Two Numbers That Add to the Middle and Multiply to the End
Learn the identity for the product of two binomials and the square of three terms, build a factorisation with algebra tiles, split the middle term of a quadratic, and multiply 52 by 53 in your head.
What two numbers turn a quadratic into two brackets?
Expand term by term:
Notice where the and the came from. The is and the is . The two numbers in the brackets added to make the middle coefficient and multiplied to make the constant.
That observation runs the whole of this page, in both directions. Forwards it multiplies brackets without writing out four products. Backwards it factorises: given , look for two numbers adding to and multiplying to — and and announce themselves.
It also multiplies numbers. Since and ,
One identity, three jobs — expansion, factorisation and mental arithmetic. This page covers the second part of the CBSE Class 9 Mathematics chapter on algebraic identities: the product of two binomials, the square of a three-term expression, and factorising by splitting the middle term.
Notice where the and the came from. The is and the is . The two numbers in the brackets added to make the middle coefficient and multiplied to make the constant.
That observation runs the whole of this page, in both directions. Forwards it multiplies brackets without writing out four products. Backwards it factorises: given , look for two numbers adding to and multiplying to — and and announce themselves.
It also multiplies numbers. Since and ,
One identity, three jobs — expansion, factorisation and mental arithmetic. This page covers the second part of the CBSE Class 9 Mathematics chapter on algebraic identities: the product of two binomials, the square of a three-term expression, and factorising by splitting the middle term.
Formula
How do you expand two brackets and a three-term square?
The product of two binomials sharing a first term:
The square of a three-term expression:
The first is four products with the two middle ones collected. The second is nine products: three squares and each pair counted twice.
Worked example 1. Expand . Sum , product :
Worked example 2 — with a negative. Expand . Here and , so the sum is and the product is :
Check at : the brackets give , and the expansion gives . Correct.
Worked example 3 — both negative. Expand . Sum , product :
Worked example 4 — the three-term square. Expand . Squares first, then each pair doubled:
- squares:
- the with pair doubled:
- the with pair doubled:
- the with pair doubled:
Check at : the bracket is , and the expansion gives . Correct.
Worked example 5 — a minus inside the three-term square. Expand by treating the second term as :
The squares are always positive; only the pair terms carry signs. That is worth stating as a rule, because catches people out in exactly the way does.
The interpretation as area. is the area of a rectangle with sides and , which splits into a square , two strips and , and a corner — the same four pieces the algebra produces. For the picture is a square cut into nine regions: three along the diagonal giving , , , and six off-diagonal ones pairing up into the doubled products , and . The doubling is not a rule to memorise; it is the fact that each rectangle appears twice, once on each side of the diagonal.
The square of a three-term expression:
The first is four products with the two middle ones collected. The second is nine products: three squares and each pair counted twice.
Worked example 1. Expand . Sum , product :
Worked example 2 — with a negative. Expand . Here and , so the sum is and the product is :
Check at : the brackets give , and the expansion gives . Correct.
Worked example 3 — both negative. Expand . Sum , product :
Worked example 4 — the three-term square. Expand . Squares first, then each pair doubled:
- squares:
- the with pair doubled:
- the with pair doubled:
- the with pair doubled:
Check at : the bracket is , and the expansion gives . Correct.
Worked example 5 — a minus inside the three-term square. Expand by treating the second term as :
The squares are always positive; only the pair terms carry signs. That is worth stating as a rule, because catches people out in exactly the way does.
The interpretation as area. is the area of a rectangle with sides and , which splits into a square , two strips and , and a corner — the same four pieces the algebra produces. For the picture is a square cut into nine regions: three along the diagonal giving , , , and six off-diagonal ones pairing up into the doubled products , and . The doubling is not a rule to memorise; it is the fact that each rectangle appears twice, once on each side of the diagonal.
How do algebra tiles show you the factorisation?
Lay the tiles out and rearrange them into a rectangle. The rectangle's two sides are the factors.
The tile set has three shapes: a large square of area , a long rectangle of area , and a small unit square of area .
**Worked example 1 — .** Take one tile, seven tiles and twelve unit tiles.
Put the tile in a corner. The tiles must line the two sides of it, and the unit tiles fill the corner left over. Try splitting the seven tiles as along one side and along the other. Then the corner gap is by — needing exactly unit tiles, which is what we have.
