if it is negative
- If the constant is negative, the two numbers have opposite signs, and the one with the larger size carries the sign of the middle coefficient
So look at the constant's sign first and the middle term's sign second. That tells you what kind of pair you are hunting before you list any factors, and it means you never test a pair with the wrong signs.
And notice that this method is the identity of the previous chapter read backwards. — so factorising is the same statement used from right to left. Expansion and factorisation are one identity in two directions, which is why every factorisation can be checked by expanding.
Formula
What changes when the coefficient of x squared is not 1?
Multiply the first and last coefficients together, and look for two numbers with that product and the middle coefficient as their sum.
For you need two numbers and with
Then split the middle term and group as before. **The only change from the simple case is that the product is instead of .
Worked example 1.** Factorise .
Here , , , so . Two numbers with product and sum : the pairs are , , — and . So the numbers are and :
Check: . Agreed.
Worked example 2. Factorise .
, and the sum must be . Positive product with a negative sum means both negative: and .
Check: . Agreed.
Worked example 3. Factorise .
, and the sum must be . Negative product means opposite signs, and the larger must be positive: and .
Check: . Agreed.
**Why multiplying by is the right adjustment**, since the rule looks arbitrary. In the two brackets are with and . The middle term is , and the product of those two coefficients is . **So the two numbers you are hunting always multiply to **, whatever is — and when that reduces to the simple rule.
One practical warning. After splitting, the grouping must leave the same bracket in both halves. If it does not, you have chosen the right numbers but grouped them in the wrong order — swap which one goes with the first term and try again.
In example 1, splitting as instead gives — the same answer, reached by pairing the other way round. So both orders work here, and when one does not, the other will.
For you need two numbers and with
Then split the middle term and group as before. **The only change from the simple case is that the product is instead of .
Worked example 1.** Factorise .
Here , , , so . Two numbers with product and sum : the pairs are , , — and . So the numbers are and :
Check: . Agreed.
Worked example 2. Factorise .
, and the sum must be . Positive product with a negative sum means both negative: and .
Check: . Agreed.
Worked example 3. Factorise .
, and the sum must be . Negative product means opposite signs, and the larger must be positive: and .
Check: . Agreed.
**Why multiplying by is the right adjustment**, since the rule looks arbitrary. In the two brackets are with and . The middle term is , and the product of those two coefficients is . **So the two numbers you are hunting always multiply to **, whatever is — and when that reduces to the simple rule.
One practical warning. After splitting, the grouping must leave the same bracket in both halves. If it does not, you have chosen the right numbers but grouped them in the wrong order — swap which one goes with the first term and try again.
In example 1, splitting as instead gives — the same answer, reached by pairing the other way round. So both orders work here, and when one does not, the other will.
How do you factorise a sum or a difference of cubes?
Take the cube roots for the linear factor, then build the quadratic factor from their squares and their product — with the opposite sign.
How to build the quadratic factor, which is the same recipe in both cases:
- First term: the square of the first cube root
- Middle term: the product of the two cube roots, with the opposite sign to the linear factor
- Last term: the square of the second cube root
Note that the middle term is and not — this is not a perfect square, and the coefficient is .
Worked example 1. Factorise .
The cube roots are and , since and . It is a sum, so the linear factor is and the quadratic factor has a minus:
**Check at **: the original is , and the factors give . Correct.
Worked example 2. Factorise .
The cube roots are and . It is a difference, so the linear factor is and the quadratic factor has a plus:
**Check at **: the original is , and the factors give . Correct.
Worked example 3 — numerical. Evaluate by factorising.
which is indeed . **So the identity gives you a factorisation of for free** — it is , which would have taken some searching to find otherwise.
Now the two differences from the squares of the last chapter, both of which matter.
First, a sum of cubes DOES factorise. A sum of two squares does not — that was the whole reason completing the square existed. But has a perfectly good factorisation, and it is available immediately.
Second, the quadratic factor does not factorise further over the real numbers. cannot be split, which is why the answer stops there and why it is not a mistake to leave a three-term bracket in the final line.
So the two chapters divide neatly: with squares a difference factorises and a sum does not; with cubes both do. That is worth stating before any attempt, because it tells you at once whether the expression in front of you has a factorisation to find.
