Exploring Algebraic Identities Class 9 Mathematics Notes

Introduction

In earlier chapters, you learned about linear polynomials and how they represent real-life situations and help solve problems. You also studied linear equations and their role in describing relationships between quantities. This chapter, Exploring Algebraic Identities, builds upon that foundation by introducing algebraic identities.

An algebraic identity is a special type of equation that is true for all possible values of the variables involved. These identities act as powerful mathematical rules that simplify complex algebraic calculations and enable efficient manipulation of algebraic expressions.

Let's begin by exploring some simple patterns that hint at the usefulness of these identities.

Example 1: Pattern with Consecutive Square Numbers

Given: Three consecutive square numbers. To find: A pattern by adding the smallest and largest squares and then subtracting twice the middle square.

Solution: Consider the first set of three consecutive square numbers: $1^{2}, 2^{2}, 3^{2}$ , which are $1, 4, 9$ . Add the smallest and largest squares: $1 + 9 = 10$ . Subtract twice the middle square from this sum: $10 - (2 \times 4) = 10 - 8 = 2$ .

Consider another set of three consecutive square numbers: $3^{2}, 4^{2}, 5^{2}$ , which are $9, 16, 25$ . Add the smallest and largest squares: $9 + 25 = 34$ . Subtract twice the middle square from this sum: $34 - (2 \times 16) = 34 - 32 = 2$ .

Consider a third set: $5^{2}, 6^{2}, 7^{2}$ , which are $25, 36, 49$ . Add the smallest and largest squares: $25 + 49 = 74$ . Subtract twice the middle square from this sum: $74 - (2 \times 36) = 74 - 72 = 2$ .

Answer: The result consistently appears to be 2.

This observation, though seemingly coincidental, reveals a deeper mathematical truth that will be explained using algebra later in the chapter. Understanding the underlying algebraic principle behind such patterns is key to mastering algebraic identities.

Remember
Algebraic identities are equations that hold true for all values of the variables, making them universal tools in algebra.

Exam-smart
Questions might present a pattern with numbers and ask you to identify the underlying rule or predict the next outcome. Recognizing that algebraic identities can explain such numerical patterns is crucial for solving these competency-based problems.

Visualising Identities

The identity $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ can be visualized using geometry. Consider a square with side length $(a + b)$ . This square can be divided into four parts: a square of side $a$ (area $a^{2}$ ), a square of side $b$ (area $b^{2}$ ), and two rectangles of sides $a$ and $b$ (each with area $a b$ ). The sum of these areas is $a^{2} + b^{2} + 2 a b$ , which equals the total area of the square, $(a + b)^{2}$ .

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

Using the distributive property:

(a + b) (a + b) = a (a + b) + b (a + b)

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

Since multiplication is commutative ( $a b = b a$ ):

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

This proves the identity for all real numbers.

An algebraic identity is an equation that holds true for all possible values of the variables involved. In contrast, a regular equation is only true for specific values of the variables. For instance, $x + 2 = 5$ is an equation true only for $x = 3$ , whereas $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ is an identity true for any values of $a$ and $b$ .

The identity $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ can be used for calculations.

Example 3: Expand $(5 x + 2 y)^{2}$

Given: The expression $(5 x + 2 y)^{2}$ . To find: The expanded form of the expression. Solution: We use the identity $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ , where $a = 5 x$ and $b = 2 y$ .

(5 x + 2 y)^{2} = (5 x)^{2} + 2 (5 x) (2 y) + (2 y)^{2}

= 25 x^{2} + 20 x y + 4 y^{2}

Answer: $25 x^{2} + 20 x y + 4 y^{2}$

Example 4: Calculate $43^{2}$

Given: The number $43^{2}$ . To find: The value of $43^{2}$ . Solution: We can write $43$ as $(40 + 3)$ . Using the identity $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ , where $a = 40$ and $b = 3$ .

43^{2} = (40 + 3)^{2}

= 40^{2} + 2 \times 40 \times 3 + 3^{2}

= 1600 + 240 + 9

= 1849

Answer: $1849$

Exercise 4.1: Expansion

Using the identity $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ , expand the following: (i) $(7 x + 4 y)^{2}$ (ii) $(\frac{7}{3} x + y)^{2}$ (iii) $(2.5 p + 1.5 q)^{2}$ (iv) $(\frac{3}{4} s + 8 t)^{2}$ (v) $(\frac{1}{2} x + 2 y)^{2}$ (vi) $(\frac{1}{3} x + y)^{2}$

Answer: (i) $49 x^{2} + 56 x y + 16 y^{2}$ (ii) $\frac{49}{9} x^{2} + \frac{14}{3} x y + y^{2}$ (iii) $6.25 p^{2} + 7.5 p q + 2.25 q^{2}$ (iv) $\frac{9}{16} s^{2} + 12 s t + 64 t^{2}$ (v) $\frac{1}{4} x^{2} + 2 x y + 4 y^{2}$ (vi) $\frac{1}{9} x^{2} + \frac{2}{3} x y + y^{2}$

Using the same identity, find the values of the following: (i) $(64)^{2}$ (ii) $(105)^{2}$ (iii) $(205)^{2}$

Answer: (i) $4096$ (ii) $11025$ (iii) $42025$

Factorisation of Algebraic Expressions Using Identities

The identity $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ can be used in reverse to factorise algebraic expressions. This means we can rewrite an expression in the form $a^{2} + 2 a b + b^{2}$ as a perfect square $(a + b)^{2}$ .

To factorise an expression of the form $a^{2} + 2 a b + b^{2}$ :

Identify if the first and last terms are perfect squares. Let them be $a^{2}$ and $b^{2}$ .
Check if the middle term is twice the product of the square roots of the first and last terms, i.e., $2 a b$ .
If all conditions are met, the expression can be factorised as $(a + b)^{2}$ .