The assembled rectangle measures by , so
**Why the split has to be and .** If the tiles were split and , the corner gap would be units, and we hold — two tiles left over and no place for them. A split of and leaves a gap of , wanting six spare. Only a split whose product matches the unit count closes the rectangle, and that is the tile picture of add to the middle, multiply to the end.
**Worked example 2 — .** One tile, five tiles, six units. Split the five as and ; the corner gap is , matching:
Worked example 3 — a case with no rectangle. Try . The splits of are , giving a corner of , and , giving a corner of . Neither is , so no rectangle can be built — and indeed has no factorisation with whole numbers.
So the tiles do more than illustrate. A completed rectangle is a factorisation, and the impossibility of a rectangle is evidence that no whole-number factorisation exists. Once the reasoning is clear the tiles can be set aside, because searching splits on paper is faster — which is the next section.
One limit of the model. Tiles have positive areas, so cannot be laid out directly; it needs negative tiles or a sign argument. The algebra handles negatives without difficulty, and that is the usual relationship between a physical model and the symbols: the model builds the intuition and the symbols carry it further.
The tile set has three shapes: a large square of area , a long rectangle of area , and a small unit square of area .
**Worked example 1 — .** Take one tile, seven tiles and twelve unit tiles.
Put the tile in a corner. The tiles must line the two sides of it, and the unit tiles fill the corner left over. Try splitting the seven tiles as along one side and along the other. Then the corner gap is by — needing exactly unit tiles, which is what we have.
The assembled rectangle measures by , so
**Why the split has to be and .** If the tiles were split and , the corner gap would be units, and we hold — two tiles left over and no place for them. A split of and leaves a gap of , wanting six spare. Only a split whose product matches the unit count closes the rectangle, and that is the tile picture of add to the middle, multiply to the end.
**Worked example 2 — .** One tile, five tiles, six units. Split the five as and ; the corner gap is , matching:
Worked example 3 — a case with no rectangle. Try . The splits of are , giving a corner of , and , giving a corner of . Neither is , so no rectangle can be built — and indeed has no factorisation with whole numbers.
So the tiles do more than illustrate. A completed rectangle is a factorisation, and the impossibility of a rectangle is evidence that no whole-number factorisation exists. Once the reasoning is clear the tiles can be set aside, because searching splits on paper is faster — which is the next section.
One limit of the model. Tiles have positive areas, so cannot be laid out directly; it needs negative tiles or a sign argument. The algebra handles negatives without difficulty, and that is the usual relationship between a physical model and the symbols: the model builds the intuition and the symbols carry it further.
How do you split the middle term of a quadratic?
**Find two numbers whose product is the constant term and whose sum is the coefficient of **, then break the middle term into those two pieces and group.
The method compares the expression with the identity:
**Worked example 1 — .** Need a product of and a sum of . The factor pairs of are , , , with sums , , . The last one works:
**Worked example 2 — .** Product , sum . A positive product with a negative sum means both numbers are negative: and :
**Worked example 3 — .** Product , sum . A negative product means opposite signs. Pairs multiplying to : , , , , , , with sums , , , , , . The pair and gives sum :
Check by expanding: . Correct.
**Worked example 4 — .** Product , sum . Opposite signs with the larger one negative: and , since and :
The sign rules, stated once. Let the constant be and the middle coefficient be .
- : the two numbers have the same sign, and that sign is the sign of
- : the two numbers have opposite signs, and the larger in size takes the sign of
Applying these first cuts the search roughly in half, because it fixes the signs before any pair is tested.
Worked example 5 — when nothing works. Factorise . Product , sum . The only whole-number pair for is and , summing to . No factorisation exists over the integers, which matches the tile picture from the previous section and is a legitimate answer to write down.
Grouping is what actually finishes the job. Splitting the middle term is only step one; the factorisation comes from taking the common factor out of each pair, and the two brackets must come out identical — works because both show . If they differ, the split was wrong, and that is the built-in check: no separate verification needed, the method tells you itself.
