Class 9 Mathematics Chapter 4: Exploring Algebraic Identities
- 3 topics
- 132 MCQs
- 30 short answers
- 12 long answers
- 60 flashcards
- 12 games
- ~130 min read
- English
The Square Identities
An identity is an equation true for every value of its letters, and the square identities can all be read off a picture: cut a square of side a + b and the four pieces give a squared plus two ab rectangles plus b squared. Run the same identities backwards and they factorise; use them on round numbers and they square three-digit numbers in your head.
Full explanation
A pattern that always gives 2
Take any three consecutive square numbers, say $1$, $4$ and $9$. Add the outer two and subtract twice the middle one:
$$1 + 9 - (2 \times 4) = 10 - 8 = 2$$
Try it again with $25$, $36$, $49$:
$$25 + 49 - (2 \times 36) = 74 - 72 = 2$$
Always 2. Try as many as you like; it never fails. Why?
You could keep checking examples for ever and never know. Algebra answers it in one line, and that is what this chapter is for.
What an identity is
An identity is an equation that is true for every value of its letters.
That word every is the whole difference. Compare:
- $x + 3 = 7$ is an equation. It is true for one value of $x$, namely 4, and false for all the rest. There is something to solve.
- $(a + b)^2 = a^2 + 2ab + b^2$ is an identity. Put in any numbers you like, positive, negative or fractional, and both sides agree. There is nothing to solve.
An identity is a tool. Once established, it may be used in either direction — left to right to expand, and right to left to factorise.
The square of a sum, seen as an area
Draw a square of side $a + b$ and cut it with one horizontal and one vertical line:
- a square of side $a$, area $a^2$
- a square of side $b$, area $b^2$
- two rectangles, each $a$ by $b$, area $ab$ each
The whole square has area $(a+b)^2$, and the four pieces make it up:
$$\boxed{(a + b)^2 = a^2 + 2ab + b^2}$$
With $a = 3$ and $b = 2$: $9 + 6 + 6 + 4 = 25$, and $5^2 = 25$. It works.
Why the middle term is doubled
The picture answers the commonest question in the chapter. There are two rectangles, not one — one above the $a$-square and one beside it. That is where the 2 in $2ab$ comes from.
And it explains the commonest error too:
$$(a + b)^2 \neq a^2 + b^2$$
Writing $a^2 + b^2$ is claiming the two rectangles have no area at all. Test it: $(3 + 2)^2 = 25$, but $3^2 + 2^2 = 13$. Not close.
The square of a difference
Change the sign of $b$ in the first identity. Wherever $b$ appears an odd number of times, the sign flips:
$$\boxed{(a - b)^2 = a^2 - 2ab + b^2}$$
The middle term turns negative because it holds one $b$; the last term stays positive because $(-b)^2 = b^2$. A square is never negative, so that last sign is not a matter of choice.
The difference of two squares
Multiply a sum by the matching difference and the middle terms destroy each other:
$$(a + b)(a - b) = a^2 - ab + ab - b^2 = a^2 - b^2$$
$$\boxed{(a + b)(a - b) = a^2 - b^2}$$
This is the most useful identity in the chapter, because it is easy to spot in both directions. Anything of the form square minus square factorises instantly.
The square of a sum of three
What if there are three terms? Replace $b + c$ by a single letter $d$ and use the first identity:
$$(a + d)^2 = a^2 + 2ad + d^2$$
Now put $d = b + c$ back, expand, and collect:
$$\boxed{(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca}$$
The area picture matches: a square of side $a+b+c$ cuts into nine pieces — three squares and three matching pairs of rectangles. With $a=3$, $b=2$, $c=1$: $9 + 4 + 1 + 12 + 4 + 6 = 36 = 6^2$.
Learn it as three squares, then twice each pair. Six terms, and every pair appears exactly once.
Squaring numbers in your head
These identities were built for calculation long before they were built for algebra.
Find $103^2$. Write it as $(100 + 3)^2$:
$$100^2 + 2(100)(3) + 3^2 = 10000 + 600 + 9 = 10609$$
Find $98^2$. Here the difference form is easier — $(100 - 2)^2$:
$$10000 - 400 + 4 = 9604$$
Find $102 \times 98$. Neither is a square, but the pair is a sum and a difference about 100:
$$(100 + 2)(100 - 2) = 10000 - 4 = 9996$$
Choose the identity that puts a round number in place of $a$. That is the whole skill.
And with three terms: $119^2 = (100 + 10 + 9)^2$, which the third identity turns into $10000 + 100 + 81 + 2000 + 1800 + 180 = 14161$.
Back to the pattern
Now the opening puzzle. Let the middle square number be $n^2$. The three consecutive squares are $(n-1)^2$, $n^2$ and $(n+1)^2$, so the calculation is:
$$(n-1)^2 + (n+1)^2 - 2n^2$$
Expand with the first two identities:
$$(n^2 - 2n + 1) + (n^2 + 2n + 1) - 2n^2$$
The $-2n$ and $+2n$ cancel, and the $2n^2$ is taken away, leaving
$$1 + 1 = 2$$
Always 2, for every $n$, and now you know why rather than merely that. This is what algebra is for: replacing "it keeps happening" with "it must happen".
Running the identities backwards
Read each identity right to left and it becomes a factorising tool.
Factorise $x^2 + 10x + 25$. The ends are squares, $x^2$ and $5^2$, and the middle is $2 \times x \times 5$. So it matches $a^2 + 2ab + b^2$ with $a = x$ and $b = 5$:
$$x^2 + 10x + 25 = (x + 5)^2$$
Factorise $4x^2 - 12x + 9$. Again the ends are squares, $(2x)^2$ and $3^2$, and the middle is $-2(2x)(3)$:
$$4x^2 - 12x + 9 = (2x - 3)^2$$
Factorise $9x^2 - 49$. Square minus square, so the difference identity applies at once:
$$9x^2 - 49 = (3x + 7)(3x - 7)$$
Which identity, and how to tell
A short checklist, and it settles almost every question:
| What you see | What to use |
|---|---|
| Square minus square, two terms | $(a+b)(a-b)$ |
| Three terms, both ends perfect squares, middle $= 2ab$ | $(a \pm b)^2$ |
| A number near a round one, to be squared | $(a \pm b)^2$ with $a$ round |
| Two numbers equally above and below a round one | $(a+b)(a-b)$ |
Check the middle term before committing. $x^2 + 7x + 9$ has square ends but $7x \neq 2 \times x \times 3$, so it is not a perfect square and none of these identities applies to it.
Slips that cost marks
- Writing $(a+b)^2 = a^2 + b^2$ and losing the two rectangles.
- Making the last term of $(a-b)^2$ negative. $(-b)^2$ is positive.
- Forgetting a pair in $(a+b+c)^2$ — there are six terms, not four.
- Calling $x^2 + 7x + 9$ a perfect square because both ends are squares. Test the middle.
- Applying the difference of squares to $a^2 + b^2$. A sum of two squares does not factorise this way.
What to write in the exam
Name the identity you are using and write it out before substituting — that line is usually worth a mark on its own. For mental arithmetic, show the split, as in $98^2 = (100 - 2)^2$, before expanding. And for a factorisation, check your answer by expanding it back; it takes ten seconds and catches every sign error.
Watch out for these
Practice MCQs with answers (46)
- A $x^2 + 12x + 36$
- B $x^2 + 36$
- C $x^2 + 6x + 36$
- D $x^2 + 12x + 6$
Show answer
Why:
Using $(a+b)^2 = a^2 + 2ab + b^2$ with $b = 6$ gives a middle term of $2 \times x \times 6 = 12x$.
- A $x^2 - 81$
- B $x^2 - 18x - 81$
- C $x^2 - 18x + 81$
- D $x^2 + 18x + 81$
Show answer
Why:
The middle term is $-2 \times x \times 9$, and the last is $(-9)^2 = 81$, which is positive.
- A $2x^2 + 20x + 25$
- B $4x^2 + 20x + 25$
- C $4x^2 + 10x + 25$
- D $4x^2 + 25$
Show answer
Why:
Here $a = 2x$, so $a^2 = 4x^2$ and $2ab = 2(2x)(5) = 20x$.
- A $9x^2 - 6x + 1$
- B $9x^2 - 1$
- C $3x^2 - 6x + 1$
- D $9x^2 - 6x - 1$
Show answer
Why:
$a = 3x$ gives $9x^2$, the middle is $-2(3x)(1) = -6x$, and the last is $(-1)^2 = 1$.
- A $x^2 + 121$
- B $x^2 - 121$
- C $x^2 - 22x - 121$
- D $x^2 - 22x + 121$
Show answer
Why:
A sum times the matching difference gives $a^2 - b^2$, and the middle terms cancel.
- A $25x^2 + 16$
- B $25x^2 - 16$
- C $5x^2 - 16$
- D $25x^2 - 40x - 16$
Show answer
Why:
With $a = 5x$ and $b = 4$, the result is $a^2 - b^2 = 25x^2 - 16$.
- A 11449
- B 11409
- C 10049
- D 14900
Show answer
Why:
$(100 + 7)^2 = 10000 + 1400 + 49 = 11449$.