Example 5: Factorise $x^{2} + 4 x + 4$

Given: The algebraic expression $x^{2} + 4 x + 4$ . To find: The factors of the expression. Solution: We observe that: The first term is $x^{2}$ , so $a^{2} = x^{2} ⟹ a = x$ . The last term is $4$ , so $b^{2} = 4 ⟹ b = 2$ . The middle term is $4 x$ . Let's check if it matches $2 a b$ : $2 a b = 2 (x) (2) = 4 x$ . Since the middle term matches $2 a b$ , we can use the identity $a^{2} + 2 a b + b^{2} = (a + b)^{2}$ .

x^{2} + 4 x + 4 = x^{2} + 2 (x) (2) + 2^{2} = (x + 2)^{2}

Thus, $(x + 2)$ is a factor of $x^{2} + 4 x + 4$ . Answer: $(x + 2)^{2}$

Example 6: Factorise $36 x^{2} + 12 x + 1$

Given: The algebraic expression $36 x^{2} + 12 x + 1$ . To find: The factors of the expression. Solution: We observe that: The first term is $36 x^{2}$ , so $a^{2} = 36 x^{2} ⟹ a = 6 x$ . The last term is $1$ , so $b^{2} = 1 ⟹ b = 1$ . The middle term is $12 x$ . Let's check if it matches $2 a b$ : $2 a b = 2 (6 x) (1) = 12 x$ . Since the middle term matches $2 a b$ , we can use the identity $a^{2} + 2 a b + b^{2} = (a + b)^{2}$ .

36 x^{2} + 12 x + 1 = (6 x)^{2} + 2 (6 x) (1) + 1^{2} = (6 x + 1)^{2}

Thus, $(6 x + 1)$ is a factor of $36 x^{2} + 12 x + 1$ . Answer: $(6 x + 1)^{2}$

Example 7: Factorise $50 p^{2} + 60 p q + 18 q^{2}$

Given: The algebraic expression $50 p^{2} + 60 p q + 18 q^{2}$ . To find: The factors of the expression. Solution: First, we look for a common factor in all terms. The greatest common divisor of $50, 60,$ and $18$ is $2$ .

50 p^{2} + 60 p q + 18 q^{2} = 2 (25 p^{2} + 30 p q + 9 q^{2})

Now, let's focus on the expression inside the parenthesis: $25 p^{2} + 30 p q + 9 q^{2}$ . We observe that: The first term is $25 p^{2}$ , so $a^{2} = 25 p^{2} ⟹ a = 5 p$ . The last term is $9 q^{2}$ , so $b^{2} = 9 q^{2} ⟹ b = 3 q$ . The middle term is $30 p q$ . Let's check if it matches $2 a b$ : $2 a b = 2 (5 p) (3 q) = 30 p q$ . Since the middle term matches $2 a b$ , we can use the identity $a^{2} + 2 a b + b^{2} = (a + b)^{2}$ .

25 p^{2} + 30 p q + 9 q^{2} = (5 p)^{2} + 2 (5 p) (3 q) + (3 q)^{2} = (5 p + 3 q)^{2}

Combining this with the common factor we took out:

50 p^{2} + 60 p q + 18 q^{2} = 2 (5 p + 3 q)^{2}

Answer: $2 (5 p + 3 q)^{2}$

The identity $(a - b)^{2} = a^{2} - 2 a b + b^{2}$ is also an identity. This can be derived by replacing $b$ with $- b$ in the $(a + b)^{2}$ identity: $(a + (- b))^{2} = a^{2} + 2 a (- b) + (- b)^{2} = a^{2} - 2 a b + b^{2}$ .

This identity can be visualized by considering a square of side $a$ . If we remove a strip of width $b$ from two adjacent sides, the remaining area will be $(a - b)^{2}$ . This area can be calculated as the total area $a^{2}$ minus the areas of the two rectangles of size $a b$ and the small square of size $b^{2}$ that has been removed twice.

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

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

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

Example 8: Calculate $29^{2}$

Given: The number $29^{2}$ . To find: The value of $29^{2}$ . Solution: We can write $29$ as $(30 - 1)$ . Using the identity $(a - b)^{2} = a^{2} - 2 a b + b^{2}$ , where $a = 30$ and $b = 1$ .

29^{2} = (30 - 1)^{2}

= 30^{2} - 2 \times 30 \times 1 + 1^{2}

= 900 - 60 + 1

= 841

Answer: $841$

Exercise 4.2: Factorisation and Values

Factor completely: (i) $9 x^{2} + 24 x y + 16 y^{2}$ (ii) $4 s^{2} + 20 s t + 25 t^{2}$ (iii) $49 x^{2} + 28 x y + 4 y^{2}$ (iv) $64 p^{2} + \frac{1}{2} p q + \frac{1}{256} q^{2}$ (Hint: $\frac{1}{256} = (\frac{1}{16})^{2}$ ) (v) $12 a^{2} + 8 a b + \frac{4}{3} b^{2}$ (Hint: $4 a^{2}$ is a factor) (vi) $3 s^{2} + 6 s v + 3 v^{2}$ (Hint: $3$ is a common factor)

Answer: (i) $(3 x + 4 y)^{2}$ (ii) $(2 s + 5 t)^{2}$ (iii) $(7 x + 2 y)^{2}$ (iv) $(8 p + \frac{1}{16} q)^{2}$ (v) $4 (\sqrt{3} a + \frac{1}{\sqrt{3}} b)^{2}$ or $4 (3 a^{2} + 2 a b + b^{2} / 3)$ which doesn't look like a perfect square. (Note: For (v) and (vi) and similar problems, ensure the question implies a common factor can be taken out first. If not, direct factorisation into $(a + b)^{2}$ might not be possible with simple integers. For (v), if we take $4 / 3$ as a common factor, we get $(4 / 3) (9 a^{2} + 6 a b + b^{2}) = (4 / 3) (3 a + b)^{2}$ . If the question intends for $a^{2} = 12 a^{2}$ , then $a = \sqrt{12} a = 2 \sqrt{3} a$ , and $b^{2} = 4 / 3 b^{2}$ , $b = \sqrt{4 / 3} b = 2 / \sqrt{3} b$ . $2 a b = 2 (2 \sqrt{3} a) (2 / \sqrt{3} b) = 8 a b$ . So, $(2 \sqrt{3} a + 2 / \sqrt{3} b)^{2}$ . The expression implies simpler numbers were expected.) (vi) $3 (s + v)^{2}$

Find the values of the following using the identity $(a – b)^{2} = a^{2} – 2 a b + b^{2}$ : (i) $(79)^{2}$ (ii) $(193)^{2}$ (iii) $(299)^{2}$

Answer: (i) $6241$ (ii) $37249$ (iii) $89401$

Remember
Visualising identities with geometry helps understand why they are true, not just what they are. The algebraic proof using the distributive property is essential for all numbers, not just lengths.