The method compares the expression with the identity:
**Worked example 1 — .** Need a product of and a sum of . The factor pairs of are , , , with sums , , . The last one works:
**Worked example 2 — .** Product , sum . A positive product with a negative sum means both numbers are negative: and :
**Worked example 3 — .** Product , sum . A negative product means opposite signs. Pairs multiplying to : , , , , , , with sums , , , , , . The pair and gives sum :
Check by expanding: . Correct.
**Worked example 4 — .** Product , sum . Opposite signs with the larger one negative: and , since and :
The sign rules, stated once. Let the constant be and the middle coefficient be .
- : the two numbers have the same sign, and that sign is the sign of
- : the two numbers have opposite signs, and the larger in size takes the sign of
Applying these first cuts the search roughly in half, because it fixes the signs before any pair is tested.
Worked example 5 — when nothing works. Factorise . Product , sum . The only whole-number pair for is and , summing to . No factorisation exists over the integers, which matches the tile picture from the previous section and is a legitimate answer to write down.
Grouping is what actually finishes the job. Splitting the middle term is only step one; the factorisation comes from taking the common factor out of each pair, and the two brackets must come out identical — works because both show . If they differ, the split was wrong, and that is the built-in check: no separate verification needed, the method tells you itself.
How do you multiply 52 by 53 in your head?
Write both numbers with the same round part, then use with the round number as .
**Worked example 1 — .** Take , , :
Three easy pieces: the square of the round number, the sum times the round number, and the small product.
**Worked example 2 — .** Take , , . The sum is and the product is :
Worked example 3 — mixed signs. Compute with , , . Sum , product :
Check the last digit: ends in , and ends in . Correct.
Worked example 4 — near 40. Compute with , , . Sum , product :
Worked example 5 — near 60. Compute with , , . Sum , product :
Why this is more general than the difference of squares. The previous part of this chapter handled , where the two numbers are equally spaced either side of , so the middle term vanishes and only remains. This identity needs no such symmetry — it works for too:
**So is the special case where , and recognising which situation you are in decides whether the middle term survives.
Choose the round number that makes both adjustments small.** For , using gives adjustments of and ; using would give and , still correct but with a harder product of and a harder sum of . The method works from any base and is only worth doing from a well-chosen one.
**Worked example 1 — .** Take , , :
Three easy pieces: the square of the round number, the sum times the round number, and the small product.
**Worked example 2 — .** Take , , . The sum is and the product is :
Worked example 3 — mixed signs. Compute with , , . Sum , product :
Check the last digit: ends in , and ends in . Correct.
Worked example 4 — near 40. Compute with , , . Sum , product :
Worked example 5 — near 60. Compute with , , . Sum , product :
Why this is more general than the difference of squares. The previous part of this chapter handled , where the two numbers are equally spaced either side of , so the middle term vanishes and only remains. This identity needs no such symmetry — it works for too:
**So is the special case where , and recognising which situation you are in decides whether the middle term survives.
Choose the round number that makes both adjustments small.** For , using gives adjustments of and ; using would give and , still correct but with a harder product of and a harder sum of . The method works from any base and is only worth doing from a well-chosen one.
Exam tip
Exam tip: fix the signs before you start searching for the pair
Read the sign of the constant first. If both numbers share the sign of ; if they have opposite signs and the larger takes the sign of . That halves the search before you test anything.
List the factor pairs with their sums. For : , , — then pick. Written out, it is one line and it cannot go wrong.
Show the split and the grouping: . Marks are awarded for the middle steps.
The two brackets after grouping must be identical. If they are not, the split was wrong — that is the method's own check.
"Cannot be factorised over the integers" is a valid answer. For , show the pair search failing.
**For , write the three squares then the three doubled pairs** — — and remember all three squares are positive even when a term is negative.
**Check every expansion by substituting for each letter.** at gives , and .
For mental products, pick the base that makes both adjustments small — for , not .
Verify the last digit against the ordinary product: must end in .
And for tile questions, draw the rectangle with its two sides labelled and ; the diagram is the answer, not a decoration for it.
List the factor pairs with their sums. For : , , — then pick. Written out, it is one line and it cannot go wrong.
Show the split and the grouping: . Marks are awarded for the middle steps.
The two brackets after grouping must be identical. If they are not, the split was wrong — that is the method's own check.
