Exploring Algebraic Identities: Class 9 Ganita Manjari · Chapter 4

Squares, strips, tiles and cubes that explain algebra

Ganita Manjari Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 28 Sep 2026 · Updated 28 Sep 2026

107 min
30 practice questions
31 MCQs

What you will learn in Exploring Algebraic Identities

Tell an identity from an equation, test a claimed identity with numbers and prove it with the distributive law
Explain (a + b)², (a − b)², (a + b + c)² and a² − b² with area models, and use them to expand expressions and square numbers mentally
Factorise perfect-square trinomials, including those with a common factor or fractional coefficients
Factorise quadratics with algebra tiles and by splitting the middle term, and multiply linear binomials quickly
Derive and apply the cube identities (a ± b)³, x³ ± y³ and x³ + y³ + z³ − 3xyz
Simplify rational expressions by cancelling non-zero common factors, and solve simple problems that lead to quadratics

Original companion text aligned to NCERT Mathematics Class 9, Chapter 4, pp. 68–91. Not affiliated with or endorsed by NCERT.

4.1 Introduction

Here is a trick to try on a friend. Ask them to choose three whole numbers in a row, such as 11, 12 and 13, and to tell you only the middle one. Then say: “Square the largest and the smallest, subtract, and you will get 48.” It works every time, because the answer is always four times the middle number: $13^{2} - 11^{2} = 169 - 121 = 48 = 4 \times 12$ .

Example 1NCERT Example 1 · p. 68

A pattern hiding in squares

Take three consecutive numbers. Square the largest and the smallest, and subtract. Do this for 6, 7, 8, then for 19, 20, 21, and then for 49, 50, 51. Compare each answer with the middle number.

Try it yourself first. Then reveal the solution one step at a time.

Three trials, three matches. Fig. 4.1 shows why it happens for 5, 6 and 7. Lay the smaller square of tiles on the larger one: the tiles left uncovered can be rearranged into 4 rows of 6, and 6 is the middle number.

4/4Fold the strip in half: 4 rows of 6, and 6 is the middle number.

Fig. 4.1The border left over when

5^{2}

is laid on

7^{2}

rearranges into 4 rows of 6.

Step 1 shows a 7 by 7 square of blue tiles labelled 49 and a separate 5 by 5 square of green tiles labelled 25. Step 2 places the green square in the bottom-left corner of the blue one; the 24 uncovered tiles, two rows across the top and two columns down the right, turn orange. Step 3 lifts the orange border off, leaving a dashed outline, and lays it out as a strip 2 tiles tall and 12 tiles long, since 12 = 5 + 7. Step 4 folds the strip into a block of 4 rows of 6 tiles, labelled 7 squared minus 5 squared equals 4 times 6, where 6 is the middle number.

Still, three examples and one picture do not prove that the pattern works for every set of three numbers, say 1000, 1001 and 1002, or $- 3$ , $- 2$ and $- 1$ . To be certain we need algebra, and in particular the special equalities called identities (सर्वसमिका). We will prove this pattern in Section 4.3.

This chapter builds a small toolkit of identities. Each one comes with a picture made of squares, rectangles or cubes, and a short algebraic proof. You will use them to square numbers in your head, to break expressions into factors, and to simplify fractions built from polynomials.

Quick check

1 of 3

Use the pattern to find

31^{2} - 29^{2}

without squaring.

Show answer

Answer 120

The three numbers are 29, 30, 31, so the answer is

4 \times 30 = 120

. Check:

961 - 841 = 120

4.2 Visualising Identities

Think of a square garden plot whose side is made of two parts, a long stretch and a short one. The plot splits naturally into beds, and adding up the beds is algebra in disguise. When letters stand for lengths, $a b$ is the area of a rectangle and $a^{2}$ is the area of a square. An identity then becomes a claim about cutting shapes up and fitting them back together.

Fig. 4.2 grows such a square one piece at a time. It begins with a square of side $a$ , widens it by $b$ , raises it by $b$ , and finally plugs the gap left in the corner. By the end, every side measures $a + b$ .

6/7The side is $a + b$ , so $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ .

Fig. 4.2Counting the same area as one square and as four pieces gives

(a + b)^{2} = a^{2} + 2 a b + b^{2}

.NCERT Fig. 4.2 · p. 69

Step 1 shows a blue segment of length a and a green segment of length b joined into a line of length a plus b. Step 2 draws a blue square of side a, area a squared. Step 3 slides an orange strip of width b and area ab against its right edge. Step 4 slides a second orange strip of area ab onto its top edge. Step 5 grows a small green square of area b squared in the top-right corner. Step 6 outlines the whole square of side a plus b and states that its area is a squared plus 2ab plus b squared. Step 7 removes the two strips, leaving dashed holes with question marks, to show that a squared plus b squared misses 2ab.

The big square's area can be counted in two ways. As a single square it is $(a + b)^{2}$ . Piece by piece it is $a^{2} + a b + a b + b^{2}$ . One area counted twice must give one answer, so

Square of a sum

(a + b)^{2} = a^{2} + 2 a b + b^{2}

Now try to break the formula. Feed it values that no drawing could handle: a negative number, a pair of fractions, a pair of decimals.A side of a square is never negative, so the picture cannot vouch for a value such as $a = - 5$ . If the formula survives them all, we have good reason to trust it beyond the world of lengths.

Example 2NCERT Example 2 · p. 70

Testing beyond lengths

Set up a small test lab. Check whether $(a + b)^{2}$ and $a^{2} + 2 a b + b^{2}$ agree when (i) $a = - 5$ , $b = 2$ ; (ii) $a = \frac{1}{2}$ , $b = - \frac{1}{3}$ ; (iii) $a = 0.3$ , $b = 0.7$ .

Try it yourself first. Then reveal the solution one step at a time.

The formula survived every test. Yet however many pairs we try, infinitely many stay unchecked: checking a few mangoes from a crate cannot promise that none of the others is bruised. We need an argument that covers every $a$ and every $b$ at once, and the distributive law supplies it.

TheoremSquare of a sum

For all numbers $a$ and $b$ , $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ .

Definition

Identity

An equality between two algebraic expressions that stays true whatever values we give to the letters in it.

Example $x (x + 3) = x^{2} + 3 x$ is an identity: try $x = 2$ , $x = - 1$ or $x = \frac{1}{2}$ .

Compare this with an ordinary equation (समीकरण). The equation $3 x - 5 = 16$ is true only when $x = 7$ ; every other value of $x$ makes it false. Solving an equation means finding the few values that make it true. An identity needs no solving, because every value works.

Compare

Equation or identity?

Equation

True only for particular values: $x^{2} + x = 6$ holds for $x = 2$ and $x = - 3$ alone
We solve it to find those values
Example: $2 x + 1 = 11$ (only $x = 5$ )

Identity

True for every value of every letter
We use it to expand, factorise or simplify
Example: $2 (x + 3) = 2 x + 6$

A very common slip is to write $(a + b)^{2} = a^{2} + b^{2}$ . The last step of Fig. 4.2 shows what goes wrong: $a^{2} + b^{2}$ leaves out the two strips and so misses $2 a b$ . With $a = 9$ and $b = 4$ , for instance, $(a + b)^{2} = 169$ while $a^{2} + b^{2} = 97$ , short by exactly $2 \times 9 \times 4 = 72$ .

The identity is a recipe for expanding. Whatever stands in the places of $a$ and $b$ (a number, a letter or a whole term), the recipe is the same. Square the first term, add twice the product of the two terms, then add the square of the second term.

Example 3NCERT Example 3 · p. 71

Expanding a binomial square

Expand $(6 u + 5 v)^{2}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example

Fractions inside the brackets

Expand ${(\frac{2}{3} p + \frac{3}{2} q)}^{2}$ .

Try it yourself first. Then reveal the solution one step at a time.

The same identity turns awkward squares into mental arithmetic. Split the number into a round part and a small part, and add three easy pieces.

Example 4NCERT Example 4 · p. 71

Area of a square courtyard

A square temple courtyard has sides of 62 m. Find its area without long multiplication.

Try it yourself first. Then reveal the solution one step at a time.

InteractiveArea model lab

An interactive area model with four tabs: (a + b) squared, (a minus b) squared, a squared minus b squared, and (a + b + c) squared. Sliders and plus/minus buttons set a from 1 to 60, b from 0 to 30 and c from 0 to 20. The square redraws to scale with its pieces coloured and labelled with their areas, and a panel below works out both sides of the identity with the chosen numbers. In the a squared minus b squared tab, a Rearrange button turns the leftover L-shape into a rectangle of sides a + b and a minus b. A note under the panel explains special cases such as b = 0 or a = b.

Fig. 4.3Choose an identity, then change

a

b

(and

c

): the pieces, their areas and both sides of the identity update together.