- A 9616
- B 9216
- C 9016
- D 8836
Show answer
Why:
$(100 - 4)^2 = 10000 - 800 + 16 = 9216$.
- A 10025
- B 9950
- C 9975
- D 10000
Show answer
Why:
The two numbers straddle 100, so $(100+5)(100-5) = 10000 - 25 = 9975$.
- A 40016
- B 39984
- C 39900
- D 40000
Show answer
Why:
They straddle 200, so $(200+4)(200-4) = 40000 - 16 = 39984$.
- A 2501
- B 2601
- C 2621
- D 2401
Show answer
Why:
$(50 + 1)^2 = 2500 + 100 + 1 = 2601$.
- A 998801
- B 999001
- C 980001
- D 998001
Show answer
Why:
$(1000 - 1)^2 = 1000000 - 2000 + 1 = 998001$.
- A 1936
- B 4400
- C 5184
- D 2000
Show answer
Why:
Factorising gives $(72+28)(72-28) = 100 \times 44 = 4400$.
- A 12321
- B 12221
- C 13321
- D 11121
Show answer
Why:
$(100 + 10 + 1)^2 = 10000 + 100 + 1 + 2000 + 20 + 200 = 12321$.
- A 81
- B 29
- C 49
- D 77
Show answer
Why:
$4 + 9 + 16 + 12 + 24 + 16 = 81$, and $9^2 = 81$ confirms it.
- A Six terms: $a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$
- B Three terms: $a^2 + b^2 + c^2$
- C Four terms: $a^2 + b^2 + c^2 + 2abc$
- D Nine terms, one for each piece of the square
Show answer
Why:
Three squares and three doubled products make six terms in all.
- A $-56x$
- B $-28x$
- C $+56x$
- D $-14x$
Show answer
Why:
The middle term is $-2(4x)(7) = -56x$.
- A An equation that is true for exactly one value of its letter
- B Any equation containing a square term
- C An equation that is true for every value of its letters
- D An equation that cannot be solved
Show answer
Why:
The word "every" is what separates an identity from an ordinary equation.
- A It leaves out the two rectangles of area $ab$ in the area picture
- B It is right only when $a$ and $b$ are both positive
- C The squares should be added rather than multiplied
- D It is right; the middle term cancels out
Show answer
Why:
Testing with $a = 3$ and $b = 2$ gives 25 against 13, and the $2ab$ accounts for the gap.
- A Because the identity was written for positive values only
- B Because two minus signs are always dropped
- C Because it is $(-b)^2$, and squaring always gives a positive result
- D Because $b$ must be larger than $a$
Show answer
Why:
A square is never negative, so the sign there is forced rather than chosen.
- A $25x^2 + 36$
- B $25x^2 - 36$
- C $25x^2 - 36x$
- D $25x + 36$
Show answer
Why:
Both terms must be perfect squares with a minus between them, which only the first satisfies.
- A Yes, as $(a+b)(a+b)$
- B Yes, as $(a+b)(a-b)$
- C No, only a difference of two squares factorises
- D Yes, as $(a+b)^2$ with the middle term removed
Show answer
Why:
The difference identity requires a minus sign between the squares, and a sum has none.
- A Yes, because both $x^2$ and 9 are perfect squares
- B Yes, because it has three terms
- C No, because 9 is not a perfect square
- D No, because $7x$ is not twice the product of $x$ and 3
Show answer
Why:
Square ends are not enough; the middle term must equal $2ab$, and here it does not.
- A The middle term must equal the sum of the two end terms
- B The middle term must be a perfect square as well
- C The middle term must be twice the product of the two square roots
- D The two end terms must be equal
Show answer
Why:
That is exactly the condition $2ab$ imposes in the identity.
- A $(a - b)^2$, with $a = 100$ and $b = 3$
- B $(a + b)^2$, with $a = 90$ and $b = 7$
- C $(a+b)(a-b)$, with $a = 97$
- D $(a+b+c)^2$, with $a = 90$, $b = 7$ and $c = 0$
Show answer
Why:
The split that puts a round number in place of $a$ makes the arithmetic easiest.
- A $(a+b)^2$, since both numbers are close together
- B $(a-b)^2$, since one number is smaller
- C None; the identities do not help with products
- D $(a+b)(a-b)$, since the numbers straddle 60
Show answer
Why:
They sit equally above and below 60, so the product is $60^2 - 3^2$.
- A Six
- B Four
- C Three
- D Nine
Show answer
Why:
Three squares and six rectangles, which group into six terms.
- A A different expression results each time
- B The identity can only be used in one direction
- C The result is always a perfect square
- D The original expression comes back, which is why factorising can be checked by expanding
Show answer
Why:
An identity holds in both directions, which is what makes the check reliable.
- A They are equal, since addition may be done in either order
- B They differ by $2ab$
- C They are equal only when $a = b$
- D They differ in sign
Show answer
Why:
The identity is symmetric in $a$ and $b$, so swapping them changes nothing.
- A $(x + 14)^2$
- B $(x + 7)(x - 7)$
- C $(x + 7)^2$
- D $(x + 49)(x + 1)$
Show answer
Why:
The ends are $x^2$ and $7^2$, and $14x = 2 \times x \times 7$.
- A $(x - 8)^2$
- B $(x + 8)^2$
- C $(x - 8)(x + 8)$
- D $(x - 16)(x - 4)$
Show answer
Why:
The ends are $x^2$ and $8^2$, and the middle $-16x$ is $-2 \times x \times 8$.
- A $(4x - 9)^2$
- B $(16x + 81)(16x - 81)$
- C $(4x + 9)^2$
- D $(4x + 9)(4x - 9)$
Show answer
Why:
This is $(4x)^2 - 9^2$, a difference of two squares.
- A $(x - 11)^2$
- B $(x + 11)(x - 11)$
- C $(x + 11)^2$
- D $(x - 121)(x + 1)$
Show answer
Why:
Square minus square, with $b = 11$.
- A $(x + 8)^2$
- B $(x + 4)(x - 4)$
- C $(x + 16)(x + 1)$
- D $(x + 4)^2$
Show answer
Why:
The ends are $x^2$ and $4^2$, and $8x = 2 \times x \times 4$.
- A $(2x + 3)^2$
- B $(2x - 3)^2$
- C $(2x - 3)(2x + 3)$
- D $(4x - 3)(x - 3)$
Show answer
Why:
The ends are $(2x)^2$ and $3^2$, and $-12x = -2(2x)(3)$.
- A $(a + b)$ and $(a + b)$
- B $(a + b)$ and $(a - b)$
- C $(a - b)$ and $(a - b)$
- D $(a^2 + b)$ and $(a^2 - b)$
Show answer
Why:
A sum times the matching difference is exactly what produces a difference of squares.
- A $(3x - 5)^2$
- B $(9x + 5)^2$
- C $(3x + 5)^2$
- D $(3x + 25)^2$
Show answer
Why:
The ends give $3x$ and 5, and $2(3x)(5) = 30x$ matches the middle term.
- A Because the examples might be chosen badly
- B Because squares are hard to calculate
- C Because no finite number of examples covers every possible case
- D It does prove it, once enough examples are tried
Show answer
Why:
Algebra settles every case at once, which is exactly what examples cannot do.
- A $2n$
- B $2n^2$
- C 0
- D 2
Show answer
Why:
The expansions give $2n^2 + 2$ before the $2n^2$ is removed, and the $2n$ terms cancel.
- A Six, each once
- B Two, each doubled
- C One, tripled
- D Three, each doubled
Show answer
Why:
The pairs are $ab$, $bc$ and $ca$, and each appears twice in the expansion.
- A One
- B Four
- C Two
- D None
Show answer
Why:
One sits above the $a$-square and one beside it, which is why the middle term is doubled.
- A Exactly one
- B Exactly two
- C Only the positive ones
- D All of them
Show answer
Why:
Holding for every value is the defining property of an identity.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
Both are true, and R is exactly what the missing $2ab$ accounts for.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
Both statements are true, but the general definition of an identity is not what makes this particular factorisation work; the difference-of-squares identity is.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
A is true, since $(-b)^2 = b^2$. R is false: squaring a negative number gives a positive result.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
A is false — expanding $(x+6)(x-6)$ gives $x^2 - 36$, not $x^2 + 36$. R is true, and it is why A fails.
Important short questions (10)
Using $(a+b)^2 = a^2 + 2ab + b^2$ with $a = 3x$ and $b = 8$: the first term is $9x^2$, the middle is $2(3x)(8) = 48x$, and the last is $64$. So the expansion is $9x^2 + 48x + 64$.
Writing $94 = 100 - 6$ and using $(a-b)^2 = a^2 - 2ab + b^2$ gives $10000 - 2(100)(6) + 36 = 10000 - 1200 + 36 = 8836$.
The two numbers sit equally above and below 200, so they are $200 + 6$ and $200 - 6$. By the difference of squares, the product is $200^2 - 6^2 = 40000 - 36 = 39964$.
Both end terms are perfect squares, $x^2$ and $10^2$, and the middle term is $2 \times x \times 10 = 20x$. So the expression is $(x + 10)^2$.