"Cannot be factorised over the integers" is a valid answer. For , show the pair search failing.
**For , write the three squares then the three doubled pairs** — — and remember all three squares are positive even when a term is negative.
**Check every expansion by substituting for each letter.** at gives , and .
For mental products, pick the base that makes both adjustments small — for , not .
Verify the last digit against the ordinary product: must end in .
And for tile questions, draw the rectangle with its two sides labelled and ; the diagram is the answer, not a decoration for it.
Did you know
Why some quadratics refuse to become two brackets
splits into without trouble. refuses, and no amount of searching helps — the pairs multiplying to are and , and they sum to , not .
There is a way to see the refusal rather than just discover it. Rewrite the expression by completing the square:
Check at : on the left, on the right. Correct.
Now the obstruction is visible. is never negative, so the whole expression is never less than . **It has no value of making it zero** — and a factorisation would hand you two such values immediately, namely and .
So the expression cannot be factorised because its graph never touches the -axis. It sits above the axis at every point, dipping to at and rising on both sides.
Compare , which the same treatment turns into — a square minus something, which does reach zero, at and . The sign of that leftover constant is the whole difference between a quadratic that factorises and one that does not.
That leftover has a name in the next class: the discriminant. But the idea is available now, and it converts "I could not find the pair" into "no pair exists, and here is why" — a much better sentence to write in an answer.
There is a way to see the refusal rather than just discover it. Rewrite the expression by completing the square:
Check at : on the left, on the right. Correct.
Now the obstruction is visible. is never negative, so the whole expression is never less than . **It has no value of making it zero** — and a factorisation would hand you two such values immediately, namely and .
So the expression cannot be factorised because its graph never touches the -axis. It sits above the axis at every point, dipping to at and rising on both sides.
Compare , which the same treatment turns into — a square minus something, which does reach zero, at and . The sign of that leftover constant is the whole difference between a quadratic that factorises and one that does not.
That leftover has a name in the next class: the discriminant. But the idea is available now, and it converts "I could not find the pair" into "no pair exists, and here is why" — a much better sentence to write in an answer.
Exam relevance
How does splitting the middle term feed into JEE Main?
Because factorising a quadratic is a step inside an enormous number of later problems, and the identities here are the case of a theorem examined directly.
This is the foundation for Class 10 Quadratic Equations and Class 11 Mathematics Complex Numbers and Quadratic Equations and Binomial Theorem, all examined in JEE Main. The identity is the source of the relationships between roots and coefficients: if the roots are and then the sum is and the product is — which is exactly adds to the middle, multiplies to the end, written in the language of roots. Every question about the sum or product of roots is this page's identity.
The square of three terms is used constantly in that chapter and in Class 11 Straight Lines. The rearrangement is a standard move for evaluating symmetric expressions when only the sum and the pairwise sum are given — a recurring JEE Main question type with a one-line solution for a student who knows it.
Completing the square, met in the previous section as a way to see why some quadratics do not factorise, becomes the main technique later. It converts into standard form for the quadratic formula, puts a circle's equation into centre-radius form in Class 11 Conic Sections, and is how the maximum or minimum of a quadratic is found without calculus.
Where the impossibility matters. The discriminant of Class 10 and Class 11 answers the question this page raises — whether a quadratic factorises over the integers, over the reals, or only over the complex numbers. Assertion-reason questions on the nature of the roots are standard, and they rest on the observation made here that a factorisation hands you the zeros.
What the questions look like. For board work, expect factorise by splitting the middle term with the grouping shown, expand using an identity, the algebra-tile rectangle with labelled sides, and evaluate a product mentally — each self-contained. For JEE Main, factorisation is never the question; it is how a rational expression gets simplified, a denominator gets cleared, or an inequality gets solved by the sign of each bracket.
How board and competitive emphasis differ. A board paper rewards the written split and grouping. A competitive paper rewards spotting the pair instantly and, more importantly, knowing when not to bother — using the sum and product of roots directly instead of finding them.
The single trap that costs the most marks. Getting the signs of the pair the wrong way round when the constant is negative. For the numbers are and , not and — both pairs multiply to , but only one sums to . The product test passes either way, so the sum is the test that matters, and checking it takes one addition.