Practice · 3 questions

Practice 4.1: questions with answers

1
Short answerEasyNCERT Exercise Set 4.1, Q1
Expand using $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ .
1. $(8 a + 3 b)^{2}$
  Show solution
  Answer $64 a^{2} + 48 a b + 9 b^{2}$
  $(8 a)^{2} + 2 (8 a) (3 b) + (3 b)^{2} = 64 a^{2} + 48 a b + 9 b^{2}$ .
2. ${(\frac{4}{5} m + \frac{5}{2} n)}^{2}$
  Show solution
  Answer $\frac{16}{25} m^{2} + 4 m n + \frac{25}{4} n^{2}$
  The middle term is $2 \times \frac{4}{5} \times \frac{5}{2} m n = 4 m n$ , so the answer is $\frac{16}{25} m^{2} + 4 m n + \frac{25}{4} n^{2}$ .
3. $(1.2 x + 0.5 y)^{2}$
  Show solution
  Answer $1.44 x^{2} + 1.2 x y + 0.25 y^{2}$
  ${1.2}^{2} = 1.44$ , $2 \times 1.2 \times 0.5 = 1.2$ and ${0.5}^{2} = 0.25$ , giving $1.44 x^{2} + 1.2 x y + 0.25 y^{2}$ .
4. ${(\frac{2}{3} k + 9)}^{2}$
  Show solution
  Answer $\frac{4}{9} k^{2} + 12 k + 81$
  ${(\frac{2}{3} k)}^{2} + 2 \cdot \frac{2}{3} k \cdot 9 + 81 = \frac{4}{9} k^{2} + 12 k + 81$ .
5. ${(y + \frac{1}{3 z})}^{2}$
  Show solution
  Answer $y^{2} + \frac{2 y}{3 z} + \frac{1}{9 z^{2}}$
  $y^{2} + 2 \cdot y \cdot \frac{1}{3 z} + \frac{1}{9 z^{2}} = y^{2} + \frac{2 y}{3 z} + \frac{1}{9 z^{2}}$ .
6. ${(\frac{1}{p} + \frac{2}{q})}^{2}$
  Show solution
  Answer $\frac{1}{p^{2}} + \frac{4}{p q} + \frac{4}{q^{2}}$
  $\frac{1}{p^{2}} + 2 \cdot \frac{1}{p} \cdot \frac{2}{q} + \frac{4}{q^{2}} = \frac{1}{p^{2}} + \frac{4}{p q} + \frac{4}{q^{2}}$ .
2
NumericalEasyNCERT Exercise Set 4.1, Q2
Use the same identity to find each square.
1. $73^{2}$
  Show answer
  Answer 5329
  $(70 + 3)^{2} = 4900 + 420 + 9 = 5329$ .
2. $102^{2}$
  Show answer
  Answer 10404
  $(100 + 2)^{2} = 10000 + 400 + 4 = 10404$ .
3. $306^{2}$
  Show answer
  Answer 93636
  $(300 + 6)^{2} = 90000 + 3600 + 36 = 93636$ .
Quick answerEasy
Which of these is an identity?
$x^{2} = 4 x$ $(y + 2)^{2} = y^{2} + 4$ $(m + n)^{2} = m^{2} + 2 m n + n^{2}$ $3 t = t + 6$
Show answer
Answer (C) $(m + n)^{2} = m^{2} + 2 m n + n^{2}$
Only the third holds for every value of the letters. The first and last are true only for particular values ( $x = 0$ or $x = 4$ ; $t = 3$ ), and the second is false whenever $y \neq 0$ .

4.3 Factorisation of Algebraic Expressions Using Identities

An identity can be read in either direction. From left to right, $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ expands a square. From right to left, it says that an expression shaped like $a^{2} + 2 a b + b^{2}$ is really the square of $a + b$ . Rewriting an expression as a product is called factorisation (गुणनखंडन), and each part of the product is a factor.

How do we recognise a perfect-square trinomial (पूर्ण वर्ग)? Run three checks. Is the first term a square? Is the last term a square? Is the middle term exactly twice the product of their square roots? Fig. 4.4 runs the checks on $49 m^{2} + 42 m + 9$ , and then on a look-alike that fails.

4/5So $49 m^{2} + 42 m + 9 = (7 m + 3)^{2}$ .

Fig. 4.4A trinomial is a perfect square only if its end terms are squares and its middle term is twice the product of their roots.

Step 1 shows the expression 49m squared plus 42m plus 9. Step 2 boxes the first and last terms and notes that they are the squares of 7m and 3. Step 3 boxes the middle term and checks that 2 times 7m times 3 equals 42m, with a tick. Step 4 writes the result: 49m squared plus 42m plus 9 equals (7m + 3) squared. Step 5 shows the look-alike 49m squared plus 40m plus 9, whose middle term should be 42m but is 40m, so it is not a perfect square.

Example 5NCERT Example 5 · p. 72

The simplest case

Factorise $y^{2} + 10 y + 25$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 6NCERT Example 6 · p. 72

A coefficient in front

Factorise $16 t^{2} + 56 t + 49$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 7NCERT Example 7 · p. 72

Take out a common factor first

Factorise $3 x^{3} + 24 x^{2} + 48 x$ .

Try it yourself first. Then reveal the solution one step at a time.

Example

Fractions are no obstacle

Factorise $4 x^{2} + 2 x + \frac{1}{4}$ .

Try it yourself first. Then reveal the solution one step at a time.

TheoremSquare of a difference

For all numbers $a$ and $b$ , $(a - b)^{2} = a^{2} - 2 a b + b^{2}$ .

An area picture confirms it (Fig. 4.5). Start with a square of side $a$ and cut away two strips of width $b$ : one down the side and one across the top. The strips overlap in a small $b \times b$ corner, so that corner is removed twice. Giving it back once leaves exactly a square of side $a - b$ .

4/5Put the corner square $b^{2}$ back once: $a^{2} - 2 a b + b^{2}$ .

Fig. 4.5Two strips of area

a b

come off, but their shared corner was counted twice, so

b^{2}

is added back.NCERT Fig. 4.3 · p. 74

Step 1 shows a blue square of side a, area a squared. Step 2 slides a strip of width b off its right side; the strip turns orange and is labelled ab, leaving a squared minus ab. Step 3 marks the top strip, also of area ab, in orange; its right end lies over the corner that has already gone, shown hatched, so the corner has been taken away twice. Step 4 puts a green square b squared back into that corner, giving a squared minus 2ab plus b squared. Step 5 shows only the blue square that is left, with sides a minus b, and the identity (a minus b) squared equals a squared minus 2ab plus b squared.

With both square identities in hand, the trick from the start of the chapter needs only one line of algebra. Everything is written in terms of the middle number, $n$ .

TheoremWhy the pattern always works

For every number $n$ , $(n + 1)^{2} - (n - 1)^{2} = 4 n$ .

The square of a difference is handy for numbers just below a round number.

Example 8NCERT Example 8 · p. 73

Just below a hundred

Find $98^{2}$ mentally.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 2 questions

Practice 4.2: questions with answers

1
Short answerMediumNCERT Exercise Set 4.2, Q1
Factorise completely.
1. $25 x^{2} + 30 x y + 9 y^{2}$
  Show solution
  Answer $(5 x + 3 y)^{2}$
  $(5 x)^{2} + 2 (5 x) (3 y) + (3 y)^{2} = (5 x + 3 y)^{2}$ .
2. $16 m^{2} + 8 m n + n^{2}$
  Show solution
  Answer $(4 m + n)^{2}$
  $16 m^{2} = (4 m)^{2}$ and $2 \times 4 m \times n = 8 m n$ , so $(4 m + n)^{2}$ .
3. $81 p^{2} + 36 p q + 4 q^{2}$
  Show solution
  Answer $(9 p + 2 q)^{2}$
  $(9 p)^{2} + 2 (9 p) (2 q) + (2 q)^{2} = (9 p + 2 q)^{2}$ .
4. $36 a^{2} + 16 a b + \frac{16}{9} b^{2}$
  Show solution
  Answer ${(6 a + \frac{4}{3} b)}^{2}$
  $\frac{16}{9} b^{2} = {(\frac{4}{3} b)}^{2}$ and $2 \times 6 a \times \frac{4}{3} b = 16 a b$ , so the answer is ${(6 a + \frac{4}{3} b)}^{2}$ .
5. $5 x^{2} + 20 x y + 20 y^{2}$
  Show solution
  Answer $5 (x + 2 y)^{2}$
  $5 x^{2} + 20 x y + 20 y^{2} = 5 (x^{2} + 4 x y + 4 y^{2}) = 5 (x + 2 y)^{2}$ .
6. $\frac{2}{3} p^{2} + 4 p q + 6 q^{2}$
  Show solution
  Answer $\frac{2}{3} (p + 3 q)^{2}$
  Take out $\frac{2}{3}$ : $\frac{2}{3} (p^{2} + 6 p q + 9 q^{2}) = \frac{2}{3} (p + 3 q)^{2}$ . Check: $\frac{2}{3} \times 6 = 4$ and $\frac{2}{3} \times 9 = 6$ .
2
NumericalEasyNCERT Exercise Set 4.2, Q2
Use $(a - b)^{2} = a^{2} - 2 a b + b^{2}$ to find each square.
1. $88^{2}$
  Show answer
  Answer 7744
  $(90 - 2)^{2} = 8100 - 360 + 4 = 7744$ .
2. $197^{2}$
  Show answer
  Answer 38809
  $(200 - 3)^{2} = 40000 - 1200 + 9 = 38809$ .
3. $495^{2}$
  Show answer
  Answer 245025
  $(500 - 5)^{2} = 250000 - 5000 + 25 = 245025$ .

4.4 More Identities

Squares of two-term sums are now familiar. For a three-term sum such as $(a + b + c)^{2}$ , multiplying out all nine pairs of terms would work, but a shortcut exists: bundle $b + c$ into one number and reuse the two-term identity.

DerivationSquare of a three-term sum

Let $d = b + c$ . Then $a + b + c = a + d$ , and the square of a sum gives

(a + b + c)^{2} = (a + d)^{2} = a^{2} + 2 a d + d^{2}

Now put $d = b + c$ back and expand each piece:

= a^{2} + 2 a (b + c) + (b + c)^{2} = a^{2} + 2 a b + 2 a c + b^{2} + 2 b c + c^{2}

Grouped so that it is easy to remember:

Square of a three-term sum

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a

In words: square each of the three terms, then add twice the product of every pair. Three terms make three pairs, $a b$ , $b c$ and $c a$ . Fig. 4.6 makes the same substitution with areas.