The expression is $(7x)^2 - 5^2$, a difference of two squares, so it factorises as $(7x + 5)(7x - 5)$.
Numerically, take $a = 3$ and $b = 2$. Then $(a+b)^2 = 5^2 = 25$, while $a^2 + b^2 = 9 + 4 = 13$. The two differ by 12, which is exactly $2ab$. The area picture shows why: a square of side $a + b$ divides into a square of side $a$, a square of side $b$, and two rectangles each measuring $a$ by $b$. Writing $a^2 + b^2$ counts only the two squares and throws the two rectangles away.
The expression is a difference of two squares, so it factorises as $(85 + 15)(85 - 15) = 100 \times 70 = 7000$. The method works because the identity $a^2 - b^2 = (a+b)(a-b)$ holds for every value of $a$ and $b$, and here it replaces two large squarings with one easy multiplication.
Both end terms are perfect squares, since $x^2 = (x)^2$ and $9 = 3^2$. For the expression to be a perfect square the middle term would have to be $2 \times x \times 3 = 6x$. The middle term given is $7x$, which is not $6x$, so the expression is not a perfect square. The general test is that both ends must be perfect squares and the middle term must equal twice the product of their square roots.
With $a = 1$, $b = 2$ and $c = 3$: the squares give $1 + 4 + 9 = 14$, and the doubled products give $2(1)(2) + 2(2)(3) + 2(3)(1) = 4 + 12 + 6 = 22$. The total is $14 + 22 = 36$. Checking directly, $1 + 2 + 3 = 6$ and $6^2 = 36$, which agrees.
An equation is a statement that two expressions are equal, and it is generally true only for particular values of the letter. For instance $x + 3 = 7$ holds when $x = 4$ and fails otherwise, so there is something to solve. An identity is an equation that holds for every value of its letters, such as $(a+b)^2 = a^2 + 2ab + b^2$; substituting any numbers at all makes both sides agree, so there is nothing to solve and the statement may instead be used as a tool.
Long answer questions (4)
(i) Using $(a+b)^2 = a^2 + 2ab + b^2$ with $a = 5x$ and $b = 2$: $25x^2 + 20x + 4$.
(ii) Using $(a-b)^2 = a^2 - 2ab + b^2$ with $a = x$ and $b = 12$: $x^2 - 24x + 144$.
(iii) Using $(a+b)(a-b) = a^2 - b^2$ with $a = 8x$ and $b = 3$: $64x^2 - 9$.
(iv) The ends are $(3x)^2$ and $4^2$, and the middle term is $2(3x)(4) = 24x$, which matches. So the factorisation is $(3x + 4)^2$.
(v) Expanding $(3x+4)^2$ gives $9x^2 + 2(3x)(4) + 16 = 9x^2 + 24x + 16$, which is the original expression, so the factorisation is correct.
(i) $103^2 = (100 + 3)^2 = 10000 + 600 + 9 = 10609$.
(ii) $97^2 = (100 - 3)^2 = 10000 - 600 + 9 = 9409$.
(iii) The two numbers straddle 100, so $103 \times 97 = (100+3)(100-3) = 10000 - 9 = 9991$.
(iv) Squaring a single number needs the square identity, with the sign chosen to match whether the number is above or below the round value. Multiplying the two together is a different operation, and because one is as far above 100 as the other is below, it is exactly a difference of two squares. The identity to reach for is decided by the operation, not by the numbers.
(v) $119^2 = (100 + 10 + 9)^2 = 10000 + 100 + 81 + 2(100)(10) + 2(10)(9) + 2(9)(100) = 10000 + 100 + 81 + 2000 + 180 + 1800 = 14161$.
(i) Add the smallest and largest of the three squares, then subtract twice the middle one.
(ii) $16 + 36 - 2(25) = 52 - 50 = 2$, as expected.
(iii) The three squares are $(n-1)^2$, $n^2$ and $(n+1)^2$, so the expression is $(n-1)^2 + (n+1)^2 - 2n^2$.
(iv) Expanding, $(n^2 - 2n + 1) + (n^2 + 2n + 1) - 2n^2$. The $-2n$ and $+2n$ cancel, and the two $n^2$ terms are removed by the $-2n^2$, leaving $1 + 1 = 2$.
(v) The letter $n$ stands for any whole number at all, so the single calculation covers every possible set of three consecutive squares at once. Examples, however many are tried, only ever confirm the cases actually tested and can never rule out an untested one.
(i) $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
(ii) A square of side $a + b + c$ is cut by two horizontal and two vertical lines into nine pieces: three squares of sides $a$, $b$ and $c$, and three matching pairs of rectangles measuring $a$ by $b$, $b$ by $c$ and $c$ by $a$. The three squares give the first three terms and each pair of rectangles gives one doubled product.
(iii) With $a = x$, $b = 2y$ and $c = 3$: $x^2 + 4y^2 + 9 + 2(x)(2y) + 2(2y)(3) + 2(3)(x) = x^2 + 4y^2 + 9 + 4xy + 12y + 6x$.
(iv) The difference of two squares identity requires a minus sign between the two square terms. Here the two squares are added, and no pair of brackets with real numbers multiplies out to a sum of two squares, so the expression does not factorise by these identities.
(v) Expanding gives $(x^2 + 4x + 4) - (x^2 - 4x + 4)$. The $x^2$ and the $4$ cancel, leaving $4x + 4x = 8x$.
Flashcards (20)
Play this topic
- Match each expression to its factorisation
- True or false about identities
- Complete the identity
- Unscramble the vocabulary
Factorising Quadratic Expressions
Lay the tiles of a quadratic out as a rectangle and its two side lengths are its factors. Without tiles the same thing is done by finding two numbers that multiply to the constant and add to the middle coefficient, and when the first term carries a coefficient, by splitting the middle term and factorising in pairs.
Full explanation
Build the rectangle and read off the answer
Suppose you are given the expression $x^2 + 5x + 6$ and asked to write it as two brackets multiplied together.
Here is a way to see the answer rather than guess it. Think of the expression as an area, and try to lay it out as a rectangle:
- one square of side $x$, area $x^2$
- five rectangles of size $x$ by 1, area $x$ each
- six unit squares, area 1 each
Arrange all of them into one complete rectangle with no gaps and no overlaps. It can be done, and when it is, the rectangle turns out to measure $(x + 2)$ by $(x + 3)$.
$$x^2 + 5x + 6 = (x + 2)(x + 3)$$
Those pieces are called algebra tiles, and the rectangle they form displays the factorisation directly: its two side lengths are the two factors.
What the picture is really saying
Look at where each part of the expression ends up:
- the $x^2$ tile fills one corner
- the six unit squares fill the opposite corner, in a $2$ by $3$ block
- the five $x$-tiles split into a row of 3 and a column of 2
So the numbers 2 and 3 do two jobs at once. They multiply to give 6, because they form the block of unit squares. And they add to give 5, because together they account for all five $x$-tiles.
That double role is the entire method, and it works without any tiles at all.
The identity behind it
Multiply out $(x + a)(x + b)$ and you see why:
$$(x + a)(x + b) = x^2 + bx + ax + ab$$
$$\boxed{(x + a)(x + b) = x^2 + (a + b)x + ab}$$
So to factorise $x^2 + bx + c$, look for two numbers that add to $b$ and multiply to $c$.
The method, and a worked case
Factorise $x^2 + 7x + 12$.
Find two numbers multiplying to 12 and adding to 7. List the factor pairs of 12 and check their sums:
| Pair | Product | Sum |
|---|---|---|
| 1 and 12 | 12 | 13 |
| 2 and 6 | 12 | 8 |
| 3 and 4 | 12 | 7 |
So $x^2 + 7x + 12 = (x + 3)(x + 4)$.
Always check the product first and the sum second. There are few factor pairs of a number but endless pairs adding to a given total, so starting from the product narrows the search immediately.
When the signs change
The signs of $b$ and $c$ tell you the signs of the two numbers before you begin looking.
| $c$ | $b$ | The two numbers |
|---|---|---|
| positive | positive | both positive |
| positive | negative | both negative |
| negative | either | one of each sign |
Both negative. $x^2 - 7x + 12$: the product 12 is positive so the signs match, and the sum $-7$ is negative so both are negative. The pair is $-3$ and $-4$, giving $(x - 3)(x - 4)$.
Opposite signs. $x^2 + x - 12$: the product $-12$ is negative, so one number is positive and one negative. The pair multiplying to $-12$ and adding to $+1$ is $4$ and $-3$, giving $(x + 4)(x - 3)$.
And the other way round. $x^2 - x - 12$ needs a sum of $-1$, so the larger number takes the minus: $-4$ and $3$, giving $(x - 4)(x + 3)$.
When the constant is negative, the larger of the two numbers carries the sign of the middle term. That one sentence settles most sign errors.
A coefficient in front of the $x^2$
When the first term is not simply $x^2$, the same idea works with one extra step. The identity is now
$$(ax + b)(cx + d) = acx^2 + (ad + bc)x + bd$$
and the practical method is called splitting the middle term.
Factorise $2x^2 + 7x + 3$.
- Multiply the first coefficient by the last: $2 \times 3 = 6$.