How to build the quadratic factor, which is the same recipe in both cases:
- First term: the square of the first cube root
- Middle term: the product of the two cube roots, with the opposite sign to the linear factor
- Last term: the square of the second cube root
Note that the middle term is and not — this is not a perfect square, and the coefficient is .
Worked example 1. Factorise .
The cube roots are and , since and . It is a sum, so the linear factor is and the quadratic factor has a minus:
**Check at **: the original is , and the factors give . Correct.
Worked example 2. Factorise .
The cube roots are and . It is a difference, so the linear factor is and the quadratic factor has a plus:
**Check at **: the original is , and the factors give . Correct.
Worked example 3 — numerical. Evaluate by factorising.
which is indeed . **So the identity gives you a factorisation of for free** — it is , which would have taken some searching to find otherwise.
Now the two differences from the squares of the last chapter, both of which matter.
First, a sum of cubes DOES factorise. A sum of two squares does not — that was the whole reason completing the square existed. But has a perfectly good factorisation, and it is available immediately.
Second, the quadratic factor does not factorise further over the real numbers. cannot be split, which is why the answer stops there and why it is not a mistake to leave a three-term bracket in the final line.
So the two chapters divide neatly: with squares a difference factorises and a sum does not; with cubes both do. That is worth stating before any attempt, because it tells you at once whether the expression in front of you has a factorisation to find.
How do you factorise an expression completely?
Keep applying methods until no factor can be factorised any further — and start with the common factor.
"Factorise completely" means exactly that: the answer is finished only when every bracket is beyond further factorisation.
The order to work in.
- First take out the HCF
- Then look at what is left — is it a difference of squares, a sum or difference of cubes, a trinomial, or a candidate for grouping?
- Then check each resulting factor again, because a factor can itself factorise
Worked example 1 — a difference of squares twice over. Factorise .
But is itself a difference of squares:
And is a sum of squares, so it stops there.
**Check at **: the original is , and the factors give . Correct.
Worked example 2 — HCF then squares. Factorise .
**Check at **: , and . Correct.
Worked example 3 — HCF then splitting. Factorise .
Now split the middle term of : product , sum , so and :
**Check at **: , and . Correct.
Worked example 4 — squares then cubes. Factorise .
Treat it as a difference of squares first, since :
Now apply both cube identities:
**Check at , **: the original is . The factors give . Correct.
Notice that example 4 could have been started the other way. Treating as a difference of cubes first, since , gives — and then splits, and the quartic needs the completing-the-square method of the previous part.
Both routes reach a correct factorisation, and the squares-first route is much easier. So when an expression can be seen as either a difference of squares or a difference of cubes, take the squares first — it produces smaller pieces and avoids the harder step.
And the general habit that this section is really teaching. After every factorisation, look at each bracket again and ask whether it has a factorisation of its own. Most lost marks in "factorise completely" questions are for stopping one step early, and one extra glance at each bracket is the whole remedy.
"Factorise completely" means exactly that: the answer is finished only when every bracket is beyond further factorisation.
The order to work in.
- First take out the HCF
- Then look at what is left — is it a difference of squares, a sum or difference of cubes, a trinomial, or a candidate for grouping?
- Then check each resulting factor again, because a factor can itself factorise
Worked example 1 — a difference of squares twice over. Factorise .
But is itself a difference of squares:
And is a sum of squares, so it stops there.
**Check at **: the original is , and the factors give . Correct.
Worked example 2 — HCF then squares. Factorise .
**Check at **: , and . Correct.
Worked example 3 — HCF then splitting. Factorise .
Now split the middle term of : product , sum , so and :
**Check at **: , and . Correct.
Worked example 4 — squares then cubes. Factorise .
Treat it as a difference of squares first, since :
Now apply both cube identities:
**Check at , **: the original is . The factors give . Correct.
Notice that example 4 could have been started the other way. Treating as a difference of cubes first, since , gives — and then splits, and the quartic needs the completing-the-square method of the previous part.