3/3Three squares and three pairs of equal rectangles: $2 a b$ , $2 b c$ and $2 c a$ .

Fig. 4.6Splitting

d = b + c

turns four pieces into nine: three squares and three matching pairs.NCERT Fig. 4.4 · p. 76

Step 1 shows a square of side a plus d divided into four pieces: a blue a squared, two orange pieces ad and a green d squared, with the sides labelled a and d. Step 2 draws dashed lines that split each side d into b and c, so the square has nine pieces labelled a squared, ab, ca, ab, b squared, bc, ca, bc and c squared. Step 3 recolours the pieces: the three squares a squared, b squared and c squared on the diagonal, and the equal pairs ab, bc and ca in matching colours, with the identity written beside the square.

Example 9NCERT Example 9 · p. 76

Squaring a three-digit number

Find $213^{2}$ by splitting 213 into three parts.

Try it yourself first. Then reveal the solution one step at a time.

Our toolkit now holds three identities:

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$
$(a - b)^{2} = a^{2} - 2 a b + b^{2}$
$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$

Choosing the right one matters. For $198^{2}$ , the split $200 - 2$ needs one easy subtraction, while $100 + 90 + 8$ needs six products. Look for a split into a round number and a small leftover.

Practice · 4 questions

Practice 4.3: questions with answers

1
NumericalEasyNCERT Exercise Set 4.3, Q1
Find each square. First decide which identity makes the work easiest.
1. $122^{2}$
  Show answer
  Answer 14884
  $(120 + 2)^{2} = 14400 + 480 + 4 = 14884$ .
2. $69^{2}$
  Show answer
  Answer 4761
  $(70 - 1)^{2} = 4900 - 140 + 1 = 4761$ .
3. $395^{2}$
  Show answer
  Answer 156025
  $(400 - 5)^{2} = 160000 - 4000 + 25 = 156025$ .
4. $312^{2}$
  Show answer
  Answer 97344
  $(300 + 10 + 2)^{2} = 90000 + 100 + 4 + 6000 + 40 + 1200 = 97344$ .
5. $1006^{2}$
  Show answer
  Answer 1012036
  $(1000 + 6)^{2} = 1000000 + 12000 + 36 = 1012036$ .
6. $2030^{2}$
  Show answer
  Answer 4120900
  $(2000 + 30)^{2} = 4000000 + 120000 + 900 = 4120900$ .
2
Short answerMediumNCERT Exercise Set 4.3, Q2
Factorise using a suitable identity.
1. $25 k^{2} - 40 k + 16$
  Show solution
  Answer $(5 k - 4)^{2}$
  $(5 k)^{2} - 2 (5 k) (4) + 4^{2} = (5 k - 4)^{2}$ .
2. $\frac{4}{9} a^{2} + 4 a b + 9 b^{2}$
  Show solution
  Answer ${(\frac{2}{3} a + 3 b)}^{2}$
  $\frac{4}{9} a^{2} = {(\frac{2}{3} a)}^{2}$ , $9 b^{2} = (3 b)^{2}$ and $2 \times \frac{2}{3} a \times 3 b = 4 a b$ .
3. $\frac{x^{2}}{4} + \frac{y^{2}}{9} + 4 z^{2} + \frac{x y}{3} - \frac{4 y z}{3} - 2 z x$
  Show solution
  Answer ${(\frac{x}{2} + \frac{y}{3} - 2 z)}^{2}$
  The squares come from $\frac{x}{2}$ , $\frac{y}{3}$ and $2 z$ . The cross terms $- \frac{4 y z}{3}$ and $- 2 z x$ are negative while $+ \frac{x y}{3}$ is positive, so the $z$ -term carries the minus sign: $2 \cdot \frac{x}{2} \cdot \frac{y}{3} = \frac{x y}{3}$ , $2 \cdot \frac{y}{3} \cdot (- 2 z) = - \frac{4 y z}{3}$ and $2 \cdot (- 2 z) \cdot \frac{x}{2} = - 2 z x$ .
4. $\frac{x^{2}}{9} + 2 + \frac{9}{x^{2}}$
  Show solution
  Answer ${(\frac{x}{3} + \frac{3}{x})}^{2}$
  $2 \times \frac{x}{3} \times \frac{3}{x} = 2$ , so the expression is ${(\frac{x}{3} + \frac{3}{x})}^{2}$ .
5. $4 p^{2} + 9 q^{2} + r^{2} + 12 p q - 4 p r - 6 q r$
  Show solution
  Answer $(2 p + 3 q - r)^{2}$
  Roots $2 p$ , $3 q$ , $r$ . Both cross terms with $r$ are negative, so take $- r$ : $2 (2 p) (3 q) = 12 p q$ , $2 (3 q) (- r) = - 6 q r$ , $2 (- r) (2 p) = - 4 p r$ .
3
Short answerEasyNCERT Exercise Set 4.3, Q3
Expand using the square of a three-term sum.
1. $(2 a + b + 5 c)^{2}$
  Show solution
  Answer $4 a^{2} + b^{2} + 25 c^{2} + 4 a b + 10 b c + 20 c a$
  Squares: $4 a^{2}$ , $b^{2}$ , $25 c^{2}$ . Pairs: $2 (2 a) (b) = 4 a b$ , $2 (b) (5 c) = 10 b c$ , $2 (5 c) (2 a) = 20 c a$ .
2. $(4 m - 3 n - 2 k)^{2}$
  Show solution
  Answer $16 m^{2} + 9 n^{2} + 4 k^{2} - 24 m n + 12 n k - 16 k m$
  Squares: $16 m^{2}$ , $9 n^{2}$ , $4 k^{2}$ . Pairs: $2 (4 m) (- 3 n) = - 24 m n$ , $2 (- 3 n) (- 2 k) = 12 n k$ , $2 (- 2 k) (4 m) = - 16 k m$ .
4
Short answerMediumNCERT Exercise Set 4.3, Q4
Is each statement an identity? Answer True if it holds for all values of the letters.
1. $(a + b)^{2} + (a - b)^{2} = 2 a^{2} + 2 b^{2}$
  TrueFalse
  Show answer
  Answer True
  Adding the two expansions, the $2 a b$ and $- 2 a b$ cancel, leaving $2 a^{2} + 2 b^{2}$ .
2. $(a + b + c)^{2} - (a - b - c)^{2} = 4 a b + 4 a c$
  TrueFalse
  Show answer
  Answer True
  With $d = b + c$ this is $(a + d)^{2} - (a - d)^{2} = 4 a d = 4 a b + 4 a c$ .
3. $(a - b - c)^{2} = a^{2} - b^{2} - c^{2} - 2 a b - 2 b c - 2 c a$
  TrueFalse
  Show answer
  Answer False
  All three squared terms come out positive: $(a - b - c)^{2} = a^{2} + b^{2} + c^{2} - 2 a b + 2 b c - 2 c a$ . With $a = 1$ , $b = 2$ , $c = 3$ the left side is 16 but the right side is $- 34$ .

How would you square 47 in your head? The product $50 \times 44$ is easy, and it falls short of $47^{2}$ by only 9. The reason is the difference of two squares from Class 8, $a^{2} - b^{2} = (a + b) (a - b)$ , with $b^{2}$ moved to the other side:

a^{2} = (a + b) (a - b) + b^{2}

TheoremDifference of two squares

For all numbers $a$ and $b$ , $a^{2} - b^{2} = (a + b) (a - b)$ .

Fig. 4.7 shows the rearranged form with areas. Cut a $b \times b$ corner out of a square of side $a$ . Then cut the leftover L-shape into two strips and turn one of them, so that together they make a rectangle $(a + b)$ long and $(a - b)$ tall.

4/5Turn the thin strip and put it beside the wide one: a rectangle $(a + b)$ long and $(a - b)$ tall.

Fig. 4.7The square

a^{2}

becomes a rectangle

(a + b) (a - b)

plus a small square

b^{2}

.NCERT Fig. 4.5 · p. 77

Step 1 shows a blue square of side a. Step 2 lifts a green corner square of side b off the top-right and parks it to one side, leaving an L-shape of area a squared minus b squared. Step 3 marks a cut along the top of the wide part, separating a thin orange strip, (a minus b) by b, from a wide blue strip, a by (a minus b). Step 4 turns the thin strip through a right angle and slides it against the right end of the wide strip, making one rectangle with sides a plus b and a minus b. Step 5 places the small square beside the rectangle and states that a squared equals (a + b)(a minus b) plus b squared.

Example

Friendly partners

Use $a^{2} = (a + b) (a - b) + b^{2}$ to find (i) $96^{2}$ and (ii) $38^{2}$ .

Try it yourself first. Then reveal the solution one step at a time.

Think about itNCERT Think and Reflect · p. 78

Read an identity from a picture

Four equal rectangles, each $a$ by $b$ , surround a small square to make one big square (Fig. 4.8). Find the big square’s area in two different ways. Which identity do you get? Check it with $a = 5$ and $b = 2$ .

2/3Each rectangle is $a$ by $b$ , and the middle square has side $a - b$ .

Fig. 4.8One square, counted two ways, gives

(a + b)^{2} = (a - b)^{2} + 4 a b

.NCERT Fig. 4.6 · p. 78

Step 1 shows a big square with sides marked a and b, made of four orange rectangles arranged like a pinwheel around a small purple square, and asks which identity it shows. Step 2 pulls the four rectangles slightly apart and labels each one ab, with a question mark in the middle square, whose side is a minus b. Step 3 puts the pieces back, labels the middle square (a minus b) squared and writes (a + b) squared equals (a minus b) squared plus 4ab.