- Find two numbers multiplying to 6 and adding to the middle coefficient 7. They are 6 and 1.
- Split the middle term using them: $2x^2 + 6x + x + 3$.
- Factorise in pairs: $2x(x + 3) + 1(x + 3)$.
- The bracket $(x+3)$ is common, so take it out: $(2x + 1)(x + 3)$.
Step 5 is the one that feels like magic the first time. It is not — the pairs were chosen precisely so that the same bracket would appear twice.
Another, with a leading coefficient of 3
Factorise $3x^2 + 10x + 8$.
The product is $3 \times 8 = 24$ and the sum is 10, so the numbers are 6 and 4:
$$3x^2 + 6x + 4x + 8 = 3x(x + 2) + 4(x + 2) = (3x + 4)(x + 2)$$
And one with negatives throughout
Factorise $6x^2 - x - 2$.
The product is $6 \times (-2) = -12$ and the sum is $-1$, so the numbers are $-4$ and $3$:
$$6x^2 - 4x + 3x - 2 = 2x(3x - 2) + 1(3x - 2) = (3x - 2)(2x + 1)$$
Notice the $+1$ in the second pairing. When the second pair has no obvious common factor, the common factor is 1, and it must still be written — leaving it out is how the final bracket gets lost.
Always expand it back
Every factorisation can be checked in ten seconds, and it should be.
$$(3x - 2)(2x + 1) = 6x^2 + 3x - 4x - 2 = 6x^2 - x - 2 ;\checkmark$$
This catches sign errors, which are by far the commonest mistake here, and it needs no memory of the method at all.
When it simply does not factorise
Not every quadratic breaks into brackets with whole numbers.
Try $x^2 + 3x + 5$. The pairs multiplying to 5 are $1 \times 5$ and $(-1) \times (-5)$, with sums 6 and $-6$. Neither is 3, and there are no other pairs.
So this expression does not factorise over the integers. That is a perfectly good answer, and writing "cannot be factorised with integer factors" is worth more than forcing a wrong pair of brackets.
Slips that cost marks
- Searching from the sum rather than the product. The product has few pairs; the sum has endless ones.
- Getting the signs the wrong way round when the constant is negative.
- Dropping the $+1$ when a pair has no visible common factor.
- Taking out different brackets from the two pairs, which means the split was wrong — go back and re-choose.
- Forcing brackets onto a quadratic that does not factorise.
What to write in the exam
For a simple quadratic, state the two numbers and both conditions — "3 and 4, since $3 \times 4 = 12$ and $3 + 4 = 7$" — before writing the brackets. For a leading coefficient, show the product $ac$, the split of the middle term, and the pairing, as separate lines. Then expand your answer back. The working carries most of the marks.
Watch out for these
Practice MCQs with answers (44)
- A $(x + 2)(x + 5)$
- B $(x + 1)(x + 10)$
- C $(x - 2)(x - 5)$
- D $(x + 3)(x + 4)$
Show answer
Why:
The pair multiplying to 10 and adding to 7 is 2 and 5.
- A $(x + 2)(x + 3)$
- B $(x - 1)(x - 6)$
- C $(x + 7)(x + 1)$
- D $(x + 1)(x + 6)$
Show answer
Why:
The pair multiplying to 6 and adding to 7 is 1 and 6.
- A $(x + 2)(x + 5)$
- B $(x - 2)(x - 5)$
- C $(x - 1)(x - 10)$
- D $(x - 2)(x + 5)$
Show answer
Why:
A positive constant with a negative sum means both numbers are negative: $-2$ and $-5$.
- A $(x - 6)(x + 2)$
- B $(x + 6)(x - 2)$
- C $(x + 4)(x - 3)$
- D $(x + 12)(x - 1)$
Show answer
Why:
A negative constant needs opposite signs, and the larger number follows the middle term: 6 and $-2$.
- A $(x + 6)(x - 2)$
- B $(x - 6)(x + 2)$
- C $(x - 4)(x + 3)$
- D $(x - 12)(x + 1)$
Show answer
Why:
The sum is $-4$, so the larger number carries the minus: $-6$ and 2.
- A $(x + 1)(x + 8)$
- B $(x + 2)(x + 4)$
- C $(x + 3)(x + 6)$
- D $(x - 1)(x - 8)$
Show answer
Why:
1 and 8 multiply to 8 and add to 9.
- A $(x - 2)(x - 9)$
- B $(x - 3)(x - 6)$
- C $(x + 2)(x + 9)$
- D $(x - 1)(x - 18)$
Show answer
Why:
$-2$ and $-9$ multiply to 18 and add to $-11$.
- A $(x + 7)(x - 4)$
- B $(x - 7)(x + 4)$
- C $(x + 14)(x - 2)$
- D $(x + 7)(x + 4)$
Show answer
Why:
7 and $-4$ multiply to $-28$ and add to 3, with the larger number positive.
- A $(2x + 1)(x + 3)$
- B $(2x + 3)(x + 1)$
- C $(2x + 3)(x + 3)$
- D $(x + 3)(x + 1)$
Show answer
Why:
$ac = 6$, and 3 and 2 give $2x^2 + 3x + 2x + 3 = x(2x+3) + 1(2x+3)$.
- A $(3x + 2)(x + 1)$
- B $(x + 1)(x + 2)$
- C $(3x + 7)(x + 2)$
- D $(3x + 1)(x + 2)$
Show answer
Why:
$ac = 6$, and 6 and 1 give $3x^2 + 6x + x + 2 = 3x(x+2) + 1(x+2)$.
- A $(5x + 1)(x + 2)$
- B $(5x + 2)(x + 2)$
- C $(5x + 2)(x + 1)$
- D $(x + 2)(x + 1)$
Show answer
Why:
$ac = 10$, and 5 and 2 give $5x^2 + 5x + 2x + 2 = 5x(x+1) + 2(x+1)$.
- A $(4x - 1)(x - 3)$
- B $(2x - 1)(2x - 3)$
- C $(2x + 1)(2x + 3)$
- D $(4x - 3)(x - 1)$
Show answer
Why:
$ac = 12$, and $-2$ and $-6$ give $4x^2 - 2x - 6x + 3 = 2x(2x-1) - 3(2x-1)$.
- A $(2x - 5)(3x + 1)$
- B $(2x + 5)(3x - 1)$
- C $(6x + 5)(x - 1)$
- D $(3x + 5)(2x - 1)$
Show answer
Why:
$ac = -30$, and 15 and $-2$ give $6x^2 + 15x - 2x - 5 = 3x(2x+5) - 1(2x+5)$.
- A $(x - 9)^2$
- B $(x + 9)^2$
- C $(x - 81)(x + 1)$
- D $(x + 9)(x - 9)$
Show answer
Why:
A difference of two squares, with $b = 9$.
- A $(x + 5)(x - 5)$
- B $(x + 10)(x + 5)$
- C $(x + 25)(x + 1)$
- D $(x + 5)(x + 5)$
Show answer
Why:
5 and 5 multiply to 25 and add to 10, so both brackets are the same.
- A 7
- B 3
- C 6
- D 2
Show answer
Why:
Multiply the coefficient of $x^2$ by the constant: $2 \times 3 = 6$.
- A 5 and 5
- B 8 and 2
- C 3 and 7
- D 6 and 4
Show answer
Why:
They must multiply to $ac = 24$ and add to 10, which only 6 and 4 do.
- A The two solutions of the quadratic
- B The coefficient of $x$ and the constant term
- C The two factors of the quadratic
- D The area and the perimeter of the figure
Show answer
Why:
The rectangle has the quadratic as its area, so its sides multiply to give it.
- A They add to $c$ and multiply to $b$
- B They add to $b$ and multiply to $c$
- C They are both factors of $b$
- D Their difference is $b$ and their sum is $c$
Show answer
Why:
That is exactly what expanding $(x+a)(x+b)$ produces.
- A The product is always the smaller number
- B A number has only a few factor pairs, while endless pairs share a given sum
- C The sum is not part of the condition
- D Products are easier to calculate than sums
Show answer
Why:
Starting from the product cuts the search down to a handful of cases.
- A Both numbers are positive
- B One is positive and one negative
- C The expression does not factorise
- D Both numbers are negative
Show answer
Why:
A positive product means matching signs, and a negative sum makes them both negative.
- A One is positive and one is negative
- B Both are negative
- C Both are positive
- D They must be equal in size
Show answer
Why:
A negative product can only come from opposite signs.
- A The larger of the two
- B The smaller of the two
- C Both of them
- D Neither of them
Show answer
Why:
The larger number decides whether the sum comes out positive or negative.
- A Nothing; leave that pair alone
- B 1, and it must be written down
- C The whole bracket from the other pair
- D The variable $x$
Show answer
Why:
The 1 becomes a term of the final bracket, so omitting it loses part of the answer.
- A The expression does not factorise
- B The two numbers chosen were the wrong pair
- C The brackets should be added together
- D One of the pairs should be discarded
Show answer
Why:
A correct split always leaves the same bracket twice; a mismatch means going back.
- A By substituting $x = 0$ only
- B By checking that the brackets look similar
- C By expanding the brackets and comparing with the original
- D By counting the number of terms
Show answer
Why:
Expanding takes ten seconds and catches every sign error.