Both routes reach a correct factorisation, and the squares-first route is much easier. So when an expression can be seen as either a difference of squares or a difference of cubes, take the squares first — it produces smaller pieces and avoids the harder step.
And the general habit that this section is really teaching. After every factorisation, look at each bracket again and ask whether it has a factorisation of its own. Most lost marks in "factorise completely" questions are for stopping one step early, and one extra glance at each bracket is the whole remedy.
Exam tip
Exam tip: check the sign pattern before hunting for numbers
**For , find two numbers with product and sum .** For , the product is and the sum is still .
Read the signs before listing any factors. A positive constant means the two numbers share a sign; a negative constant means opposite signs, with the larger carrying the sign of the middle term.
After splitting, the two groups must leave the SAME bracket. If they do not, pair the other way round — both orders give the same answer when one of them works.
For cubes, the sign in the quadratic factor is the OPPOSITE of the sign in the linear factor. Sum of cubes gives ; difference of cubes gives .
**The middle term of the quadratic factor is , not — it is not a perfect square.
Verify a cube factorisation by expanding and checking that the four middle terms cancel in pairs. If they do not, the sign is wrong.
A sum of CUBES factorises; a sum of SQUARES does not. Check which you have before starting.
The quadratic factor from a cube does not factorise further — leaving a three-term bracket is the correct final answer.
"Factorise completely" means take out the HCF first, then apply an identity, then re-examine every bracket.
When an expression is both a difference of squares and a difference of cubes, take the SQUARES first** — is far easier that way.
Verify every answer numerically. Substitute , or into the original and into your factors — one line and the whole answer is confirmed.
And if the question says completely, glance at each final bracket once more before writing the answer down.
Read the signs before listing any factors. A positive constant means the two numbers share a sign; a negative constant means opposite signs, with the larger carrying the sign of the middle term.
After splitting, the two groups must leave the SAME bracket. If they do not, pair the other way round — both orders give the same answer when one of them works.
For cubes, the sign in the quadratic factor is the OPPOSITE of the sign in the linear factor. Sum of cubes gives ; difference of cubes gives .
**The middle term of the quadratic factor is , not — it is not a perfect square.
Verify a cube factorisation by expanding and checking that the four middle terms cancel in pairs. If they do not, the sign is wrong.
A sum of CUBES factorises; a sum of SQUARES does not. Check which you have before starting.
The quadratic factor from a cube does not factorise further — leaving a three-term bracket is the correct final answer.
"Factorise completely" means take out the HCF first, then apply an identity, then re-examine every bracket.
When an expression is both a difference of squares and a difference of cubes, take the SQUARES first** — is far easier that way.
Verify every answer numerically. Substitute , or into the original and into your factors — one line and the whole answer is confirmed.
And if the question says completely, glance at each final bracket once more before writing the answer down.
Did you know
Why factorising 1027 by hand is easier than it looks
Suppose somebody asks you to factorise . Nothing about it looks helpful. It is odd, so is out. Its digits sum to , so is out. It does not end in or , so is out. You would go on to , , , and so on.
Now notice that , and that both parts are perfect cubes — and .
The factorisation appears in one line, and the trial division was never needed.
The same thing works wherever a number happens to sit near a cube or a square. . And .
Or take a difference of squares. — a number that looks stubbornly prime and is not.
And .
There is a general lesson here worth more than the trick. An algebraic identity does not know whether its letters are symbols or numbers, so every identity in this chapter is simultaneously a fact about algebra and a fact about arithmetic.
Which means the reverse is true as well. A number that is a difference of two squares is never prime unless the two squares are consecutive — because hands you two factors, and the only way one of them can be is if and differ by one.
So cannot be prime, and you know that before doing any division at all. The identity is not merely a shortcut to the factors; it is a proof that factors exist.
Now notice that , and that both parts are perfect cubes — and .
The factorisation appears in one line, and the trial division was never needed.
The same thing works wherever a number happens to sit near a cube or a square. . And .
Or take a difference of squares. — a number that looks stubbornly prime and is not.
And .
There is a general lesson here worth more than the trick. An algebraic identity does not know whether its letters are symbols or numbers, so every identity in this chapter is simultaneously a fact about algebra and a fact about arithmetic.