A bigger challenge: expand $(a + b + c)^{2} + (a + b - c)^{2} + (a - b + c)^{2} + (a - b - c)^{2}$ and add. Nearly every term cancels. Can you draw the short answer as a set of squares?

4.5 Factorisation Using Algebra Tiles

Algebra tiles turn multiplication and factorisation into a puzzle you can hold. A set has three kinds of piece: a big square of side $x$ (area $x^{2}$ ), a long strip $x$ by 1 (area $x$ ), and a small square 1 by 1 (area 1). The length $x$ is deliberately not a whole number of units, so a strip can never be swapped for a row of small squares.

An expression such as $x^{2} + 6 x + 8$ becomes a pile of tiles: one big square, six strips and eight small squares. Factorising it means fitting all of them, with no gaps and no overlaps, into one rectangle. The two sides of that rectangle are the factors.

5/5The rectangle is $x + 4$ wide and $x + 2$ tall, so $x^{2} + 6 x + 8 = (x + 2) (x + 4)$ .

Fig. 4.9Six strips split as 4 and 2 leave a 4 by 2 gap, exactly right for the 8 unit tiles.NCERT Fig. 4.7 · p. 79

Step 1 shows a loose pile of tiles: one blue x-squared tile, six orange x-strips standing upright and eight small green unit tiles. Step 2 moves the x-squared tile into place on the right. Step 3 stands four strips side by side to its left and lays two strips flat underneath it. Step 4 fills the 4 by 2 gap in the bottom-left corner with the eight unit tiles exactly; a row of four unit tiles is visibly longer than the side x, because x is not a whole number of units. Step 5 labels the width x plus 4 and the height x plus 2 and writes x squared plus 6x plus 8 equals (x + 2)(x + 4).

Why did it work? The strips beside the big square and the strips under it fix the size of the empty corner. With 4 on one side and 2 on the other, the corner is 4 by 2, and that holds exactly the 8 unit tiles in the pile. Any other sharing of the strips would leave tiles over or a gap. The finished rectangle tells two stories at once:

multiplying its sides gives its area: $(x + 2) (x + 4) = x^{2} + 6 x + 8$ ;
reading its sides off the tiles factorises the area: $x^{2} + 6 x + 8 = (x + 2) (x + 4)$ .

InteractiveAlgebra tile board

An algebra tile board. A menu offers expressions such as x squared + 6x + 8, x squared + 4x + 5 and 2x squared + 5x + 2. The x-squared tiles sit in the corner; plus and minus buttons set how many unit columns go on the right and how many rows go underneath. The board draws the strips and unit tiles this shape needs, shows empty dashed spaces when tiles are missing and a left-over pile when tiles are spare, and labels the two side lengths. Cards count the x-strips and unit tiles used against those given, and a message announces when the rectangle is complete and states the factorisation. A button lists the factor pairs of the constant term as a hint.

Fig. 4.10Pick an expression, then add unit columns and rows until every tile fits into one rectangle.

Try thisNCERT Think and Reflect · p. 79

At the tile board

Predict first

Before building it: how many unit tiles does the product

(x + 1) (x + 5)

need?

56110

Choose $x^{2} + 7 x + 10$ and build the rectangle. What are its sides?
Choose $x^{2} + 9 x + 18$ . Try 1, then 2, then 3 unit columns on the right, each time adding rows until all 9 strips are used. Which choice also uses up all 18 unit tiles?
Choose $x^{2} + 4 x + 5$ . Can you make a rectangle? Press Show factor pairs to explain what you find.

Compare the numbers in $(x + 2) (x + 4) = x^{2} + 6 x + 8$ with those in the tile-board rectangle $(x + 3) (x + 6) = x^{2} + 9 x + 18$ . In each case the middle coefficient is the sum of the two numbers in the brackets, and the last term is their product. Multiplying out the brackets shows why this always happens.

TheoremTwo binomials with the same first term

For all numbers $x$ , $a$ and $b$ , $(x + a) (x + b) = x^{2} + (a + b) x + a b$ .

Tiles also cope with a first coefficient bigger than 1. To factorise $2 x^{2} + 5 x + 2$ , stack the two big squares one above the other. Fig. 4.11 shows how the five strips and two unit tiles complete the rectangle.

4/4 $2 x^{2} + 5 x + 2 = (2 x + 1) (x + 2)$ ; the strips count $1 + 2 \times 2 = 5$ .

Fig. 4.11The tiles for

2 x^{2} + 5 x + 2

fit a rectangle

x + 2

wide and

2 x + 1

tall.NCERT Fig. 4.8 · p. 80

Step 1 shows the tiles in a pile: two blue x-squared tiles, five orange strips and two green unit tiles. Step 2 stacks the two big squares one above the other and lays one strip flat underneath them, making a column 2x plus 1 tall. Step 3 builds two more columns to its right, each made of two upright strips with one unit tile at the bottom, so each is also 2x plus 1 tall. Step 4 labels the width x plus 2 and the height 2x plus 1 and writes 2x squared plus 5x plus 2 equals (2x + 1)(x + 2), noting that the strips count 1 plus 2 times 2 equals 5.

Where did the 5 come from? One strip under the big squares, plus two strips in each of the two side columns: $1 + 2 \times 2 = 5$ . Multiplying out any two linear brackets shows the general pattern.

TheoremTwo general linear binomials

For all numbers $p$ , $q$ , $a$ , $b$ and $x$ , $(p x + a) (q x + b) = p q x^{2} + (p b + q a) x + a b$ .

4.6 Factorisation Without Using Algebra Tiles

Tiles get tiresome for large numbers: nobody wants to lay out $x^{2} + 25 x + 144$ . Luckily the tiles have already shown us what to look for. By the product rule for two binomials, $x^{2} + b x + c$ factorises as $(x + p) (x + q)$ exactly when we can find two numbers $p$ and $q$ with

p + q = b and p q = c

The product condition is the stricter one, so list the factor pairs of $c$ and check each pair's sum. This method is called splitting the middle term: the pair that works tells us how to break $b x$ into $p x + q x$ . Fig. 4.12 runs the search for $x^{2} + 9 x + 20$ .

4/4So $9 x = 4 x + 5 x$ and $x^{2} + 9 x + 20 = (x + 4) (x + 5)$ .

Fig. 4.12Search the factor pairs of the constant for the one whose sum is the middle coefficient.

Step 1 shows the expression x squared plus 9x plus 20 and the goal: two numbers with product 20 and sum 9, above an empty table with columns pair, product and sum. Step 2 fills in the pairs 1 and 20, with sum 21, and 2 and 10, with sum 12, each marked with a cross. Step 3 adds the pair 4 and 5, with sum 9, highlighted with a tick. Step 4 splits 9x as 4x plus 5x and writes x squared plus 9x plus 20 equals (x + 4)(x + 5).

Example 10NCERT Example 10 · p. 81

Sum 9, product 20

Factorise $x^{2} + 9 x + 20$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 11NCERT Example 11 · p. 81

Right product, wrong sum

Factorise $x^{2} + 13 x + 36$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 12NCERT Example 12 · p. 81

A negative middle term

Factorise $x^{2} - 9 x + 14$ .

Try it yourself first. Then reveal the solution one step at a time.

Example

A negative constant term

Factorise $t^{2} + 2 t - 35$ .

Try it yourself first. Then reveal the solution one step at a time.

Table 4.1Reading the signs of

p

and

q

from

x^{2} + b x + c

$c$ is	$b$ is	$p$ and $q$ are	Example
positive	positive	both positive	$x^{2} + 7 x + 6 = (x + 1) (x + 6)$
positive	negative	both negative	$x^{2} - 8 x + 12 = (x - 2) (x - 6)$
negative	positive	opposite signs; the one farther from 0 is positive	$x^{2} + 3 x - 10 = (x + 5) (x - 2)$
negative	negative	opposite signs; the one farther from 0 is negative	$x^{2} - 3 x - 10 = (x - 5) (x + 2)$

When the first coefficient is not 1, one change is needed. Multiply the first and last coefficients, then look for two numbers with that product whose sum is the middle coefficient. The reason comes from the general product: the two parts of the middle coefficient, $p b$ and $q a$ , multiply to $p q \times a b$ , the first coefficient times the last.

Example

First coefficient 3

Factorise $3 x^{2} + 10 x + 8$ .

Try it yourself first. Then reveal the solution one step at a time.