- A Choose the closest pair and use it anyway
- B Divide every term by the leading coefficient
- C State that it cannot be factorised with integer factors, showing the pairs tested
- D Change the sign of the constant term
Show answer
Why:
Saying so, with evidence, scores more than forcing wrong brackets.
- A $x^2 + 4x + 7$
- B $x^2 + 4x + 3$
- C $x^2 + 4x + 4$
- D $x^2 + 4x - 5$
Show answer
Why:
The pairs multiplying to 7 are 1 and 7, and $-1$ and $-7$, with sums 8 and $-8$; neither is 4.
- A The product of $a$ and $c$
- B The sum of $b$ and $d$
- C The sum of the two cross products $ad$ and $bc$
- D The product of $b$ and $d$
Show answer
Why:
Multiplying out gives $adx$ and $bcx$, which combine into the middle term.
- A $x^2 - 4x - 12$
- B $x^2 + 8x - 12$
- C $x^2 + 4x - 12$
- D $x^2 + 4x + 12$
Show answer
Why:
The middle term is $-2x + 6x = 4x$ and the constant is $6 \times (-2) = -12$.
- A $2x^2 + 7x + 3$
- B $2x^2 + 5x + 3$
- C $2x^2 + 4x + 3$
- D $2x^2 + 3x + 1$
Show answer
Why:
The cross terms are $6x$ and $x$, which combine into $7x$.
- A $x^2 + 9x + 20$
- B $x^2 + 20x + 9$
- C $x^2 - 9x + 20$
- D $x^2 + 9x - 20$
Show answer
Why:
The sum becomes the middle coefficient and the product becomes the constant.
- A $x^2 + 20x + 9$
- B $2x + 9$
- C $x^2 + 9x + 20$
- D $x^2 + 9x + 9$
Show answer
Why:
Multiplying the sides gives $x^2 + 5x + 4x + 20$.
- A $(2x - 5)(3x + 1)$
- B $(6x + 5)(x - 1)$
- C $(3x + 5)(2x - 1)$
- D $(2x + 5)(3x - 1)$
Show answer
Why:
Pairing gives $3x(2x+5) - 1(2x+5)$, so the shared bracket is $(2x+5)$.
- A $-2$ and $-6$
- B $-4$ and $-4$
- C 2 and 6
- D $-1$ and $-3$
Show answer
Why:
They must multiply to $ac = 12$ and add to $-8$, which only $-2$ and $-6$ do.
- A Twelve
- B Six
- C One
- D Three
Show answer
Why:
The pairs are 1 and 12, 2 and 6, and 3 and 4.
- A Because any four terms can be paired into equal brackets
- B Because the brackets are always $(x + 1)$
- C Because the two numbers were chosen so that the four terms pair up that way
- D It does not; matching brackets are a coincidence
Show answer
Why:
The conditions on $ac$ and the middle coefficient are exactly what force the match.
- A One
- B Four
- C Infinitely many
- D Two
Show answer
Why:
A prime $c$ gives 1 and $c$, or $-1$ and $-c$, and nothing else.
- A Three
- B Two
- C Five
- D Four
Show answer
Why:
The three original terms become four once the middle one is written as two pieces.
- A It has no constant term
- B It does not really factorise
- C It is a perfect square trinomial
- D Its leading coefficient must be 1
Show answer
Why:
Two equal brackets multiply to a squared bracket, which is what a perfect square is.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
Both are true, and R is the rule that produces the pair 3 and 4.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
Both statements are true, but the tile picture is not what determines this particular pair of signs; the sum and product conditions are.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
A is true, since only opposite signs give a negative product. R is false: two negatives multiply to a positive.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
A is false — $x^2 + 3x + 5$ has no integer factorisation. R is true and is the standard check.
Important short questions (10)
Two numbers are needed that multiply to 36 and add to 13. The pair is 4 and 9, since $4 \times 9 = 36$ and $4 + 9 = 13$. So the factorisation is $(x + 4)(x + 9)$.
The constant 40 is positive and the middle term is negative, so both numbers are negative. The pair is $-5$ and $-8$, since $(-5)(-8) = 40$ and $-5 - 8 = -13$. So the expression is $(x - 5)(x - 8)$.
The constant is negative, so the two numbers have opposite signs, and the middle term is positive, so the larger one is positive. The pair is 9 and $-5$, since $9 \times (-5) = -45$ and $9 - 5 = 4$. So the expression is $(x + 9)(x - 5)$.
A given number has only a small number of factor pairs, so listing them takes a moment and cannot miss a case. By contrast, endlessly many pairs of numbers add to any given total, so starting from the sum gives no way to work systematically through the possibilities.
Multiplying out gives $2x^2 - 4x + 5x - 10$, which collects to $2x^2 + 3x - 10$. So $2x^2 + 3x - 10$ is a quadratic factorising as $(2x + 5)(x - 2)$.
First $ac = 4 \times 5 = 20$. Two numbers are needed multiplying to 20 and adding to 12, and they are 10 and 2. Splitting gives $4x^2 + 10x + 2x + 5$. Factorising in pairs gives $2x(2x + 5) + 1(2x + 5)$, and the shared bracket comes out as $(2x + 1)(2x + 5)$.
Here $ac = 3 \times 8 = 24$, and two numbers are needed multiplying to 24 and adding to $-14$. Since the product is positive and the sum negative, both are negative: $-12$ and $-2$. Splitting gives $3x^2 - 12x - 2x + 8$, and pairing gives $3x(x - 4) - 2(x - 4)$, so the factorisation is $(3x - 2)(x - 4)$.
The two numbers are chosen so that they multiply to $ac$ and add to the middle coefficient. Those two conditions are exactly what is needed for the four terms to separate into two pairs sharing a common bracket. The first pair contributes the leading coefficient and one of the numbers, the second contributes the other number and the constant, and the ratio within each pair comes out the same. If the brackets do not match, the two numbers were not the right pair and the choice must be made again.
Two integers are needed that multiply to 11 and add to 5. Since 11 is prime, the only integer pairs multiplying to it are 1 and 11, whose sum is 12, and $-1$ and $-11$, whose sum is $-12$. Neither sum is 5, and no other pairs exist, so the expression cannot be factorised with integer factors.
The check is to expand the brackets and compare with the original expression. Multiplying out gives $x^2 + 3x - 5x - 15$, which collects to $x^2 - 2x - 15$. That is exactly the original, so the factorisation is correct. The check takes only a few seconds and catches sign errors, which are the commonest mistake in these questions.
Long answer questions (4)
(i) Two numbers are needed multiplying to 36 and adding to 12; they are 6 and 6. So the factorisation is $(x + 6)(x + 6)$, which may be written $(x + 6)^2$.
(ii) The constant is negative so the signs differ, and the middle term is negative so the larger number is negative. The pair multiplying to $-40$ and adding to $-3$ is $-8$ and 5, giving $(x - 8)(x + 5)$.
(iii) Here $ac = 2 \times (-14) = -28$. Two numbers multiplying to $-28$ and adding to 3 are 7 and $-4$. Splitting gives $2x^2 + 7x - 4x - 14$, and pairing gives $x(2x + 7) - 2(2x + 7)$, so the factorisation is $(2x + 7)(x - 2)$.
(iv) Both brackets are the same, so the expression is a perfect square trinomial and can be written as a single squared bracket.
(v) Expanding $(2x + 7)(x - 2)$ gives $2x^2 - 4x + 7x - 14 = 2x^2 + 3x - 14$, which matches the original.
(i) $(ax + b)(cx + d) = acx^2 + (ad + bc)x + bd$. The middle coefficient is the sum of the two cross products, which is why the middle term can be split into two pieces.
(ii) Multiply the leading coefficient by the constant to get $ac$; find two numbers multiplying to $ac$ and adding to the middle coefficient; split the middle term into those two pieces; factorise the four terms in pairs; take out the bracket both pairs share.
(iii) $ac = 6 \times (-5) = -30$. Two numbers multiplying to $-30$ and adding to 13 are 15 and $-2$. Splitting gives $6x^2 + 15x - 2x - 5$, and pairing gives $3x(2x + 5) - 1(2x + 5)$, so the factorisation is $(2x + 5)(3x - 1)$.
(iv) Different brackets mean the two numbers chosen were not the correct pair. The right course is to go back and re-check that they multiply to $ac$ and add to the middle coefficient, then try again — not to press on.
(v) Because the 1 becomes a term of the final bracket. In $3x(2x+5) - 1(2x+5)$ the second coefficient is $-1$, and it appears in the answer as the $-1$ of $(3x - 1)$. Omitting it loses a term of the factorisation entirely.
(i) The two numbers must add to $b$ and multiply to $c$.
(ii) If $c$ is positive the two numbers share a sign, and that shared sign matches the sign of $b$ — both positive when $b$ is positive, both negative when $b$ is negative. If $c$ is negative the two numbers have opposite signs, and the larger of the two carries the sign of $b$.
(iii) The constant is negative, so the signs differ, and the middle term is negative, so the larger number is negative. The pair multiplying to $-27$ and adding to $-6$ is $-9$ and 3, giving $(x - 9)(x + 3)$.