Which means the reverse is true as well. A number that is a difference of two squares is never prime unless the two squares are consecutive — because hands you two factors, and the only way one of them can be is if and differ by one.
So cannot be prime, and you know that before doing any division at all. The identity is not merely a shortcut to the factors; it is a proof that factors exist.
Exam relevance
Why does JEE Main assume you can factorise instantly?
Because factorisation is never the question in Class 11 — it is the step you take on the way to the question, and time spent on it is time lost.
This is the foundation for Class 11 Mathematics Complex Numbers and Quadratic Equations and Limits and Derivatives, and Class 12 Integrals, examined in JEE Main. Splitting the middle term is how a quadratic is solved without the formula, and **the relation , is the same information as the sum and product of the roots** — since for the roots multiply to and add to .
Limits depend on it completely. Evaluating requires factorising the numerator and cancelling; requires the difference of cubes identity from this page, giving . That second limit is a standard result in JEE Main, and it is this identity used at speed.
Integration by partial fractions needs the denominator factorised first. Class 12 splits into partial fractions only after writing the denominator as , and a cubic denominator needs the cube identities. An integral that cannot be started is very often a factorisation that was not spotted.
The cube identities appear in their own right. Class 11 uses and the sum and difference of cubes in algebraic manipulation, and **Class 12 Determinants produces exactly these expressions** when a symmetric determinant is expanded — so recognising as a factorisable form saves a great deal of expansion.
Complete factorisation matters for roots. A polynomial's roots are read directly off its factors, so shows two real roots at and, over the complex numbers, two more from giving . Questions asking for the number of real roots of a polynomial are answered by factorising completely, and stopping early gives the wrong count.
The prime-testing observation is examinable in olympiad-style questions. The fact that a difference of two squares is never prime unless the squares are consecutive is the kind of reasoning tested in NTSE and in mathematics olympiads, and it follows directly from the identity.
Trigonometric and exponential expressions use the same identities. Factorising uses the difference of cubes with trigonometric letters, and factorises as — **the same identity with , which is exactly the point made on the previous page about an identity's letters standing for anything.
What the questions look like. For board work, expect factorise a trinomial by splitting the middle term, factorise with , factorise a sum or difference of cubes, and factorise a given expression completely. The splitting should be visible in the working. For JEE Main, expect quadratic roots, limits requiring cancellation, partial fractions and root counting.
How board and competitive emphasis differ. A board paper rewards the split middle term shown and the identity named. A competitive paper never asks for the factorisation — it asks for a limit, an integral or a root count that depends on it being done quickly and correctly.
The single trap that costs the most marks. Putting the wrong sign in the quadratic factor of a cube. The sum** of cubes gives and the difference gives — the quadratic factor always takes the opposite sign. The defence is not to memorise it but to multiply out for two seconds: with the opposite sign the four middle terms cancel in pairs and leave exactly , and with the same sign they do not cancel at all. A factorisation that fails to expand back is a factorisation that is wrong, and that check takes less time than recalling which version you were taught.
This is the foundation for Class 11 Mathematics Complex Numbers and Quadratic Equations and Limits and Derivatives, and Class 12 Integrals, examined in JEE Main. Splitting the middle term is how a quadratic is solved without the formula, and **the relation , is the same information as the sum and product of the roots** — since for the roots multiply to and add to .
Limits depend on it completely. Evaluating requires factorising the numerator and cancelling; requires the difference of cubes identity from this page, giving . That second limit is a standard result in JEE Main, and it is this identity used at speed.
Integration by partial fractions needs the denominator factorised first. Class 12 splits into partial fractions only after writing the denominator as , and a cubic denominator needs the cube identities. An integral that cannot be started is very often a factorisation that was not spotted.
The cube identities appear in their own right. Class 11 uses and the sum and difference of cubes in algebraic manipulation, and **Class 12 Determinants produces exactly these expressions** when a symmetric determinant is expanded — so recognising as a factorisable form saves a great deal of expansion.
Complete factorisation matters for roots. A polynomial's roots are read directly off its factors, so shows two real roots at and, over the complex numbers, two more from giving . Questions asking for the number of real roots of a polynomial are answered by factorising completely, and stopping early gives the wrong count.