Practice · 3 questions

Practice 4.4: questions with answers

1
Short answerMediumNCERT Exercise Set 4.4, Q1
Complete each identity.
1. $k^{2} - 13 k + 40 = (□) (□)$
  $(k - 5) (k - 8)$ $(k + 5) (k + 8)$ $(k - 4) (k - 10)$ $(k - 5) (k + 8)$
  Show answer
  Answer (A) $(k - 5) (k - 8)$
  We need product 40 and sum $- 13$ : $- 5$ and $- 8$ .
2. $(□) (x + 2) = 2 x^{2} + x - 6$
  $2 x - 3$ $2 x + 3$ $x - 3$ $2 x - 6$
  Show answer
  Answer (A) $2 x - 3$
  $(2 x - 3) (x + 2) = 2 x^{2} + 4 x - 3 x - 6 = 2 x^{2} + x - 6$ .
3. $6 x^{2} + □ x - 15 = (2 x + 3) (3 x - 5)$ . What number goes in the box?
  Show answer
  Answer -1
  $(2 x + 3) (3 x - 5) = 6 x^{2} - 10 x + 9 x - 15 = 6 x^{2} - x - 15$ , so the box holds $- 1$ .
4. $4 x^{2} + 8 x + 3 = (□) (□)$
  Show solution
  Answer $(2 x + 1) (2 x + 3)$
  $4 \times 3 = 12$ ; numbers with product 12 and sum 8 are 2 and 6. So $4 x^{2} + 2 x + 6 x + 3 = 2 x (2 x + 1) + 3 (2 x + 1) = (2 x + 1) (2 x + 3)$ .
2
NumericalEasyNCERT Exercise Set 4.4, Q2
Choose a suitable identity and find each value without multiplying directly.
1. $52^{2}$
  Show answer
  Answer 2704
  $(50 + 2)^{2} = 2500 + 200 + 4 = 2704$ .
2. $39^{2}$
  Show answer
  Answer 1521
  $(40 - 1)^{2} = 1600 - 80 + 1 = 1521$ .
3. $26 \times 14$
  Show answer
  Answer 364
  $(20 + 6) (20 - 6) = 400 - 36 = 364$ .
4. $115^{2}$
  Show answer
  Answer 13225
  $(100 + 15)^{2} = 10000 + 3000 + 225 = 13225$ .
5. $94^{2}$
  Show answer
  Answer 8836
  $(100 - 6)^{2} = 10000 - 1200 + 36 = 8836$ .
6. $19 \times 26$
  Show answer
  Answer 494
  $(20 - 1) (20 + 6) = 400 + (6 - 1) \times 20 - 6 = 494$ , using $(x + a) (x + b)$ .
7. $37 \times 44$
  Show answer
  Answer 1628
  $(40 - 3) (40 + 4) = 1600 + 1 \times 40 - 12 = 1628$ .
8. $305^{2}$
  Show answer
  Answer 93025
  $(300 + 5)^{2} = 90000 + 3000 + 25 = 93025$ .
3
Short answerMediumNCERT Exercise Set 4.4, Q3
Factorise.
1. $4 x^{2} + y^{2} + 9 z^{2} + 4 x y - 12 x z - 6 y z$
  Show solution
  Answer $(2 x + y - 3 z)^{2}$
  Roots $2 x$ , $y$ , $3 z$ ; the two cross terms containing $z$ are negative, so use $- 3 z$ : $2 (2 x) (y) = 4 x y$ , $2 (2 x) (- 3 z) = - 12 x z$ , $2 (y) (- 3 z) = - 6 y z$ .
2. $9 m^{2} + 16 n^{2} - 24 m n$
  Show solution
  Answer $(3 m - 4 n)^{2}$
  $(3 m)^{2} - 2 (3 m) (4 n) + (4 n)^{2} = (3 m - 4 n)^{2}$ .
3. $t^{2} + 3 t - 40$
  Show solution
  Answer $(t + 8) (t - 5)$
  Product $- 40$ , sum 3: 8 and $- 5$ .
4. $25 p^{2} + 20 p q + 4 q^{2}$
  Show solution
  Answer $(5 p + 2 q)^{2}$
  $2 (5 p) (2 q) = 20 p q$ , so $(5 p + 2 q)^{2}$ .
5. $36 a^{2} + 4 b^{2} + 49 c^{2} - 24 a b + 84 a c - 28 b c$
  Show solution
  Answer $(6 a - 2 b + 7 c)^{2}$
  Roots $6 a$ , $2 b$ , $7 c$ . The cross terms with $b$ are negative, so take $- 2 b$ : $2 (6 a) (- 2 b) = - 24 a b$ , $2 (6 a) (7 c) = 84 a c$ , $2 (- 2 b) (7 c) = - 28 b c$ .

4.7 Finding New Identities

Squares have given us plenty. What happens with cubes? Multiply $(a + b)^{2}$ by one more factor $a + b$ :

DerivationCube of a sum

(a + b)^{3} = (a + b) (a^{2} + 2 a b + b^{2})

= a^{3} + 2 a^{2} b + a b^{2} + a^{2} b + 2 a b^{2} + b^{3}

Collecting like terms:

Cube of a sum

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

The volume model uses a cube of edge $a + b$ , cut along every edge where $a$ meets $b$ (Fig. 4.13). That gives eight blocks. Sort them by how many of their three edges have length $b$ . With no $b$ -edges there is one block, $a^{3}$ . With exactly one there are three, since the $b$ -edge can point in three directions: $3 a^{2} b$ . With exactly two there are again three: $3 a b^{2}$ . With three $b$ -edges there is one block, $b^{3}$ .

6/6 $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ .

Fig. 4.13Eight blocks make the cube:

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

.NCERT Fig. 4.10 · p. 83

Step 1 shows a cube drawn in three dimensions with every edge made of a length a and a length b. Step 2 pulls the eight blocks apart: a large blue cube at the front bottom, a small pink cube at the back top, three orange slabs and three green rods. Step 3 highlights the two cubes, a cubed and b cubed. Step 4 highlights the three orange slabs, each a by a by b, labelled a squared b. Step 5 highlights the three green rods, each a by b by b, labelled a b squared. Step 6 shows all eight blocks with the identity (a + b) cubed equals a cubed plus 3a squared b plus 3a b squared plus b cubed.

TheoremCube of a difference

For all numbers $a$ and $b$ , $(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$ .

Example 13NCERT Example 13 · p. 84

Recognising a perfect cube

Show that $27 a^{3} + 27 a^{2} b + 9 a b^{2} + b^{3}$ is a perfect cube. Then use it to find $31^{3}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 14NCERT Example 14 · p. 84

Is it the cube of a difference?

Decide whether $64 x^{3} - 144 x^{2} y + 108 x y^{2} - 27 y^{3}$ is the cube of a binomial. If it is, find the binomial and check with numbers.

Try it yourself first. Then reveal the solution one step at a time.

Example

Cubing a number near 100

Find $102^{3}$ .

Try it yourself first. Then reveal the solution one step at a time.

Some products collapse in a surprising way. Multiply $x - y$ by $x^{2} + x y + y^{2}$ with a grid, putting one term from each bracket into every cell (Fig. 4.14).

4/4 $(x - y) (x^{2} + x y + y^{2}) = x^{3} - y^{3}$ .

Fig. 4.14In the grid the diagonal pairs cancel, leaving only

x^{3}

and

- y^{3}

Step 1 shows an empty grid: rows headed x and minus y, columns headed x squared, xy and y squared. Step 2 fills the six cells with the products x cubed, x squared y, x y squared, minus x squared y, minus x y squared and minus y cubed. Step 3 joins the two diagonal pairs x squared y with minus x squared y and x y squared with minus x y squared, fades them because each pair adds to zero, and highlights the corner cells x cubed and minus y cubed. Step 4 writes the result: (x minus y)(x squared plus xy plus y squared) equals x cubed minus y cubed.

TheoremDifference of two cubes

For all numbers $x$ and $y$ , $x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$ .

Now swap the signs: what do you expect $(x + y) (x^{2} - x y + y^{2})$ to collapse to? Make a guess, test it with $x = 2$ and $y = 1$ , and then expand to be sure.

TheoremSum of two cubes

For all numbers $x$ and $y$ , $x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})$ .

One more identity links three cubes with their product. It looks long, but the proof is just careful bookkeeping.

TheoremThree cubes

For all numbers $x$ , $y$ and $z$ , $x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$ .

Example 15NCERT Example 15 · p. 85

Cubes from sums and products

Three numbers have sum 9. The sum of their products in pairs is 26, and their product is 24. Find the sum of their cubes without finding the numbers.

Try it yourself first. Then reveal the solution one step at a time.

4.8 Simplifying Rational Expressions

A fraction such as $\frac{18}{24}$ is simplified by spotting a common factor: $\frac{18}{24} = \frac{3 \times 6}{4 \times 6} = \frac{3}{4}$ . A rational expression (परिमेय व्यंजक) is a fraction whose numerator and denominator are polynomials, such as $\frac{x^{2} + 5 x + 6}{x^{2} + 4 x + 3}$ . It is simplified in the same way, and our identities are what reveal the common factors.

Definition

Rational expression

A fraction whose numerator and denominator are polynomials, defined only where the denominator is not zero.

Example $\frac{x + 2}{x - 5}$ , for $x \neq 5$ .

Fig. 4.15 gives a picture of one such fraction. Both tile rectangles have the same height, $x + 3$ , so their areas are in the same ratio as their widths:

3/3 $\frac{(x + 2) (x + 3)}{(x + 1) (x + 3)} = \frac{x + 2}{x + 1}$ .

Fig. 4.15Same height, so the ratio of the areas is the ratio of the widths.

Step 1 shows two rectangles made of algebra tiles, labelled x squared plus 5x plus 6 and x squared plus 4x plus 3. Step 2 marks their widths, x plus 2 and x plus 1, and highlights their equal left sides, each of height x plus 3. Step 3 writes the fraction of the two areas, factorises top and bottom with the common factor x plus 3 picked out in colour, and cancels it to leave (x + 2) over (x + 1).

\frac{x^{2} + 5 x + 6}{x^{2} + 4 x + 3} = \frac{(x + 2) (x + 3)}{(x + 1) (x + 3)} = \frac{x + 2}{x + 1}

There is one condition. We may divide top and bottom by $x + 3$ only if $x + 3 \neq 0$ , because dividing by zero has no meaning. The original fraction also needs a non-zero denominator, so $x \neq - 1$ and $x \neq - 3$ . With those two values set aside, both sides agree.