Exam-smart
Questions might present an expression like $4 x^{2} + 12 x + 9$ and ask to factorise it, or provide a numerical calculation like $102^{2}$ and ask for its value using identities. You need to identify which identity to use and apply it correctly. Sometimes, a common factor needs to be extracted first, as in Example 7.

More Identities

We extend algebraic identities to include squares of the sum of three numbers and explore their applications in calculations and factorisation.

Identity for the Square of a Sum of Three Numbers

The identity for the square of the sum of three numbers $(a + b + c)^{2}$ can be derived by considering $(a + d)^{2}$ where $d = b + c$ .

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

This identity states that the square of the sum of three terms is equal to the sum of the squares of each term plus twice the product of each pair of terms.

Example 9: Using the identity $(a + b + c)^{2}$

Given: Calculate $119^{2}$ using the identity $(a + b + c)^{2}$ . To find: The value of $119^{2}$ . Solution: Let $119 = 100 + 10 + 9$ . Using the identity $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$ , with $a = 100$ , $b = 10$ , and $c = 9$ :

119^{2} = (100 + 10 + 9)^{2}

119^{2} = 100^{2} + 10^{2} + 9^{2} + 2 (100) (10) + 2 (10) (9) + 2 (9) (100)

119^{2} = 10000 + 100 + 81 + 2000 + 180 + 1800

119^{2} = 14161

Answer: $119^{2} = 14161$ .

Remember
The identity $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$ is particularly useful for squaring numbers that can be easily broken down into three convenient parts, like multiples of 100, 10, and single digits.

Exam-smart
Problems might ask you to calculate squares of large numbers like $214^{2}$ or $1104^{2}$ . You would need to identify suitable values for $a, b, c$ (e.g., $214 = 200 + 10 + 4$ ) and apply the identity.

Summary of Key Identities

The following three algebraic identities have been established and are useful for calculations and manipulation:

Identity Number	Identity	Use
1	$(a + b)^{2} = a^{2} + 2 a b + b^{2}$	Expanding squares of binomials
2	$(a - b)^{2} = a^{2} - 2 a b + b^{2}$	Expanding squares of binomials
3	$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$	Expanding squares of trinomials

Exercise $4.3$ Applications

1. Find the following squares using one of the above identities.

Example: $78^{2}$

Given: $78^{2}$ . To find: The value of $78^{2}$ using an identity. Solution: We can use the identity $(a - b)^{2} = a^{2} - 2 a b + b^{2}$ . Let $78 = 80 - 2$ , so $a = 80$ and $b = 2$ .

78^{2} = (80 - 2)^{2}

78^{2} = 80^{2} - 2 (80) (2) + 2^{2}

78^{2} = 6400 - 320 + 4

78^{2} = 6084

Answer: $78^{2} = 6084$ .

Try this: Find the value of $198^{2}$ using an identity. Answer: $39204$

2. Factor using suitable identities:

Example: Factorise $16 y^{2} - 24 y + 9$

Given: $16 y^{2} - 24 y + 9$ . To find: The factors of the expression. Solution: We recognise that $16 y^{2} = (4 y)^{2}$ and $9 = 3^{2}$ . The middle term is $- 24 y$ , which is $- 2 \times (4 y) \times 3$ . This matches the form of the identity $(a - b)^{2} = a^{2} - 2 a b + b^{2}$ , where $a = 4 y$ and $b = 3$ . Therefore, $16 y^{2} - 24 y + 9 = (4 y - 3)^{2}$ . Answer: $(4 y - 3)^{2}$ .

Try this: Factorise $4 p^{2} - 12 p q + 9 q^{2}$ . Answer: $(2 p - 3 q)^{2}$

3. Expand the following using the identity $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$ :

Example: Expand $(p + 3 q + 7 r)^{2}$

Given: $(p + 3 q + 7 r)^{2}$ . To find: The expanded form of the expression. Solution: Using the identity $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$ , with $a = p$ , $b = 3 q$ , and $c = 7 r$ :

(p + 3 q + 7 r)^{2} = p^{2} + (3 q)^{2} + (7 r)^{2} + 2 (p) (3 q) + 2 (3 q) (7 r) + 2 (7 r) (p)

(p + 3 q + 7 r)^{2} = p^{2} + 9 q^{2} + 49 r^{2} + 6 p q + 42 q r + 14 r p

Answer: $p^{2} + 9 q^{2} + 49 r^{2} + 6 p q + 42 q r + 14 r p$ .

Try this: Expand $(3 x - 2 y + 4 z)^{2}$ . Answer: $9 x^{2} + 4 y^{2} + 16 z^{2} - 12 x y + 24 y z - 16 x z$

Exploring the Identity $a^{2} - b^{2} = (a + b) (a - b)$

The identity $a^{2} - b^{2} = (a + b) (a - b)$ is a fundamental tool for factorisation. It can also be rewritten as $a^{2} = (a + b) (a - b) + b^{2}$ , which can be used for quick calculation of squares.

Example: Calculating $55^{2}$ using $a^{2} = (a + b) (a - b) + b^{2}$

Given: Calculate $55^{2}$ . To find: The value of $55^{2}$ using the identity. Solution: Let $a = 55$ . We can choose a value for $b$ that makes $(a + b)$ and $(a - b)$ easy to multiply. Let $b = 5$ . Then $a + b = 55 + 5 = 60$ and $a - b = 55 - 5 = 50$ . Using the identity $a^{2} = (a + b) (a - b) + b^{2}$ :

55^{2} = (55 + 5) (55 - 5) + 5^{2}

55^{2} = (60) (50) + 25

55^{2} = 3000 + 25

55^{2} = 3025

Answer: $55^{2} = 3025$ .