(iv) The constant is positive and the middle term negative, so both numbers are negative. The pair is $-3$ and $-7$, giving $(x - 3)(x - 7)$.
(v) Integer pairs multiplying to 6 are 1 and 6 (sum 7), 2 and 3 (sum 5), $-1$ and $-6$ (sum $-7$), and $-2$ and $-3$ (sum $-5$). None of these sums is 2, so there is no integer factorisation.
(i) The tiles have total area $x^2 + 7x + 10$.
(ii) Two numbers are needed multiplying to 10 and adding to 7; they are 2 and 5. So the rectangle measures $(x + 2)$ by $(x + 5)$.
(iii) The two numbers do both jobs at once. They multiply to 10 because they form the block of ten unit squares in one corner, and they add to 7 because together they account for all seven $x$-strips, which split into a row and a column.
(iv) Two numbers multiplying to 12 and adding to 8 are 2 and 6, so the rectangle measures $(x + 2)$ by $(x + 6)$.
(v) That set has area $x^2 + 4x + 7$. Integer pairs multiplying to 7 are 1 and 7 (sum 8) and $-1$ and $-7$ (sum $-8$), and neither sum is 4. Since no whole-number pair satisfies both conditions, the tiles cannot be closed into a rectangle with whole-number side parts.
Flashcards (20)
Play this topic
- Match each quadratic to its factorisation
- True or false about factorising
- Complete the statement
- Unscramble the vocabulary
Cube Identities and Rational Expressions
Multiplying a square by one more copy of its bracket produces the cube identities, and two of them factorise: unlike a sum of two squares, a sum of two cubes breaks into brackets. Those factorisations, together with everything earlier in the chapter, are what make it possible to cancel in a rational expression — provided the excluded value is stated.
Full explanation
What happens if you cube a bracket?
The last two topics squared brackets. The natural next question is what $(a + b)^3$ comes to.
There is no need to guess. A cube is a square multiplied by one more copy:
$$(a + b)^3 = (a + b)(a + b)^2 = (a + b)(a^2 + 2ab + b^2)$$
Multiply through carefully:
$$= a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3$$
and collect the like terms:
$$\boxed{(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3}$$
A new identity, found rather than remembered. That is worth noticing: identities are not a list handed down, they are produced by multiplying out and collecting.
The cube of a difference
Replace $b$ by $-b$. Terms holding an odd number of $b$s change sign; terms holding an even number do not:
$$\boxed{(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3}$$
The signs alternate: plus, minus, plus, minus. That is much easier to remember than four separate rules.
The coefficients have a pattern
Look at the numbers in front:
$$(a+b)^1: \quad 1, ; 1$$ $$(a+b)^2: \quad 1, ; 2, ; 1$$ $$(a+b)^3: \quad 1, ; 3, ; 3, ; 1$$
Each row is symmetric, and each inner number is the sum of the two above it. And in every term the powers of $a$ and $b$ add to the power of the bracket — in $3a^2b$ the powers are 2 and 1, adding to 3.
Use that as a check. If a term's powers do not add to 3, it has been written wrongly.
Cubing a number in your head
Just like the square identities, these work on round numbers.
Find $103^3$. Write it as $(100 + 3)^3$:
$$1000000 + 3(10000)(3) + 3(100)(9) + 27 = 1000000 + 90000 + 2700 + 27 = 1092727$$
The sum and difference of two cubes
Two more identities, and these are the most useful of the four because they factorise:
$$\boxed{x^3 + y^3 = (x + y)(x^2 - xy + y^2)}$$
$$\boxed{x^3 - y^3 = (x - y)(x^2 + xy + y^2)}$$
Check the first by expanding: $x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3$, and everything in the middle cancels, leaving $x^3 + y^3$.
Note the contrast with squares: a sum of two squares does not factorise, but a sum of two cubes does.
Getting the signs right
The pattern is worth stating carefully, because the two identities look alike and the middle sign is easy to swap.
- The first bracket copies the sign of the original: $x^3 + y^3$ gives $(x + y)$, and $x^3 - y^3$ gives $(x - y)$.
- The middle term of the second bracket takes the opposite sign.
- The outer terms of the second bracket are always positive.
So $x^3 + y^3$ gives $(x + y)(x^2 - xy + y^2)$: plus outside, minus in the middle.
Using them
Factorise $8x^3 + 27$. Both parts are cubes: $8x^3 = (2x)^3$ and $27 = 3^3$. So with $x \to 2x$ and $y \to 3$:
$$8x^3 + 27 = (2x + 3)(4x^2 - 6x + 9)$$
Find $27^3 - 23^3$ without cubing either number. Using the difference identity with $x = 27$ and $y = 23$:
$$(27 - 23)(729 + 621 + 529) = 4 \times 1879 = 7516$$
One more, with three letters
The book's last identity brings three variables together:
$$\boxed{x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)}$$
It looks forbidding, but it earns its place through one special case. If $x + y + z = 0$, the first bracket on the right is zero, so the whole right-hand side is zero, and therefore
$$x^3 + y^3 + z^3 = 3xyz \qquad \text{whenever } x + y + z = 0$$
For instance, with $x = 5$, $y = -3$ and $z = -2$, which add to zero: $125 - 27 - 8 = 90$, and $3xyz = 3(5)(-3)(-2) = 90$. They agree.
Rational expressions
A rational expression is one algebraic expression divided by another — a fraction with polynomials in it, such as
$$\frac{x^2 - 9}{x^2 + 7x + 12}$$
Ordinary numerical fractions are simplified by cancelling a common factor. These are simplified the same way, but the common factor has to be found first, and that is what the whole chapter has been building towards.
Simplifying: factorise first, always
Simplify $\dfrac{x^2 - 9}{x^2 + 7x + 12}$.
Factorise the top and the bottom separately:
- top: a difference of two squares, $(x + 3)(x - 3)$
- bottom: two numbers multiplying to 12 and adding to 7, so $(x + 3)(x + 4)$
$$\frac{(x + 3)(x - 3)}{(x + 3)(x + 4)} = \frac{x - 3}{x + 4}$$
The factor $(x + 3)$ appears top and bottom, so it cancels.
The condition that comes with it
Cancelling $(x + 3)$ means dividing by it, and division by zero is not allowed. So the simplification holds only when $x + 3 \neq 0$, that is, when $x \neq -3$.
A full answer says so:
$$\frac{x^2 - 9}{x^2 + 7x + 12} = \frac{x - 3}{x + 4}, \qquad x \neq -3$$
Stating the excluded value is part of the answer, not an optional extra. It is the difference between a simplification that is always true and one that quietly fails at a point.
A cube identity doing the same job
Simplify $\dfrac{x^3 + 125}{x + 5}$.
The top is a sum of two cubes, since $125 = 5^3$:
$$\frac{(x + 5)(x^2 - 5x + 25)}{x + 5} = x^2 - 5x + 25, \qquad x \neq -5$$
Without the identity there is nothing to cancel and the expression looks stuck. With it, the whole thing collapses in one line.
What you may never cancel
This is the single most punished error in the topic.
$$\frac{x + 3}{x + 4} \neq \frac{3}{4}$$
You may only cancel a factor of the whole top and the whole bottom. In $x + 3$ the $x$ and the 3 are being added, not multiplied, so neither is a factor of the whole thing and neither may be struck out.
Test it with $x = 1$: the left side is $\tfrac{4}{5}$ and the right is $\tfrac{3}{4}$. Not equal.
Cancelling is for factors, never for terms.
Slips that cost marks
- Cancelling terms rather than factors, as in $\frac{x+3}{x+4}$.
- Leaving out the excluded value after cancelling.
- Swapping the middle sign in the sum or difference of two cubes.
- Giving $(a-b)^3$ signs that do not alternate.
- Trying to factorise a sum of two squares by analogy with a sum of two cubes.
What to write in the exam
For a cube expansion, check that every term's powers add to 3 before moving on. For a factorisation of cubes, name the identity and write down what $x$ and $y$ stand for. And for a rational expression, factorise top and bottom fully on separate lines, cancel, and then state the excluded value — that last line is worth a mark of its own.
Watch out for these
Practice MCQs with answers (42)
- A $x^3 + 64$
- B $x^3 + 4x^2 + 4x + 64$
- C $x^3 + 12x^2 + 48x + 64$
- D $x^3 + 12x^2 + 12x + 64$
Show answer
Why:
With $a = x$ and $b = 4$: $3a^2b = 12x^2$ and $3ab^2 = 48x$.
- A $x^3 - 6x^2 + 12x - 8$
- B $x^3 - 8$
- C $x^3 - 6x^2 - 12x - 8$
- D $x^3 + 6x^2 + 12x + 8$
Show answer
Why:
The signs alternate: $3a^2b = -6x^2$, $3ab^2 = +12x$, and the last term is $-8$.
- A $8x^3 + 1$
- B $8x^3 + 12x^2 + 6x + 1$
- C $2x^3 + 12x^2 + 6x + 1$
- D $8x^3 + 6x^2 + 12x + 1$
Show answer
Why:
With $a = 2x$: $a^3 = 8x^3$, $3a^2b = 12x^2$ and $3ab^2 = 6x$.