Example 16NCERT Example 16 · p. 86

Simplifying with two factorisations

Simplify $\frac{4 x^{2} - 9}{2 x^{2} + 5 x + 3}$ , assuming the denominator is not zero.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 1 question

Practice 4.5: questions with answers

1
Short answerMediumNCERT Exercise Set 4.5, Q1
Simplify, assuming no denominator is zero.
1. $\frac{2 m^{2} + 2 m n - 12 n^{2}}{m^{2} + 5 m n + 6 n^{2}}$
  Show solution
  Answer $\frac{2 (m - 2 n)}{m + 2 n}$
  Numerator: $2 (m^{2} + m n - 6 n^{2}) = 2 (m + 3 n) (m - 2 n)$ . Denominator: $(m + 2 n) (m + 3 n)$ . Cancel $m + 3 n$ to get $\frac{2 (m - 2 n)}{m + 2 n}$ .
2. $\frac{8 a^{3} - 12 a^{2} b + 6 a b^{2} - b^{3}}{4 a^{2} - b^{2}}$
  Show solution
  Answer $\frac{(2 a - b)^{2}}{2 a + b}$
  The numerator is $(2 a)^{3} - 3 (2 a)^{2} b + 3 (2 a) b^{2} - b^{3} = (2 a - b)^{3}$ , and the denominator is $(2 a - b) (2 a + b)$ . Cancel $2 a - b$ once to leave $\frac{(2 a - b)^{2}}{2 a + b}$ .
3. $\frac{a^{3} + 8 b^{3} + 27 c^{3} - 18 a b c}{a + 2 b + 3 c}$
  Show solution
  Answer $a^{2} + 4 b^{2} + 9 c^{2} - 2 a b - 6 b c - 3 c a$
  Use the three-cubes identity with $x = a$ , $y = 2 b$ and $z = 3 c$ . Then $3 x y z = 18 a b c$ , so the numerator is $(a + 2 b + 3 c) (a^{2} + 4 b^{2} + 9 c^{2} - 2 a b - 6 b c - 3 c a)$ . Dividing by $a + 2 b + 3 c$ leaves the second bracket.
4. $\frac{16 a^{2} - 24 a b + 9 b^{2}}{16 a^{2} - 9 b^{2}}$
  Show solution
  Answer $\frac{4 a - 3 b}{4 a + 3 b}$
  $\frac{(4 a - 3 b)^{2}}{(4 a - 3 b) (4 a + 3 b)} = \frac{4 a - 3 b}{4 a + 3 b}$ .
5. $\frac{(y^{2} + 2 y - 15) (y^{2} - 4)}{(y^{2} + 3 y - 10) (y^{2} - 5 y + 6)}$
  Show solution
  Answer $\frac{y + 2}{y - 2}$
  Numerator: $(y + 5) (y - 3) (y - 2) (y + 2)$ . Denominator: $(y + 5) (y - 2) (y - 2) (y - 3)$ . Cancel $y + 5$ , $y - 3$ and one $y - 2$ to leave $\frac{y + 2}{y - 2}$ .
6. $\frac{k^{4} - 81}{k^{3} + 27}$
  Show solution
  Answer $\frac{(k - 3) (k^{2} + 9)}{k^{2} - 3 k + 9}$
  $k^{4} - 81 = (k^{2} - 9) (k^{2} + 9) = (k - 3) (k + 3) (k^{2} + 9)$ and, as a sum of two cubes, $k^{3} + 27 = (k + 3) (k^{2} - 3 k + 9)$ . Cancel $k + 3$ .

Factorising also settles practical puzzles, as the next two examples show.

Example 17NCERT Example 17 · p. 87

Carpeting a floor

A carpet shop sells three sizes of mat: big squares of side $x$ metres, runners $x$ metres by 1 metre, and small squares of side 1 metre. (a) Meena's floor measures $(x + 2)$ m by $(x + 6)$ m. Which mats cover it exactly, without cutting? (b) Her neighbour owns 1 big square, 10 runners and 21 small squares. Can they cover a rectangular floor exactly? (c) What about 1 big square, 6 runners and 10 small squares?

Try it yourself first. Then reveal the solution one step at a time.

Example 18NCERT Example 18 · p. 88

Two consecutive even numbers

The product of two consecutive even numbers is 168. Find the numbers.

Try it yourself first. Then reveal the solution one step at a time.

That completes our toolkit. Each identity began as a picture or a pattern and ended as a proof that holds for every number. Together they let us expand quickly, factorise with confidence and simplify fractions of polynomials.

Recap

Exploring Algebraic Identities: key points and summary

An identity holds for every value of its letters; an equation holds only for particular values, which we find by solving (Section 4.2).
$(a + b)^{2} = a^{2} + 2 a b + b^{2}$ and $(a - b)^{2} = a^{2} - 2 a b + b^{2}$ : a square splits into two squares and two equal strips (Section 4.2, Section 4.3).
Read backwards, identities factorise. A trinomial is a perfect square when its end terms are squares and its middle term is twice the product of their roots (Section 4.3).
$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$ (Section 4.4).
$a^{2} - b^{2} = (a + b) (a - b)$ , and its rearranged form $a^{2} = (a + b) (a - b) + b^{2}$ gives Śrīdharācārya's quick squares (Section 4.4).
Identities speed up arithmetic: split a number into a round part and a small part (Section 4.2).
Algebra tiles factorise $x^{2} + b x + c$ by building a rectangle; its side lengths are the factors (Section 4.5).
$(x + a) (x + b) = x^{2} + (a + b) x + a b$ and $(p x + a) (q x + b) = p q x^{2} + (p b + q a) x + a b$ (Section 4.5).
Without tiles, split the middle term: find two numbers whose product is the first coefficient times the last and whose sum is the middle coefficient (Section 4.6).
$(a \pm b)^{3} = a^{3} \pm 3 a^{2} b + 3 a b^{2} \pm b^{3}$ and $x^{3} \pm y^{3} = (x \pm y) (x^{2} \mp x y + y^{2})$ (Section 4.7).
$x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$ , so $x^{3} + y^{3} + z^{3} = 3 x y z$ whenever $x + y + z = 0$ (Section 4.7).
A rational expression is simplified by factorising and cancelling common factors, never terms, and only factors that are not zero (Section 4.8).

Flash cards · 11

Identity flashcards: expand or factorise?

1 / 11

End-of-chapter exercises

Practice · 17 questions

Practice questions with answers (NCERT “End-of-chapter exercises”)