Try this: Evaluate $35^{2}$ using a suitable identity. Answer: $1225$

Remember
The identity $a^{2} - b^{2} = (a + b) (a - b)$ is useful for factorising expressions that are the difference of two perfect squares. Rewriting it as $a^{2} = (a + b) (a - b) + b^{2}$ is a clever way to calculate squares, especially for numbers ending in 5.

Exam-smart
You might be asked to factorise expressions like $49 x^{2} - 4 y^{2}$ or $100 - 9 p^{2}$ . You'll need to recognise these as differences of squares and apply the identity to get $(7 x)^{2} - (2 y)^{2} = (7 x + 2 y) (7 x - 2 y)$ .

Factorisation Using Algebra Tiles

Algebra tiles provide a visual method for understanding factorisation, which is the process of expressing an algebraic expression as a product of its factors.

Consider the expression $x^{2} + 7 x + 12$ . We want to find two binomials that multiply to give this trinomial.

An $x^{2}$ -tile represents $x^{2}$ .
$x$ -tiles represent $x$ .
Unit tiles represent $1$ .

To factor $x^{2} + 7 x + 12$ , we arrange the tiles to form a rectangle. The $x^{2}$ tile forms one corner. The $7 x$ tiles are distributed to fill the sides, and the $12$ unit tiles fill the remaining space. If we split the $7 x$ tiles into $4 x$ and $3 x$ tiles, and the $12$ unit tiles into a $4 \times 3$ array, we can form a rectangle.

This rectangle has sides of length $(x + 3)$ and $(x + 4)$ . Thus, $x^{2} + 7 x + 12 = (x + 3) (x + 4)$ .

This demonstrates how the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b$ works visually. The $x^{2}$ tile corresponds to $x^{2}$ , the $(a + b) x$ tiles correspond to the $a x$ and $b x$ terms, and the $a b$ unit tiles correspond to the constant term.

Example: Factorise $x^{2} + 5 x + 6$ using algebra tiles.

Given: $x^{2} + 5 x + 6$ . To find: The factors of the expression. Solution:

Start with one $x^{2}$ tile.
You need to arrange $5 x$ tiles and $6$ unit tiles around the $x^{2}$ tile to form a rectangle.
To form a rectangle, the $5 x$ tiles must be split into two groups, and the $6$ unit tiles must form a rectangular array.
If we split $5 x$ into $2 x$ and $3 x$ , and arrange the $6$ unit tiles as a $2 \times 3$ array, we can form a rectangle.
The dimensions of this rectangle will be $(x + 2)$ and $(x + 3)$ .

The $x^{2}$ tile is at the top-left.
Two $x$ tiles are placed to the right of the $x^{2}$ tile.
Three $x$ tiles are placed below the $x^{2}$ tile.
Six unit tiles are placed at the bottom-right, forming a $2 \times 3$ grid.

The sides of the rectangle are $(x + 2)$ and $(x + 3)$ . Answer: $(x + 2) (x + 3)$ .

Try this: Factorise $x^{2} + 6 x + 8$ using algebra tiles. Answer: $(x + 2) (x + 4)$

Remember
Factorisation using algebra tiles helps visualise the relationship between the expanded form of a quadratic expression and its factored binomial form. The middle term of the trinomial ( $b x$ ) is split into two parts, and the constant term ( $c$ ) is arranged into a rectangle.

Exam-smart
Questions might provide a visual representation of algebra tiles arranged in a rectangle and ask you to write down the corresponding algebraic expression in its factored and expanded forms, or vice versa. You would need to identify the lengths of the sides of the rectangle from the tile arrangement.

Factorisation Without Using Algebra Tiles

This section focuses on factorising quadratic expressions of the form $x^{2} + b x + c$ by strategically splitting the middle term ( $b x$ ), without the visual aid of algebra tiles. The core idea is to find two numbers, say $a$ and $b$ , such that their product ( $a b$ ) equals the constant term ( $c$ ) and their sum ( $a + b$ ) equals the coefficient of the middle term ( $b$ ).

Consider a quadratic expression like $x^{2} + 7 x + 12$ . We want to express this as a product of two linear factors, say $(x + p) (x + q)$ . Expanding this, we get $(x + p) (x + q) = x^{2} + q x + p x + p q = x^{2} + (p + q) x + p q$ .

By comparing this expanded form with $x^{2} + 7 x + 12$ , we can see that:

The coefficient of the $x$ term: $p + q = 7$
The constant term: $p q = 12$

We need to find two numbers whose product is 12 and whose sum is 7. By looking at the factors of 12 (1 and 12, 2 and 6, 3 and 4), we find that 3 and 4 satisfy both conditions: $3 \times 4 = 12$ and $3 + 4 = 7$ . Therefore, we can split the middle term $7 x$ into $3 x + 4 x$ .

x^{2} + 7 x + 12 = x^{2} + 3 x + 4 x + 12

Now, we can factor by grouping: Group the first two terms and the last two terms: $(x^{2} + 3 x) + (4 x + 12)$ Factor out the common factor from each group: $x (x + 3) + 4 (x + 3)$ Notice that $(x + 3)$ is a common factor in both terms. Factor it out: $(x + 3) (x + 4)$ .

This method is particularly useful when dealing with quadratic expressions where the leading coefficient (the coefficient of $x^{2}$ ) is 1.

Handling Negative Coefficients

When the middle term has a negative coefficient, or the constant term is positive while the middle term is negative, we need to consider negative numbers for $a$ and $b$ .

For an expression like $x^{2} - 5 x + 6$ , we are looking for two numbers $a$ and $b$ such that $a + b = - 5$ and $a b = 6$ . The pairs of factors for 6 are (1, 6) and (2, 3). To get a sum of -5, both numbers must be negative. Consider (-1, -6) and (-2, -3).

$(- 1) + (- 6) = - 7$ (Incorrect sum)
$(- 2) + (- 3) = - 5$ (Correct sum)
$(- 2) \times (- 3) = 6$ (Correct product)

So, we split $- 5 x$ into $- 2 x - 3 x$ .

x^{2} - 5 x + 6 = x^{2} - 2 x - 3 x + 6

Factor by grouping: $(x^{2} - 2 x) + (- 3 x + 6)$ Factor out common terms: $x (x - 2) - 3 (x - 2)$ Factor out the common binomial: $(x - 2) (x - 3)$ .