- A 1030103
- B 1003003
- C 1300301
- D 1030301
Show answer
Why:
$(100+1)^3 = 1000000 + 30000 + 300 + 1 = 1030301$.
- A 941129
- B 940192
- C 941192
- D 951192
Show answer
Why:
$(100-2)^3 = 1000000 - 60000 + 1200 - 8 = 941192$.
- A 370
- B 1000
- C 340
- D 400
Show answer
Why:
$(7+3)(49 - 21 + 9) = 10 \times 37 = 370$.
- A 1216
- B 1024
- C 64
- D 1280
Show answer
Why:
$(12-8)(144 + 96 + 64) = 4 \times 304 = 1216$.
- A 3250
- B 1000
- C 3125
- D 2250
Show answer
Why:
$(15-5)(225 + 75 + 25) = 10 \times 325 = 3250$.
- A 5
- B 25
- C 15
- D 75
Show answer
Why:
The term is $3a^2b = 3x^2(5) = 15x^2$.
- A $x - 3$, for $x \neq 3$
- B $x + 3$, with no restriction
- C $\dfrac{x+3}{1}$, for $x \neq -3$
- D $x + 3$, for $x \neq 3$
Show answer
Why:
The top is $(x+3)(x-3)$, so the $(x-3)$ cancels, and the cancelled factor must not be zero.
- A $x^2 - 2x + 4$, for $x \neq 2$
- B $x^2 + 4$, for $x \neq 2$
- C $x^2 + 2x + 4$, with no restriction
- D $x^2 + 2x + 4$, for $x \neq 2$
Show answer
Why:
The top is a difference of cubes, $(x-2)(x^2 + 2x + 4)$, and the $(x-2)$ cancels.
- A $x^2 + x + 1$, for $x \neq -1$
- B $x^2 - x + 1$, for $x \neq -1$
- C $x^2 + 1$, for $x \neq -1$
- D $x^2 - x + 1$, with no restriction
Show answer
Why:
The top is a sum of cubes, $(x+1)(x^2 - x + 1)$, so that bracket cancels.
- A $\dfrac{x + 5}{x + 2}$, with no restriction
- B $\dfrac{x + 5}{x + 2}$, for $x \neq -2$
- C $\dfrac{x + 2}{x + 5}$, for $x \neq -5$
- D $\dfrac{7x + 10}{4x + 4}$
Show answer
Why:
Factorising gives $\dfrac{(x+5)(x+2)}{(x+2)^2}$, and one $(x+2)$ cancels from each.
- A $-210$
- B 210
- C 0
- D $-90$
Show answer
Why:
The three add to zero, so the sum of cubes is $3xyz = 3(2)(-7)(5) = -210$.
- A 36
- B $-36$
- C 12
- D $-12$
Show answer
Why:
$3 \times 4 \times (-1) \times (-3) = 36$.
- A $(x + 4)(x^2 - 4x + 16)$
- B $(x - 4)(x^2 - 4x + 16)$
- C $(x - 4)^3$
- D $(x - 4)(x^2 + 4x + 16)$
Show answer
Why:
A difference of cubes: minus outside, plus in the middle, with $y = 4$.
- A $(3x + 2)(9x^2 + 6x + 4)$
- B $(3x + 2)(9x^2 - 6x + 4)$
- C $(3x - 2)(9x^2 + 6x + 4)$
- D $(3x + 2)^3$
Show answer
Why:
Here $27x^3 = (3x)^3$ and $8 = 2^3$, so it is a sum of cubes: plus outside, minus in the middle.
- A $(4x + 1)(16x^2 - 4x + 1)$
- B $(4x - 1)(16x^2 - 4x + 1)$
- C $(4x - 1)(16x^2 + 4x + 1)$
- D $(4x - 1)^3$
Show answer
Why:
$64x^3 = (4x)^3$ and $1 = 1^3$, giving a difference of cubes.
- A $(x + 1)(x^2 + x + 1)$
- B $(x - 1)(x^2 + x + 1)$
- C $(x + 1)(x^2 - x + 1)$
- D $(x + 1)^3$
Show answer
Why:
A sum of cubes with $y = 1$: plus outside, minus in the middle.
- A $x^3 + 125$
- B $x^3 - 25$
- C $x^3 - 125$
- D $(x-5)^3$
Show answer
Why:
Minus outside with plus in the middle is the difference of cubes pattern, with $y = 5$.
- A $8x^3 - 1$
- B $2x^3 + 1$
- C $(2x+1)^3$
- D $8x^3 + 1$
Show answer
Why:
Plus outside with minus in the middle is the sum of cubes pattern, with $x \to 2x$.
- A By squaring the identity for $(a+b)^2$
- B By adding $a^3$ and $b^3$ and then doubling
- C It cannot be derived and must simply be memorised
- D By multiplying $(a+b)$ by the expansion of $(a+b)^2$ and collecting terms
Show answer
Why:
A cube is a square multiplied by one more copy of the bracket.
- A Every coefficient is a multiple of 3
- B The powers of the two letters in each term add to 3
- C The signs are all positive
- D Every term contains both letters
Show answer
Why:
In $3a^2b$ the powers are 2 and 1; a term failing this has been written wrongly.
- A Both factorise in the same way
- B Neither of them factorises
- C A sum of two squares does not factorise, but a sum of two cubes does
- D A sum of two squares factorises, but a sum of two cubes does not
Show answer
Why:
$x^3 + y^3 = (x+y)(x^2 - xy + y^2)$, while $x^2 + y^2$ has no such factorisation.
- A Negative, matching the first bracket
- B Positive, the opposite of the sign in the first bracket
- C It depends on whether $x$ is larger than $y$
- D There is no middle term
Show answer
Why:
The rule is that the middle term always takes the opposite sign to the first bracket.
- A The left side becomes zero, so $xyz = 0$
- B The right side becomes zero, so $x^3 + y^3 + z^3 = 3xyz$
- C Both sides become zero, so nothing follows
- D The identity no longer applies
Show answer
Why:
The first bracket on the right is $(x+y+z)$, which is zero, so the whole product vanishes.
- A Any expression containing a fraction
- B An expression whose value is always a rational number
- C One algebraic expression divided by another
- D An expression with no variables in the denominator
Show answer
Why:
It is a fraction whose top and bottom are both polynomials.
- A Cancel any number appearing in both
- B Factorise the top and the bottom fully
- C Multiply out all the brackets
- D Substitute a value for the variable
Show answer
Why:
Nothing can be cancelled until the common factor has been exposed by factorising.
- A Yes, giving $\tfrac{3}{4}$
- B Yes, but only when $x$ is positive
- C Yes, because both appear in a fraction
- D No, because they are terms being added, not factors of the whole
Show answer
Why:
Testing at $x = 1$ gives $\tfrac45$ on the left and $\tfrac34$ on the right.
- A Because cancelling means dividing by that factor, which must not be zero
- B Because the answer would otherwise be too long
- C Because the factor might be negative
- D It need not be stated; the answer is complete without it
Show answer
Why:
The simplification fails exactly where the cancelled factor vanishes.
- A Factors are numbers and terms are letters
- B Factors appear on top and terms on the bottom
- C Factors are multiplied together, while terms are added
- D There is no difference
Show answer
Why:
Only factors may be cancelled, which is why the distinction matters here.
- A $\dfrac{(x+2)(x-1)}{(x+2)(x+5)}$
- B $\dfrac{x + 2}{x + 5}$
- C $\dfrac{x^2 + 4}{x^2 + 9}$
- D $\dfrac{x + 3}{x - 3}$
Show answer
Why:
Only the first has a bracket appearing as a factor of both the top and the bottom.
- A Every term is negative
- B Only the middle two terms are negative
- C The four terms run plus, minus, plus, minus
- D The signs depend on whether $a$ exceeds $b$
Show answer
Why:
Terms holding an odd number of $b$s change sign, which produces the alternation.
- A Four
- B Three
- C Two
- D Six
Show answer
Why:
They are $a^3$, $3a^2b$, $3ab^2$ and $b^3$.
- A 1, 2, 1
- B 1, 3, 1
- C 3, 3, 3, 3
- D 1, 3, 3, 1
Show answer
Why:
Each row of this pattern is symmetric, and each inner number is the sum of the two above.
- A 2
- B 6
- C They vary from term to term
- D 3
Show answer
Why:
The powers of $a$ and $b$ in every term add to the power of the bracket.
- A The sum of the cubes equals the product
- B The sum of the cubes is zero
- C There is no relationship between them
- D The sum of the cubes equals three times the product
Show answer
Why:
The identity gives $x^3 + y^3 + z^3 - 3xyz = 0$ when the first bracket vanishes.
- A None, since cancelling changes nothing
- B One, where the cancelled factor is zero
- C Two, one for each side of the fraction
- D As many as the degree of the numerator
Show answer
Why:
Each cancelled factor contributes the value that makes it vanish.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
Both are true, and R is exactly the sign rule that produces the factorisation in A.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
Both statements are true, but the contrast between squares and cubes has nothing to do with the signs in this expansion.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
A is true, since cancelling is division. R is false: dividing by zero is not allowed, which is precisely why the exclusion arises.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
A is false, since 5 and 7 are terms being added rather than factors. R is true, and it is why A fails.