1
Short answerMediumNCERT End-of-chapter Exercises, Q1
Expand each product with the help of an identity.
1. $(- 2 y + 5)^{2}$
  Show solution
  Answer $4 y^{2} - 20 y + 25$
  $(- 2 y)^{2} + 2 (- 2 y) (5) + 5^{2} = 4 y^{2} - 20 y + 25$ .
2. $(3 a + 8) (3 a - 8)$
  Show solution
  Answer $9 a^{2} - 64$
  Difference of squares: $(3 a)^{2} - 8^{2} = 9 a^{2} - 64$ .
3. $(\frac{k^{2}}{2} + 3) (\frac{k^{2}}{2} - 3)$
  Show solution
  Answer $\frac{k^{4}}{4} - 9$
  Difference of squares: ${(\frac{k^{2}}{2})}^{2} - 3^{2} = \frac{k^{4}}{4} - 9$ .
4. $(x + 3 y) (x^{2} - 3 x y + 9 y^{2})$
  Show solution
  Answer $x^{3} + 27 y^{3}$
  Sum of cubes with $3 y$ in place of $y$ : $x^{3} + (3 y)^{3} = x^{3} + 27 y^{3}$ .
5. ${(\frac{1}{3 t} - 3 t)}^{2}$
  Show solution
  Answer $\frac{1}{9 t^{2}} - 2 + 9 t^{2}$
  $\frac{1}{9 t^{2}} - 2 \cdot \frac{1}{3 t} \cdot 3 t + 9 t^{2} = \frac{1}{9 t^{2}} - 2 + 9 t^{2}$ .
6. $(- 2 p + q - 3 r)^{2}$
  Show solution
  Answer $4 p^{2} + q^{2} + 9 r^{2} - 4 p q - 6 q r + 12 p r$
  Squares $4 p^{2}$ , $q^{2}$ , $9 r^{2}$ ; pairs $2 (- 2 p) (q) = - 4 p q$ , $2 (q) (- 3 r) = - 6 q r$ , $2 (- 3 r) (- 2 p) = 12 p r$ .
7. ${(a - \frac{b}{2})}^{3}$
  Show solution
  Answer $a^{3} - \frac{3}{2} a^{2} b + \frac{3}{4} a b^{2} - \frac{1}{8} b^{3}$
  $a^{3} - 3 a^{2} \cdot \frac{b}{2} + 3 a \cdot \frac{b^{2}}{4} - \frac{b^{3}}{8} = a^{3} - \frac{3}{2} a^{2} b + \frac{3}{4} a b^{2} - \frac{1}{8} b^{3}$ .
8. ${(\frac{3}{2} u + \frac{2}{3} v)}^{3}$
  Show solution
  Answer $\frac{27}{8} u^{3} + \frac{9}{2} u^{2} v + 2 u v^{2} + \frac{8}{27} v^{3}$
  ${(\frac{3}{2} u)}^{3} + 3 {(\frac{3}{2} u)}^{2} (\frac{2}{3} v) + 3 (\frac{3}{2} u) {(\frac{2}{3} v)}^{2} + {(\frac{2}{3} v)}^{3} = \frac{27}{8} u^{3} + \frac{9}{2} u^{2} v + 2 u v^{2} + \frac{8}{27} v^{3}$ .
2
NumericalMediumNCERT End-of-chapter Exercises, Q2
Find each value using a suitable identity.
1. $28 \times 32$
  Show answer
  Answer 896
  $(30 - 2) (30 + 2) = 900 - 4 = 896$ .
2. $203 \times 197$
  Show answer
  Answer 39991
  $(200 + 3) (200 - 3) = 40000 - 9 = 39991$ .
3. $45 \times 35$
  Show answer
  Answer 1575
  $(40 + 5) (40 - 5) = 1600 - 25 = 1575$ .
4. $103^{3}$
  Show answer
  Answer 1092727
  $(100 + 3)^{3} = 1000000 + 90000 + 2700 + 27 = 1092727$ .
5. $99^{3}$
  Show answer
  Answer 970299
  $(100 - 1)^{3} = 1000000 - 30000 + 300 - 1 = 970299$ .
6. $201^{3}$
  Show answer
  Answer 8120601
  $(200 + 1)^{3} = 8000000 + 120000 + 600 + 1 = 8120601$ .
7. $(- 101)^{3}$
  Show answer
  Answer -1030301
  $(- 101)^{3} = - (100 + 1)^{3} = - (1000000 + 30000 + 300 + 1) = - 1030301$ .
8. $(- 198)^{3}$
  Show answer
  Answer -7762392
  $(- 198)^{3} = - (200 - 2)^{3} = - (8000000 - 240000 + 2400 - 8) = - 7762392$ .
3
Short answerMediumNCERT End-of-chapter Exercises, Q3
Factorise these two- and three-term expressions.
1. $4 p^{2} + 4 + \frac{1}{p^{2}}$
  Show solution
  Answer ${(2 p + \frac{1}{p})}^{2}$
  $2 \times 2 p \times \frac{1}{p} = 4$ , so it is ${(2 p + \frac{1}{p})}^{2}$ .
2. $25 a^{2} - \frac{1}{36 b^{2}}$
  Show solution
  Answer $(5 a - \frac{1}{6 b}) (5 a + \frac{1}{6 b})$
  Difference of squares: $(5 a)^{2} - {(\frac{1}{6 b})}^{2}$ .
3. $125 k^{3} - \frac{1}{8 k^{3}}$
  Show solution
  Answer $(5 k - \frac{1}{2 k}) (25 k^{2} + \frac{5}{2} + \frac{1}{4 k^{2}})$
  Difference of cubes with $x = 5 k$ and $y = \frac{1}{2 k}$ : $x^{2} + x y + y^{2} = 25 k^{2} + \frac{5}{2} + \frac{1}{4 k^{2}}$ .
4. $y^{2} + \frac{3 y}{10} + \frac{1}{50}$
  Show solution
  Answer $(y + \frac{1}{5}) (y + \frac{1}{10})$
  Two numbers with product $\frac{1}{50}$ and sum $\frac{3}{10}$ are $\frac{1}{5}$ and $\frac{1}{10}$ .
5. $8 m^{3} + \frac{n^{3}}{27}$
  Show solution
  Answer $(2 m + \frac{n}{3}) (4 m^{2} - \frac{2 m n}{3} + \frac{n^{2}}{9})$
  Sum of two cubes, with $2 m$ and $\frac{n}{3}$ in the places of $x$ and $y$ .
6. $16 h^{2} - 24 h + 9$
  Show solution
  Answer $(4 h - 3)^{2}$
  $(4 h)^{2} - 2 (4 h) (3) + 3^{2}$ .
4
Short answerHardNCERT End-of-chapter Exercises, Q3
Factorise these longer expressions.
1. $6 w - 8 + \frac{w^{3}}{8} - \frac{3 w^{2}}{2}$
  Show solution
  Answer ${(\frac{w}{2} - 2)}^{3}$
  Arrange by powers of $w$ : $\frac{w^{3}}{8} - \frac{3 w^{2}}{2} + 6 w - 8$ . With $A = \frac{w}{2}$ and $B = 2$ : $A^{3} = \frac{w^{3}}{8}$ , $3 A^{2} B = \frac{3 w^{2}}{2}$ , $3 A B^{2} = 6 w$ and $B^{3} = 8$ . The signs alternate, so it is $(A - B)^{3}$ .
2. $a^{3} + 8 b^{3} + c^{3} - 6 a b c$
  Show solution
  Answer $(a + 2 b + c) (a^{2} + 4 b^{2} + c^{2} - 2 a b - 2 b c - c a)$
  Three cubes with $x = a$ , $y = 2 b$ , $z = c$ : $3 x y z = 6 a b c$ .
3. $2 x^{3} - 16 y^{3} - 54 z^{3} - 36 x y z$
  Show solution
  Answer $2 (x - 2 y - 3 z) (x^{2} + 4 y^{2} + 9 z^{2} + 2 x y - 6 y z + 3 z x)$
  Take out 2: $x^{3} + (- 2 y)^{3} + (- 3 z)^{3} - 3 x (- 2 y) (- 3 z)$ . The three-cubes identity with $- 2 y$ and $- 3 z$ gives the answer.
4. $9 x^{2} + 4 y^{2} + 16 z^{2} + 12 x y + 16 y z + 24 z x$
  Show solution
  Answer $(3 x + 2 y + 4 z)^{2}$
  All cross terms are positive: $2 (3 x) (2 y) = 12 x y$ , $2 (2 y) (4 z) = 16 y z$ , $2 (4 z) (3 x) = 24 z x$ .
5. $64 p^{3} - 48 p^{2} + 12 p - 1$
  Show solution
  Answer $(4 p - 1)^{3}$
  $(4 p)^{3} - 3 (4 p)^{2} (1) + 3 (4 p) (1)^{2} - 1^{3}$ .
5
Short answerMediumNCERT End-of-chapter Exercises, Q4
Simplify, assuming the denominators are not zero.
1. $\frac{9 x^{2} - 1}{3 x^{2} + 2 x - 1}$
  Show solution
  Answer $\frac{3 x + 1}{x + 1}$
  Numerator: $(3 x - 1) (3 x + 1)$ . Denominator: the product $3 \times (- 1) = - 3$ and the sum 2 give the numbers 3 and $- 1$ , so $3 x^{2} + 3 x - x - 1 = 3 x (x + 1) - (x + 1) = (3 x - 1) (x + 1)$ . Cancel $3 x - 1$ to leave $\frac{3 x + 1}{x + 1}$ .
2. $\frac{2 x^{3} - 54 y^{3}}{x^{2} - 6 x y + 9 y^{2}}$
  Show solution
  Answer $\frac{2 (x^{2} + 3 x y + 9 y^{2})}{x - 3 y}$
  $\frac{2 (x - 3 y) (x^{2} + 3 x y + 9 y^{2})}{(x - 3 y)^{2}} = \frac{2 (x^{2} + 3 x y + 9 y^{2})}{x - 3 y}$ .
3. $\frac{m^{3} + 64 n^{3}}{m^{2} - 16 n^{2}}$
  Show solution
  Answer $\frac{m^{2} - 4 m n + 16 n^{2}}{m - 4 n}$
  $\frac{(m + 4 n) (m^{2} - 4 m n + 16 n^{2})}{(m + 4 n) (m - 4 n)} = \frac{m^{2} - 4 m n + 16 n^{2}}{m - 4 n}$ .
6
Short answerMediumNCERT End-of-chapter Exercises, Q5
Each expression below is the area of a rectangle, in square units. Factorise it to suggest a length and a breadth.
1. $9 p^{2} - 42 p q + 49 q^{2}$
  Show solution
  AnswerBoth sides $3 p - 7 q$ (it is a square).
  $9 p^{2} - 42 p q + 49 q^{2} = (3 p - 7 q)^{2}$ , so the rectangle is a square of side $3 p - 7 q$ , or $7 q - 3 p$ , whichever is positive.
2. $64 m^{2} - 25 n^{2}$
  Show solution
  Answer $8 m + 5 n$ and $8 m - 5 n$
  Difference of squares: $(8 m + 5 n) (8 m - 5 n)$ .
7
Short answerMediumNCERT End-of-chapter Exercises, Q6
Each expression below is the volume of a cuboid, in cubic units. Factorise it to suggest a length, a breadth and a height.
1. $7 m^{2} - 63 n^{2}$
  Show solution
  Answer $7$ , $m - 3 n$ and $m + 3 n$
  $7 (m^{2} - 9 n^{2}) = 7 (m - 3 n) (m + 3 n)$ .
2. $4 h r^{2} - 20 h r + 24 h$
  Show solution
  Answer $4 h$ , $r - 2$ and $r - 3$
  $4 h (r^{2} - 5 r + 6) = 4 h (r - 2) (r - 3)$ .
Short answerMediumNCERT End-of-chapter Exercises, Q7
A square sheet of tin has sides of 30 cm. A square of side $w$ cm is cut from each corner so that the edges can be folded up into an open box. Write the area of the tin that is left in terms of $w$ , and factorise it.
Show solution
Answer $900 - 4 w^{2} = 4 (15 - w) (15 + w)$ square centimetres
The tin left is $30^{2} - 4 w^{2} = 900 - 4 w^{2} = 4 (225 - w^{2}) = 4 (15 - w) (15 + w)$ square centimetres.
Short answerMediumNCERT End-of-chapter Exercises, Q8
Twice a number, minus three times its reciprocal, comes to 5. What could the number be?
Show solution
Answer $3$ or $- \frac{1}{2}$
$2 x - \frac{3}{x} = 5$ gives $2 x^{2} - 5 x - 3 = 0$ . Split $- 5 x$ using $- 6$ and $1$ (product $2 \times (- 3) = - 6$ , sum $- 5$ ): $2 x^{2} - 6 x + x - 3 = 2 x (x - 3) + (x - 3) = (2 x + 1) (x - 3) = 0$ .
So $x = 3$ or $x = - \frac{1}{2}$ . Check: $6 - 1 = 5$ , and $- 1 - (- 6) = 5$ .
Short answerMediumNCERT End-of-chapter Exercises, Q9
A rectangular solar panel has an area of $3 x^{2} + 11 x + 6$ square decimetres, and one of its sides is $x + 3$ decimetres. Find the other side.
Show solution
Answer $3 x + 2$ decimetres
$3 \times 6 = 18$ ; numbers with product 18 and sum 11 are 2 and 9. So $3 x^{2} + 2 x + 9 x + 6 = x (3 x + 2) + 3 (3 x + 2) = (x + 3) (3 x + 2)$ . The other side is $3 x + 2$ decimetres.
Short answerHardNCERT End-of-chapter Exercises, Q10
Both $x + 2$ and $3 x - 1$ are factors of $p x^{2} + q x - 4$ . Find $p$ and $q$ .
Show solution
Answer $p = 6$ , $q = 10$
$(x + 2) (3 x - 1) = 3 x^{2} + 5 x - 2$ , so $p x^{2} + q x - 4 = k (3 x^{2} + 5 x - 2)$ for some number $k$ .
The constant terms give $- 2 k = - 4$ , so $k = 2$ . Then $p = 3 k = 6$ and $q = 5 k = 10$ : the quadratic is $6 x^{2} + 10 x - 4$ .
ProofHardNCERT End-of-chapter Exercises, Q11
If $a + b + c = 7$ and $a^{2} + b^{2} + c^{2} = 21$ , prove that $a^{3} + b^{3} + c^{3} - 3 a b c = 49$ .
Show solution
$49 = (a + b + c)^{2} = 21 + 2 (a b + b c + c a)$ , so $a b + b c + c a = 14$ .
By the three-cubes identity, $a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - (a b + b c + c a)) = 7 (21 - 14) = 49$ .
ProofHardNCERT End-of-chapter Exercises, Q12
By factorising, show that $n^{3} + 3 n^{2} + 2 n$ is divisible by 6 for every natural number $n$ .
Show solution
$n^{3} + 3 n^{2} + 2 n = n (n^{2} + 3 n + 2) = n (n + 1) (n + 2)$ , the product of three consecutive natural numbers.
Among any two consecutive numbers one is even, so the product is divisible by 2. Among any three consecutive numbers one is a multiple of 3, so the product is divisible by 3. Being divisible by both 2 and 3, it is divisible by 6.
14
ProofHardNCERT End-of-chapter Exercises, Q13
Find each value.
1. $m^{3} - n^{3} - 8 - 6 m n$ , when $m - n = 2$
  Show answer
  Answer 0
  Take $x = m$ , $y = - n$ and $z = - 2$ . Then $x + y + z = m - n - 2 = 0$ and $3 x y z = 3 m (- n) (- 2) = 6 m n$ , so the expression is $x^{3} + y^{3} + z^{3} - 3 x y z = 0$ .
2. $13^{3} - 8^{3} - 5^{3}$ , without working out any cube
  Show answer
  Answer 1560
  $13 + (- 8) + (- 5) = 0$ , so the sum of the three cubes is $3 \times 13 \times (- 8) \times (- 5) = 1560$ . Check: $2197 - 512 - 125 = 1560$ .
Quick answerEasy
Which of these is not an identity?
$(x - 1)^{2} = x^{2} - 2 x + 1$ $(x - 1) (x + 1) = x^{2} - 1$ $(x - 1)^{3} = x^{3} - 1$ $x^{3} - 1 = (x - 1) (x^{2} + x + 1)$
Show answer
Answer (C) $(x - 1)^{3} = x^{3} - 1$
$(x - 1)^{3} = x^{3} - 3 x^{2} + 3 x - 1$ ; at $x = 2$ it is 1, while $x^{3} - 1 = 7$ . The other three are identities from this chapter.
16
Quick answerMedium
Match each expression with its factorised form.
Tap an item on the left, then its match on the right.
Show answer
Answer
$x^{2} - 9$ : $(x - 3) (x + 3)$
$x^{2} - 6 x + 9$ : $(x - 3)^{2}$
$x^{3} - 27$ : $(x - 3) (x^{2} + 3 x + 9)$
$x^{2} + 3 x - 18$ : $(x + 6) (x - 3)$
Difference of squares, square of a difference, difference of cubes and splitting the middle term (product $- 18$ , sum 3).
17
Quick answerMedium
Put the steps for factorising $2 x^{2} + 7 x + 6$ in order.
1. Rewrite as $2 x^{2} + 3 x + 4 x + 6$
2. Multiply the outer coefficients: $2 \times 6 = 12$
3. Find 3 and 4: product 12, sum 7
4. Group: $x (2 x + 3) + 2 (2 x + 3)$
5. Write $(2 x + 3) (x + 2)$
Show answer
Answer
Multiply the outer coefficients: $2 \times 6 = 12$
Find 3 and 4: product 12, sum 7
Rewrite as $2 x^{2} + 3 x + 4 x + 6$
Group: $x (2 x + 3) + 2 (2 x + 3)$
Write $(2 x + 3) (x + 2)$
Product first, then the pair, then the split, the grouping and finally the common bracket.