Example 10: Factorise $x^{2} + 7 x + 12$

Given: The expression $x^{2} + 7 x + 12$ . To find: The factors of the expression. Solution: We need to find two numbers $a$ and $b$ such that $a + b = 7$ and $a b = 12$ . The numbers are 3 and 4. We split the middle term $7 x$ as $3 x + 4 x$ .

x^{2} + 7 x + 12 = x^{2} + 3 x + 4 x + 12

Group the terms: $(x^{2} + 3 x) + (4 x + 12)$ Factor out common factors: $x (x + 3) + 4 (x + 3)$ Factor out the common binomial $(x + 3)$ : $(x + 3) (x + 4)$ Answer: $(x + 3) (x + 4)$

Example 11: Factorise $x^{2} + 11 x + 30$

Given: The expression $x^{2} + 11 x + 30$ . To find: The factors of the expression. Solution: We need two numbers $a$ and $b$ such that $a + b = 11$ and $a b = 30$ . Consider factors of 30: (1, 30), (2, 15), (3, 10), (5, 6). Check their sums:

1 + 30 = 31

2 + 15 = 17

3 + 10 = 13

5 + 6 = 11

The numbers are 5 and 6. We split the middle term $11 x$ as $5 x + 6 x$ .

x^{2} + 11 x + 30 = x^{2} + 5 x + 6 x + 30

Group the terms: $(x^{2} + 5 x) + (6 x + 30)$ Factor out common factors: $x (x + 5) + 6 (x + 5)$ Factor out the common binomial $(x + 5)$ : $(x + 5) (x + 6)$ Answer: $(x + 5) (x + 6)$

Example 12: Factorise $x^{2} - 5 x + 6$

Given: The expression $x^{2} - 5 x + 6$ . To find: The factors of the expression. Solution: We need two numbers $a$ and $b$ such that $a + b = - 5$ and $a b = 6$ . Since the product is positive and the sum is negative, both numbers must be negative. Consider negative factors of 6: (-1, -6), (-2, -3). Check their sums:

(- 1) + (- 6) = - 7

(- 2) + (- 3) = - 5

The numbers are -2 and -3. We split the middle term $- 5 x$ as $- 2 x - 3 x$ .

x^{2} - 5 x + 6 = x^{2} - 2 x - 3 x + 6

Group the terms: $(x^{2} - 2 x) + (- 3 x + 6)$ Factor out common factors: $x (x - 2) - 3 (x - 2)$ Factor out the common binomial $(x - 2)$ : $(x - 2) (x - 3)$ Answer: $(x - 2) (x - 3)$

Exercise 4.6.1 (i): Fill in the blanks: $s^{2} – 11 s + 24 = (_) (_)$

We need two numbers that multiply to 24 and add up to -11. Since the product is positive and the sum is negative, both numbers must be negative. Factors of 24: (1, 24), (2, 12), (3, 8), (4, 6). Negative factors: (-1, -24), (-2, -12), (-3, -8), (-4, -6). Sums:

- 1 + (- 24) = - 25

- 2 + (- 12) = - 14

- 3 + (- 8) = - 11

- 4 + (- 6) = - 10

The numbers are -3 and -8. $s^{2} - 11 s + 24 = s^{2} - 3 s - 8 s + 24 = s (s - 3) - 8 (s - 3) = (s - 3) (s - 8)$ . Answer: $(s - 3) (s - 8)$

Remember
When factorising $x^{2} + b x + c$ , if $c$ is positive and $b$ is negative, both numbers you seek for the sum and product must be negative. If $b$ is positive, both must be positive. If $c$ is negative, one number will be positive and the other negative.

Exam-smart
You might be given a product of two linear terms, like $(x + a) (x + b)$ , and asked to expand it to $x^{2} + (a + b) x + a b$ . Conversely, you might be given $x^{2} + b x + c$ and asked to factorise it, requiring you to find $a$ and $b$ such that $a + b = b$ and $a b = c$ . The challenge in competency questions is often identifying the correct pair of numbers quickly, especially when negative signs are involved.

Try this: Factorise $y^{2} - 13 y + 36$ .

Answer: $(y - 4) (y - 9)$

Finding New Identities

This section explores how to derive new algebraic identities by manipulating existing ones and through multiplication. We will focus on extending patterns and recognizing structures.

The Identity for $(a + b)^{3}$

We can derive the identity for the cube of a binomial $(a + b)^{3}$ by expanding it using the distributive property:

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

We know from previous sections that $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ . Substituting this:

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

Now, apply the distributive property:

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

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

Combine like terms:

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

This results in the identity:

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

Geometric Interpretation: Imagine a cube with side length $(a + b)$ . Its volume is $(a + b)^{3}$ . This large cube can be divided into smaller geometric shapes:

Two cubes with volumes $a^{3}$ and $b^{3}$ .
Three cuboids of dimensions $a \times a \times b$ , each with volume $a^{2} b$ .
Three cuboids of dimensions $a \times b \times b$ , each with volume $a b^{2}$ . The total volume is the sum of these parts: $a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}$ , which visually confirms the identity.

The Identity for $(a - b)^{3}$

To find the identity for $(a - b)^{3}$ , we can substitute $- b$ for $b$ in the identity for $(a + b)^{3}$ :

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

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

This gives us the identity:

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

Notice that the terms alternate in sign.

Example 1

Given: Volume of a cube is $p^{3} + 6 p^{2} q + 12 p q^{2} + 8 q^{3}$ cubic units. To find: The side length of the cube.

Solution: We compare the given volume with the identity $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ . The first term $p^{3}$ suggests $a = p$ . The last term $8 q^{3} = (2 q)^{3}$ suggests $b = 2 q$ . Let's check if the middle terms match: $3 a^{2} b = 3 (p)^{2} (2 q) = 6 p^{2} q$ . This matches. $3 a b^{2} = 3 (p) (2 q)^{2} = 3 (p) (4 q^{2}) = 12 p q^{2}$ . This also matches. So, the volume can be written as $(p + 2 q)^{3}$ . Therefore, the side length is $a + b = p + 2 q$ .