This is the foundation for Class 10 Quadratic Equations and Class 11 Mathematics Complex Numbers and Quadratic Equations and Binomial Theorem, all examined in JEE Main. The identity is the source of the relationships between roots and coefficients: if the roots are and then the sum is and the product is — which is exactly adds to the middle, multiplies to the end, written in the language of roots. Every question about the sum or product of roots is this page's identity.
The square of three terms is used constantly in that chapter and in Class 11 Straight Lines. The rearrangement is a standard move for evaluating symmetric expressions when only the sum and the pairwise sum are given — a recurring JEE Main question type with a one-line solution for a student who knows it.
Completing the square, met in the previous section as a way to see why some quadratics do not factorise, becomes the main technique later. It converts into standard form for the quadratic formula, puts a circle's equation into centre-radius form in Class 11 Conic Sections, and is how the maximum or minimum of a quadratic is found without calculus.
Where the impossibility matters. The discriminant of Class 10 and Class 11 answers the question this page raises — whether a quadratic factorises over the integers, over the reals, or only over the complex numbers. Assertion-reason questions on the nature of the roots are standard, and they rest on the observation made here that a factorisation hands you the zeros.
What the questions look like. For board work, expect factorise by splitting the middle term with the grouping shown, expand using an identity, the algebra-tile rectangle with labelled sides, and evaluate a product mentally — each self-contained. For JEE Main, factorisation is never the question; it is how a rational expression gets simplified, a denominator gets cleared, or an inequality gets solved by the sign of each bracket.
How board and competitive emphasis differ. A board paper rewards the written split and grouping. A competitive paper rewards spotting the pair instantly and, more importantly, knowing when not to bother — using the sum and product of roots directly instead of finding them.
The single trap that costs the most marks. Getting the signs of the pair the wrong way round when the constant is negative. For the numbers are and , not and — both pairs multiply to , but only one sums to . The product test passes either way, so the sum is the test that matters, and checking it takes one addition.
Key takeaways
Binomial products, trinomial squares and splitting: quick revision
- ** — the two numbers add to the middle coefficient and multiply to the constant.
- ** — three squares and each pair doubled.
- ; ; .
- , checked at as .
- — all three squares stay positive; only the pair terms carry signs.
- The doubling happens because each off-diagonal rectangle in the square picture appears twice.
- Algebra tiles: one tile, seven tiles and twelve units build an by rectangle, so .
- A split of and leaves a corner of against tiles held — only the matching split closes the rectangle.
- builds no rectangle, and has no integer factorisation.
- Splitting the middle term: find a product of and a sum of , split, then group.
- .
- ; ; .
- Sign rules: means both numbers share the sign of ; means opposite signs with the larger taking the sign of .
- The two brackets after grouping must match — if they differ, the split was wrong.
- Mental products: ; ; ; ; .
- ** is the special case where the two adjustments sum to zero**; otherwise the middle term survives, as in .
- Pick the base that keeps both adjustments small, and check the last digit against the ordinary product.
- is never zero, which is why it does not factorise; does reach zero.
Take any two numbers in the fifties, multiply them with this identity, then check with ordinary multiplication — and try to beat your own time on the second attempt.
- ** — three squares and each pair doubled.
- ; ; .
- , checked at as .
- — all three squares stay positive; only the pair terms carry signs.
- The doubling happens because each off-diagonal rectangle in the square picture appears twice.
- Algebra tiles: one tile, seven tiles and twelve units build an by rectangle, so .
- A split of and leaves a corner of against tiles held — only the matching split closes the rectangle.
- builds no rectangle, and has no integer factorisation.
- Splitting the middle term: find a product of and a sum of , split, then group.
- .
- ; ; .
- Sign rules: means both numbers share the sign of ; means opposite signs with the larger taking the sign of .
- The two brackets after grouping must match — if they differ, the split was wrong.
- Mental products: ; ; ; ; .
- ** is the special case where the two adjustments sum to zero**; otherwise the middle term survives, as in .
- Pick the base that keeps both adjustments small, and check the last digit against the ordinary product.
- is never zero, which is why it does not factorise; does reach zero.
Take any two numbers in the fifties, multiply them with this identity, then check with ordinary multiplication — and try to beat your own time on the second attempt.