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NCERT Exercise Set 4.1, Q1	Practice 4.1 · Q1
NCERT Exercise Set 4.1, Q2	Practice 4.1 · Q2
NCERT Exercise Set 4.2, Q1	Practice 4.2 · Q1
NCERT Exercise Set 4.2, Q2	Practice 4.2 · Q2
NCERT Exercise Set 4.3, Q1	Practice 4.3 · Q1
NCERT Exercise Set 4.3, Q2	Practice 4.3 · Q2
NCERT Exercise Set 4.3, Q3	Practice 4.3 · Q3
NCERT Exercise Set 4.3, Q4	Practice 4.3 · Q4
NCERT Exercise Set 4.4, Q1	Practice 4.4 · Q1
NCERT Exercise Set 4.4, Q2	Practice 4.4 · Q2
NCERT Exercise Set 4.4, Q3	Practice 4.4 · Q3
NCERT Exercise Set 4.5, Q1	Practice 4.5 · Q1
NCERT End-of-chapter Exercises, Q1	End-of-chapter exercises · Q1
NCERT End-of-chapter Exercises, Q2	End-of-chapter exercises · Q2
NCERT End-of-chapter Exercises, Q3	End-of-chapter exercises · Q3
NCERT End-of-chapter Exercises, Q3	End-of-chapter exercises · Q4
NCERT End-of-chapter Exercises, Q4	End-of-chapter exercises · Q5
NCERT End-of-chapter Exercises, Q5	End-of-chapter exercises · Q6
NCERT End-of-chapter Exercises, Q6	End-of-chapter exercises · Q7
NCERT End-of-chapter Exercises, Q7	End-of-chapter exercises · Q8
NCERT End-of-chapter Exercises, Q8	End-of-chapter exercises · Q9
NCERT End-of-chapter Exercises, Q9	End-of-chapter exercises · Q10
NCERT End-of-chapter Exercises, Q10	End-of-chapter exercises · Q11
NCERT End-of-chapter Exercises, Q11	End-of-chapter exercises · Q12
NCERT End-of-chapter Exercises, Q12	End-of-chapter exercises · Q13
NCERT End-of-chapter Exercises, Q13	End-of-chapter exercises · Q14

Quick check: MCQs on Exploring Algebraic Identities

4.1 Introduction3 questions
4.2 Visualising Identities4 questions
4.3 Factorisation of Algebraic Expressions Using Identities4 questions
4.4 More Identities4 questions
4.5 Factorisation Using Algebra Tiles4 questions
4.6 Factorisation Without Using Algebra Tiles4 questions
4.7 Finding New Identities4 questions
4.8 Simplifying Rational Expressions4 questions

Glossary · 10

Key terms (English–Hindi)

algebra tiles: Squares and strips of areas $x^{2}$ , $x$ and 1 used to model products and factors.First seen in the chapter →

coefficientगुणांकgunank: The number multiplying the letters in a term; in $7 x^{2}$ it is 7.First seen in the chapter →

difference of two squaresदो वर्गों का अंतरdo vargon ka antar: An expression $a^{2} - b^{2}$ , which factorises as $(a + b) (a - b)$ .First seen in the chapter →

equationसमीकरणsamikaran: A statement that two expressions are equal. An ordinary equation such as $3 x - 5 = 16$ holds only for particular values, found by solving it; an identity holds for all values.First seen in the chapter →

factorगुणनखंडgunankhand: An expression that divides another exactly; $x + 2$ is a factor of $x^{2} + 5 x + 6$ .First seen in the chapter →
factorisationगुणनखंडनgunankhandan: Writing an expression as a product of simpler expressions.First seen in the chapter →

identityसर्वसमिकाsarvasamika: An equality of two expressions that holds for every value of its letters, such as $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ .First seen in the chapter →

perfect-square trinomialपूर्ण वर्गpoorn varg: A three-term expression $a^{2} \pm 2 a b + b^{2}$ , equal to $(a \pm b)^{2}$ .First seen in the chapter →

rational expressionपरिमेय व्यंजकparimey vyanjak: A fraction whose numerator and denominator are polynomials, with a non-zero denominator.First seen in the chapter →

splitting the middle term: Factorising $a x^{2} + b x + c$ by writing $b x$ as two terms whose coefficients multiply to $a c$ .First seen in the chapter →

4.1 Introduction

4.2 Visualising Identities

4.3 Factorisation of Algebraic Expressions Using Identities

4.4 More Identities

4.5 Factorisation Using Algebra Tiles

4.6 Factorisation Without Using Algebra Tiles

4.7 Finding New Identities

4.8 Simplifying Rational Expressions

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