Answer: The side of the cube is $(p + 2 q)$ units.

Example 2

Given: The expression $8 n^{3} – 60 n^{2} m + 150 n m^{2} – 125 m^{3}$ . To find: The values of $a$ and $b$ if the expression is written in the form $(a - b)^{3}$ .

Solution: We compare the given expression with the identity $(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$ . The first term $8 n^{3} = (2 n)^{3}$ suggests $a = 2 n$ . The last term $125 m^{3} = (5 m)^{3}$ suggests $b = 5 m$ . Let's check the middle terms: $- 3 a^{2} b = - 3 (2 n)^{2} (5 m) = - 3 (4 n^{2}) (5 m) = - 60 n^{2} m$ . This matches. $3 a b^{2} = 3 (2 n) (5 m)^{2} = 3 (2 n) (25 m^{2}) = 150 n m^{2}$ . This also matches. So, the expression is equal to $(2 n - 5 m)^{3}$ . Therefore, $a = 2 n$ and $b = 5 m$ .

Answer: $a = 2 n$ and $b = 5 m$ .

The Difference of Cubes Identity: $x^{3} - y^{3}$

We can discover a new identity by multiplying specific binomials and trinomials. Let's expand $(x - y) (x^{2} + x y + y^{2})$ :

(x - y) (x^{2} + x y + y^{2}) = x (x^{2} + x y + y^{2}) - y (x^{2} + x y + y^{2})

= (x^{3} + x^{2} y + x y^{2}) - (x^{2} y + x y^{2} + y^{3})

= x^{3} + x^{2} y + x y^{2} - x^{2} y - x y^{2} - y^{3}

Combining like terms, we see that $x^{2} y$ cancels with $- x^{2} y$ , and $x y^{2}$ cancels with $- x y^{2}$ . This leaves us with the identity:

(x - y) (x^{2} + x y + y^{2}) = x^{3} - y^{3}

The Sum of Cubes Identity: $x^{3} + y^{3}$

By a similar process, or by substituting $- y$ for $y$ in the difference of cubes identity, we can find the sum of cubes identity. Let's expand $(x + y) (x^{2} - x y + y^{2})$ :

(x + y) (x^{2} - x y + y^{2}) = x (x^{2} - x y + y^{2}) + y (x^{2} - x y + y^{2})

= (x^{3} - x^{2} y + x y^{2}) + (x^{2} y - x y^{2} + y^{3})

= x^{3} - x^{2} y + x y^{2} + x^{2} y - x y^{2} + y^{3}

Combining like terms, $- x^{2} y$ cancels with $x^{2} y$ , and $x y^{2}$ cancels with $- x y^{2}$ . This results in the identity:

(x + y) (x^{2} - x y + y^{2}) = x^{3} + y^{3}

Example 3

Given: Evaluate $(2 a - 3 b) (4 a^{2} + 6 a b + 9 b^{2})$ . To find: The value of the expression.

Solution: We recognize the form of the expression resembles $(x - y) (x^{2} + x y + y^{2}) = x^{3} - y^{3}$ . Here, $x = 2 a$ . And $y = 3 b$ . Let's check the trinomial part: $x^{2} = (2 a)^{2} = 4 a^{2}$ . This matches the first term. $x y = (2 a) (3 b) = 6 a b$ . This matches the middle term. $y^{2} = (3 b)^{2} = 9 b^{2}$ . This matches the last term. So, the expression is in the form $(x - y) (x^{2} + x y + y^{2})$ . Using the identity, the expression simplifies to $x^{3} - y^{3}$ . Substitute back $x = 2 a$ and $y = 3 b$ : $(2 a)^{3} - (3 b)^{3} = 8 a^{3} - 27 b^{3}$ .

Answer: $8 a^{3} - 27 b^{3}$ .

The Identity for $(x + y + z)^{3}$

Consider the expression $(x + y + z) (x^{2} + y^{2} + z^{2} – x y – x z – y z)$ . By applying the distributive property extensively, we can derive a new identity:

(x + y + z) (x^{2} + y^{2} + z^{2} – x y – x z – y z)

= x (x^{2} + y^{2} + z^{2} – x y – x z – y z)

+ y (x^{2} + y^{2} + z^{2} – x y – x z – y z)

+ z (x^{2} + y^{2} + z^{2} – x y – x z – y z)

= (x^{3} + x y^{2} + x z^{2} – x^{2} y – x^{2} z – x y z)

+ (x^{2} y + y^{3} + y z^{2} – x y^{2} – x y z – y^{2} z)

+ (x^{2} z + y^{2} z + z^{3} – x y z – x z^{2} – y z^{2})

After cancelling out terms like $x y^{2}$ with $- x y^{2}$ , $x z^{2}$ with $- x z^{2}$ , $x^{2} y$ with $- x^{2} y$ , etc., we are left with:

x^{3} + y^{3} + z^{3} - 3 x y z

This leads to the identity:

(x + y + z) (x^{2} + y^{2} + z^{2} – x y – x z – y z) = x^{3} + y^{3} + z^{3} - 3 x y z

Example 4

Given: Three numbers $x, y, z$ such that $x + y + z = 10$ , $x y z = 25$ , and $x^{2} + y^{2} + z^{2} = 38$ . To find: The sum of the cubes of these numbers, i.e., $x^{3} + y^{3} + z^{3}$ .

Solution: We use the identity $(x + y + z) (x^{2} + y^{2} + z^{2} – x y – x z – y z) = x^{3} + y^{3} + z^{3} - 3 x y z$ . We are given:

x + y + z = 10

x^{2} + y^{2} + z^{2} = 38

x y z = 25

We need to find the value of $(x y + x z + y z)$ . We can use the identity $(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + x z + y z)$ . Substituting the given values:

(10)^{2} = 38 + 2 (x y + x z + y z)

100 = 38 + 2 (x y + x z + y z)

100 - 38 = 2 (x y + x z + y z)

62 = 2 (x y + x z + y z)

$x y + x z + y z = \frac{62}{2} = 31$ .