Important short questions (10)
Using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ with $a = x$ and $b = 3$: the terms are $x^3$, $3x^2(3) = 9x^2$, $3x(9) = 27x$ and $27$. So the expansion is $x^3 + 9x^2 + 27x + 27$.
With $a = 3x$ and $b = 1$ the signs alternate: $27x^3 - 3(9x^2)(1) + 3(3x)(1) - 1$. That simplifies to $27x^3 - 27x^2 + 9x - 1$.
Since $216 = 6^3$, this is a sum of two cubes. The first bracket copies the sign, and the middle term of the second takes the opposite sign, so the factorisation is $(x + 6)(x^2 - 6x + 36)$.
Here $125x^3 = (5x)^3$ and $27 = 3^3$, so this is a difference of two cubes. The factorisation is $(5x - 3)(25x^2 + 15x + 9)$.
Cancelling is only permitted for a factor of the whole top and the whole bottom. In $x + 4$ the two parts are being added rather than multiplied, so 4 is not a factor of the numerator, and the same holds for 9 below. A quick test confirms it: at $x = 1$ the expression is $\tfrac{5}{10} = \tfrac{1}{2}$, not $\tfrac{4}{9}$.
The numerator is a difference of two squares, giving $(x + 4)(x - 4)$. The denominator is a perfect square trinomial, giving $(x + 4)^2$. So the expression is $\dfrac{(x+4)(x-4)}{(x+4)(x+4)}$, and cancelling one $(x+4)$ from each leaves $\dfrac{x - 4}{x + 4}$. Since $(x+4)$ was divided out, the simplification holds only for $x \neq -4$.
Using $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$ with $x = 13$ and $y = 7$: the first bracket is $6$, and the second is $169 + 91 + 49 = 309$. So the value is $6 \times 309 = 1854$. The saving is that two separate cubings of two-digit numbers are replaced by three squarings that can be done mentally and one multiplication.
The first bracket copies the sign of the original expression; the middle term of the second bracket takes the opposite sign; and the outer terms of the second bracket are always positive. For $x^3 - 1000$, note that $1000 = 10^3$. The original sign is a minus, so the first bracket is $(x - 10)$, the middle term of the second bracket is $+10x$, and the outer terms are $x^2$ and $100$. The factorisation is $(x - 10)(x^2 + 10x + 100)$.
The identity $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$ holds for all values. If $x + y + z = 0$ then the first bracket on the right is zero, so the whole right-hand side is zero, and therefore $x^3 + y^3 + z^3 - 3xyz = 0$, that is $x^3 + y^3 + z^3 = 3xyz$. Checking with 6, $-4$ and $-2$, which add to zero: the cubes give $216 - 64 - 8 = 144$, and $3xyz = 3(6)(-4)(-2) = 144$. They agree.
In the expansion of a cubed bracket, the powers of the two letters in every term must add to 3, because each term comes from choosing one letter from each of the three brackets. In the expression given, the third term is $3x^2$, whose powers are 2 and 0, adding to 2 rather than 3. It should be $3x$, whose powers are 1 and 2 when the hidden $1^2$ is counted. The correct expansion is $x^3 + 3x^2 + 3x + 1$.
Long answer questions (4)
(i) With $a = 2x$ and $b = 3$: $8x^3 + 3(4x^2)(3) + 3(2x)(9) + 27 = 8x^3 + 36x^2 + 54x + 27$.
(ii) Since $27 = 3^3$, this is a difference of cubes: $(x - 3)(x^2 + 3x + 9)$.
(iii) Here $8x^3 = (2x)^3$ and $125 = 5^3$, so this is a sum of cubes: $(2x + 5)(4x^2 - 10x + 25)$.
(iv) $104^3 = (100 + 4)^3 = 1000000 + 3(10000)(4) + 3(100)(16) + 64 = 1000000 + 120000 + 4800 + 64 = 1124864$.
(v) Expanding $(x - 3)(x^2 + 3x + 9)$ gives $x^3 + 3x^2 + 9x - 3x^2 - 9x - 27$. Everything in the middle cancels, leaving $x^3 - 27$, which is the original.
(i) The numerator is a difference of cubes, $(x - 3)(x^2 + 3x + 9)$, so the $(x-3)$ cancels and the result is $x^2 + 3x + 9$, for $x \neq 3$.
(ii) The numerator factorises as $(x + 6)(x + 1)$ and the denominator as $(x + 1)^2$. Cancelling one $(x+1)$ from each gives $\dfrac{x + 6}{x + 1}$, for $x \neq -1$.
(iii) Cancelling a factor means dividing both top and bottom by it, and division by zero is not defined. So the simplified form is equal to the original only where that factor is non-zero, and the excluded value records exactly where the two part company.
(iv) Its numerator and denominator have no common factor. The 6 and the 1 are terms being added, not factors of the whole expressions, so there is nothing that may legitimately be struck out.
(v) Only a factor of the whole numerator and the whole denominator may be cancelled. Terms that are added together may never be cancelled individually.
(i) $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$.
(ii) If $x + y + z = 0$, the first bracket on the right is zero, so the whole right-hand side is zero. Hence $x^3 + y^3 + z^3 - 3xyz = 0$, that is $x^3 + y^3 + z^3 = 3xyz$.
(iii) The three add to $6 - 4 - 2 = 0$. Their cubes give $216 - 64 - 8 = 144$, and $3xyz = 3(6)(-4)(-2) = 144$. The two agree.
(iv) Since $7 - 5 - 2 = 0$, the shortcut applies: the sum of the cubes is $3(7)(-5)(-2) = 210$.
(v) Because $7 + 5 + 2 = 14$, which is not zero. The first bracket of the identity is then $14$ rather than $0$, so the right-hand side does not vanish and the sum of the cubes is not simply $3xyz$.
(i) A cube is a square multiplied by one more copy of the bracket, so $(a+b)^3 = (a+b)(a^2 + 2ab + b^2)$. Multiplying out gives $a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3$, and collecting like terms gives $a^3 + 3a^2b + 3ab^2 + b^3$.
(ii) $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$. The signs alternate, plus then minus then plus then minus, because terms holding an odd number of $b$s change sign while those holding an even number do not.
(iii) $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$ and $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$.
(iv) A sum of two squares does not factorise at all with real brackets, whereas a sum of two cubes does, into a linear bracket and a quadratic one. So an analogy from squares to cubes gives the wrong answer here, and each case has to be learned on its own terms.
(v) In every term of a cube expansion the powers of the two letters must add to 3, since each term takes one letter from each of the three brackets. Any term failing that test has been written wrongly.
Flashcards (20)
Play this topic
- Match each expression to its factorisation
- True or false about cubes and fractions
- Complete the identity
- Unscramble the vocabulary
Class 9 Mathematics Chapter 4 — frequently asked questions
Using $(a+b)^2 = a^2 + 2ab + b^2$ with $a = 3x$ and $b = 8$: the first term is $9x^2$, the middle is $2(3x)(8) = 48x$, and the last is $64$. So the expansion is $9x^2 + 48x + 64$.
Writing $94 = 100 - 6$ and using $(a-b)^2 = a^2 - 2ab + b^2$ gives $10000 - 2(100)(6) + 36 = 10000 - 1200 + 36 = 8836$.
The two numbers sit equally above and below 200, so they are $200 + 6$ and $200 - 6$. By the difference of squares, the product is $200^2 - 6^2 = 40000 - 36 = 39964$.
Two numbers are needed that multiply to 36 and add to 13. The pair is 4 and 9, since $4 \times 9 = 36$ and $4 + 9 = 13$. So the factorisation is $(x + 4)(x + 9)$.
The constant 40 is positive and the middle term is negative, so both numbers are negative. The pair is $-5$ and $-8$, since $(-5)(-8) = 40$ and $-5 - 8 = -13$. So the expression is $(x - 5)(x - 8)$.
The constant is negative, so the two numbers have opposite signs, and the middle term is positive, so the larger one is positive. The pair is 9 and $-5$, since $9 \times (-5) = -45$ and $9 - 5 = 4$. So the expression is $(x + 9)(x - 5)$.
Using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ with $a = x$ and $b = 3$: the terms are $x^3$, $3x^2(3) = 9x^2$, $3x(9) = 27x$ and $27$. So the expansion is $x^3 + 9x^2 + 27x + 27$.
With $a = 3x$ and $b = 1$ the signs alternate: $27x^3 - 3(9x^2)(1) + 3(3x)(1) - 1$. That simplifies to $27x^3 - 27x^2 + 9x - 1$.
Since $216 = 6^3$, this is a sum of two cubes. The first bracket copies the sign, and the middle term of the second takes the opposite sign, so the factorisation is $(x + 6)(x^2 - 6x + 36)$.
All Class 9 Mathematics chapters
All Mathematics chapters · Every Class 9 subject · The whole NCERT library
Study this chapter properly, free
Everything above is on this page to read. In the Veda app the same chapter becomes spaced-repetition flashcards, timed tests that mark themselves, Veda Bites you can swipe, and games — with your progress tracked topic by topic.
Get Veda free