The prime-testing observation is examinable in olympiad-style questions. The fact that a difference of two squares is never prime unless the squares are consecutive is the kind of reasoning tested in NTSE and in mathematics olympiads, and it follows directly from the identity.
Trigonometric and exponential expressions use the same identities. Factorising uses the difference of cubes with trigonometric letters, and factorises as — **the same identity with , which is exactly the point made on the previous page about an identity's letters standing for anything.
What the questions look like. For board work, expect factorise a trinomial by splitting the middle term, factorise with , factorise a sum or difference of cubes, and factorise a given expression completely. The splitting should be visible in the working. For JEE Main, expect quadratic roots, limits requiring cancellation, partial fractions and root counting.
How board and competitive emphasis differ. A board paper rewards the split middle term shown and the identity named. A competitive paper never asks for the factorisation — it asks for a limit, an integral or a root count that depends on it being done quickly and correctly.
The single trap that costs the most marks. Putting the wrong sign in the quadratic factor of a cube. The sum** of cubes gives and the difference gives — the quadratic factor always takes the opposite sign. The defence is not to memorise it but to multiply out for two seconds: with the opposite sign the four middle terms cancel in pairs and leave exactly , and with the same sign they do not cancel at all. A factorisation that fails to expand back is a factorisation that is wrong, and that check takes less time than recalling which version you were taught.
Key takeaways
Splitting the middle term and factorising cubes: quick revision
- **For : find two numbers with product and sum , split the middle term and group.
- ; ; ; .
- Read the signs first. A positive** constant means both numbers share the sign of ; a negative constant means opposite signs, with the larger carrying the sign of .
- **For **: the product is and the sum is still .
- ****: , sum , numbers and , giving .
- ****: , sum , numbers and , giving .
- ****: , sum , numbers and , giving .
- **Why **: the brackets are with and , so the two middle coefficients multiply to .
- If the grouping does not leave the same bracket, pair the other way round.
- ** and .
- The quadratic factor takes the OPPOSITE sign to the linear factor**, and its middle term is — **not .
- Verify by expanding: with the opposite sign the four middle terms cancel in pairs; with the same sign they do not.
- **, checked at as .
- ****, checked at as .
- ** — the identity supplies a factorisation of a number for free.
- A sum of CUBES factorises; a sum of SQUARES does not. And the quadratic factor from a cube does not factorise further.
- Factorise completely: take out the HCF first, then apply an identity, then re-examine every bracket.
- ** — the difference of squares applied twice; stops there.
- **; .
- **, checked at as .
- When an expression is both a difference of squares and of cubes, take the SQUARES first — the pieces come out smaller.
- Most lost marks are for stopping one step early — glance at each bracket again before writing the answer.
Factorise completely and check it at and — if both substitutions agree with the original, every bracket is right.
- ; ; ; .
- Read the signs first. A positive** constant means both numbers share the sign of ; a negative constant means opposite signs, with the larger carrying the sign of .
- **For **: the product is and the sum is still .
- ****: , sum , numbers and , giving .
- ****: , sum , numbers and , giving .
- ****: , sum , numbers and , giving .
- **Why **: the brackets are with and , so the two middle coefficients multiply to .
- If the grouping does not leave the same bracket, pair the other way round.
- ** and .
- The quadratic factor takes the OPPOSITE sign to the linear factor**, and its middle term is — **not .
- Verify by expanding: with the opposite sign the four middle terms cancel in pairs; with the same sign they do not.
- **, checked at as .
- ****, checked at as .
- ** — the identity supplies a factorisation of a number for free.
- A sum of CUBES factorises; a sum of SQUARES does not. And the quadratic factor from a cube does not factorise further.
- Factorise completely: take out the HCF first, then apply an identity, then re-examine every bracket.
- ** — the difference of squares applied twice; stops there.
- **; .
- **, checked at as .
- When an expression is both a difference of squares and of cubes, take the SQUARES first — the pieces come out smaller.
- Most lost marks are for stopping one step early — glance at each bracket again before writing the answer.
Factorise completely and check it at and — if both substitutions agree with the original, every bracket is right.