Now substitute all known values into the main identity:

(10) (38 - (x y + x z + y z)) = x^{3} + y^{3} + z^{3} - 3 (25)

(10) (38 - 31) = x^{3} + y^{3} + z^{3} - 75

(10) (7) = x^{3} + y^{3} + z^{3} - 75

70 = x^{3} + y^{3} + z^{3} - 75

x^{3} + y^{3} + z^{3} = 70 + 75

$x^{3} + y^{3} + z^{3} = 145$ .

Answer: The sum of the cubes of the three numbers is 145.

Remember
When deriving new identities from known ones, pay close attention to signs and combining like terms. Factorization of expressions like $x^{3} - y^{3}$ and $x^{3} + y^{3}$ is the reverse process of expansion.

Exam-smart
Questions involving these new identities often appear as "show that" problems or as application problems where you need to find a value given some conditions, like in Example 4. Recognizing the pattern of the terms is key to applying the correct identity.

Simplifying Rational Expressions

Rational algebraic expressions are fractions where the numerator and denominator are algebraic expressions (expressions involving variables, constants, and arithmetic operations). Simplifying these expressions involves cancelling out common factors between the numerator and the denominator. This process is fundamentally based on the principle of factorisation, which we have studied earlier.

The key idea is to factorise both the numerator and the denominator into their simplest forms. Once factored, any identical factors present in both can be cancelled out, provided they are not equal to zero. This is similar to simplifying numerical fractions, like $\frac{6}{9} = \frac{2 \times 3}{3 \times 3} = \frac{2}{3}$ .

To factorise quadratic expressions like $a x^{2} + b x + c$ , we often look for two numbers, say $m$ and $n$ , such that their sum $m + n = b$ and their product $m n = a c$ . This allows us to split the middle term ( $b x$ ) and then factor by grouping.

Example 16: Simplify the rational expression $\frac{x^{2} - 7 x + 12}{5 x^{2} + 5 x - 100}$ , assuming that $5 x^{2} + 5 x - 100 \neq 0$ .

Given: The rational expression $\frac{x^{2} - 7 x + 12}{5 x^{2} + 5 x - 100}$ . To find: The simplified form of the expression.

Solution:

Factorise the numerator: $x^{2} - 7 x + 12$ We need two numbers that add up to -7 and multiply to 12. These numbers are -3 and -4. So, $x^{2} - 7 x + 12 = (x - 3) (x - 4)$ .
Factorise the denominator: $5 x^{2} + 5 x - 100$ First, take out the common factor of 5: $5 (x^{2} + x - 20)$ . Now, factorise $x^{2} + x - 20$ . We need two numbers that add up to 1 and multiply to -20. These numbers are 5 and -4. So, $x^{2} + x - 20 = (x + 5) (x - 4)$ . Therefore, $5 x^{2} + 5 x - 100 = 5 (x + 5) (x - 4)$ .
Substitute the factored forms into the expression:

\frac{x^{2} - 7 x + 12}{5 x^{2} + 5 x - 100} = \frac{(x - 3) (x - 4)}{5 (x + 5) (x - 4)}

Cancel the common factor (x - 4), since we are given that the denominator is not zero, which implies $x \neq 4$ .

\frac{(x - 3) (x - 4)}{5 (x + 5) (x - 4)} = \frac{x - 3}{5 (x + 5)}

Answer: $\frac{x - 3}{5 (x + 5)}$

Think and Reflect: Try to simplify the following rational expression: $\frac{36 s^{2} - 12 s t + t^{2}}{t^{2} + 2 t s - 48 s^{2}}$ , assuming that $t^{2} + 2 t s - 48 s^{2} \neq 0$ .

Given: The rational expression $\frac{36 s^{2} - 12 s t + t^{2}}{t^{2} + 2 t s - 48 s^{2}}$ . To find: The simplified form of the expression.

Solution:

Factorise the numerator: $36 s^{2} - 12 s t + t^{2}$ This expression is a perfect square trinomial of the form $(a s - b t)^{2} = (a s)^{2} - 2 (a s) (b t) + (b t)^{2}$ . Here, $a^{2} = 36 ⟹ a = 6$ , $b^{2} = 1 ⟹ b = 1$ , and $2 (6 s) (t) = 12 s t$ . So, $36 s^{2} - 12 s t + t^{2} = (6 s - t)^{2}$ .
Factorise the denominator: $t^{2} + 2 t s - 48 s^{2}$ We need two numbers that add up to $2 t$ (considering $s$ as part of the coefficient) and multiply to $- 48 s^{2}$ . It's easier to rewrite this by swapping variables to match the standard form: $- 48 s^{2} + 2 t s + t^{2}$ . We need two numbers that multiply to $- 48 s^{2}$ and add up to $2 t s$ . Let's consider factors of $48$ : 6 and 8. To get a sum of $+ 2 t s$ , we need $+ 8 s$ and $- 6 s$ . So, $- 48 s^{2} + 8 s t - 6 s t + t^{2} = - 8 s (s - t) - t (6 s - t)$ . This does not factor easily.

Let's consider the variables in the form $A x^{2} + B x + C$ . Here, we can treat $t$ as the variable and $s$ as part of the coefficient, or vice-versa. Let's treat it as factoring a quadratic in $t$ : $t^{2} + (2 s) t - 48 s^{2}$ . We need two numbers that multiply to $- 48 s^{2}$ and add to $2 s$ . The numbers are $8 s$ and $- 6 s$ . So, $t^{2} + 8 s t - 6 s t - 48 s^{2} = t (t + 8 s) - 6 s (t + 8 s) = (t - 6 s) (t + 8 s)$ .
Substitute the factored forms into the expression:

\frac{(6 s - t)^{2}}{(t - 6 s) (t + 8 s)}

Note that $(6 s - t) = - (t - 6 s)$ . So, $(6 s - t)^{2} = (- (t - 6 s))^{2} = (t - 6 s)^{2}$ . The expression becomes: $\frac{(t - 6 s)^{2}}{(t - 6 s) (t + 8 s)}$
Cancel the common factor (t - 6s), since the denominator is not zero, implying $t \neq 6 s$ .

\frac{(t - 6 s) (t - 6 s)}{(t - 6 s) (t + 8 s)} = \frac{t - 6 s}{t + 8 s}

Answer: $\frac{t - 6 s}{t + 8 s}$

End of Chapter Exercises

Exercise 1: Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:

(i) $\frac{3 p^{2} - 3 p q - 18 q^{2}}{p^{2} + 3 p q - 10 q^{2}}$

Factorise numerator: $3 p^{2} - 3 p q - 18 q^{2} = 3 (p^{2} - p q - 6 q^{2})$ . For $p^{2} - p q - 6 q^{2}$ , we need two numbers that multiply to $- 6 q^{2}$ and add to $- p q$ . These are $- 3 q$ and $2 q$ . So, $p^{2} - p q - 6 q^{2} = (p - 3 q) (p + 2 q)$ . Numerator: $3 (p - 3 q) (p + 2 q)$ .
Factorise denominator: $p^{2} + 3 p q - 10 q^{2}$ . We need two numbers that multiply to $- 10 q^{2}$ and add to $3 p q$ . These are $5 q$ and $- 2 q$ . So, $p^{2} + 3 p q - 10 q^{2} = (p + 5 q) (p - 2 q)$ .
Simplify: $\frac{3 (p - 3 q) (p + 2 q)}{(p + 5 q) (p - 2 q)}$ There are no common factors to cancel.

Answer: $\frac{3 (p - 3 q) (p + 2 q)}{(p + 5 q) (p - 2 q)}$

(ii) $\frac{n^{3} - 3 n^{2} m + 3 n m^{2} - m^{3}}{5 m^{2} - 10 m n + 5 n^{2}}$

Factorise numerator: $n^{3} - 3 n^{2} m + 3 n m^{2} - m^{3}$ is the expansion of the cubic identity $(n - m)^{3}$ . Numerator: $(n - m)^{3}$ .
Factorise denominator: $5 m^{2} - 10 m n + 5 n^{2} = 5 (m^{2} - 2 m n + n^{2})$ . $m^{2} - 2 m n + n^{2}$ is the expansion of the identity $(m - n)^{2}$ . Denominator: $5 (m - n)^{2} = 5 (n - m)^{2}$ .
Simplify: $\frac{(n - m)^{3}}{5 (n - m)^{2}} = \frac{(n - m) \times (n - m)^{2}}{5 \times (n - m)^{2}}$ . Cancel out $(n - m)^{2}$ .

Answer: $\frac{n - m}{5}$

(iv) $\frac{(2 x^{2} + x - 6) (x^{2} - 7 x + 12)}{(x^{2} - 6 x + 8) (x^{2} - 9)}$

Factorise $2 x^{2} + x - 6$ : Need numbers that multiply to $2 \times - 6 = - 12$ and add to 1. These are 4 and -3. $2 x^{2} + 4 x - 3 x - 6 = 2 x (x + 2) - 3 (x + 2) = (2 x - 3) (x + 2)$ .
Factorise $x^{2} - 7 x + 12$ : Need numbers that multiply to 12 and add to -7. These are -3 and -4. $x^{2} - 7 x + 12 = (x - 3) (x - 4)$ .
Factorise $x^{2} - 6 x + 8$ : Need numbers that multiply to 8 and add to -6. These are -4 and -2. $x^{2} - 6 x + 8 = (x - 4) (x - 2)$ .
Factorise $x^{2} - 9$ : This is a difference of squares, $x^{2} - 3^{2} = (x - 3) (x + 3)$ .
Substitute and simplify:

\frac{(2 x - 3) (x + 2) (x - 3) (x - 4)}{(x - 4) (x - 2) (x - 3) (x + 3)}

Cancel common factors: $(x - 3)$ and $(x - 4)$ .

Answer: $\frac{(2 x - 3) (x + 2)}{(x - 2) (x + 3)}$

Exercise 5: Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.

This exercise requires factorising quadratic expressions into two factors, representing length and breadth.

(i) Area = $25 a^{2} - 30 a b + 9 b^{2}$

Factorise the expression: This is a perfect square trinomial of the form $(x a - y b)^{2} = (x a)^{2} - 2 (x a) (y b) + (y b)^{2}$ . Here, $(5 a)^{2} - 2 (5 a) (3 b) + (3 b)^{2} = (5 a - 3 b)^{2}$ . So, the area is $(5 a - 3 b) (5 a - 3 b)$ .

Possible expressions: Length = $(5 a - 3 b)$ units, Breadth = $(5 a - 3 b)$ units.

(ii) Area = $36 s^{2} - 49 t^{2}$

Factorise the expression: This is a difference of squares, $(a s)^{2} - (b t)^{2} = (a s - b t) (a s + b t)$ . Here, $(6 s)^{2} - (7 t)^{2} = (6 s - 7 t) (6 s + 7 t)$ .

Possible expressions: Length = $(6 s + 7 t)$ units, Breadth = $(6 s - 7 t)$ units (or vice-versa).

Exercise 6: Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.

This exercise requires factorising expressions into three factors, representing length, breadth, and height.

(i) Volume = $6 a^{2} - 24 b^{2}$

Factorise the expression: First, take out the common factor of 6: $6 (a^{2} - 4 b^{2})$ . The term in the parenthesis is a difference of squares: $a^{2} - (2 b)^{2} = (a - 2 b) (a + 2 b)$ . So, the volume is $6 \times (a - 2 b) \times (a + 2 b)$ .

Possible expressions: Length = 6 units, Breadth = $(a - 2 b)$ units, Height = $(a + 2 b)$ units (any permutation of these three factors).

(ii) Volume = $3 p s^{2} - 15 p s + 12 p$

Factorise the expression: First, take out the common factor of $3 p$ : $3 p (s^{2} - 5 s + 4)$ . Now, factorise the quadratic expression $s^{2} - 5 s + 4$ . We need two numbers that multiply to 4 and add to -5. These are -1 and -4. So, $s^{2} - 5 s + 4 = (s - 1) (s - 4)$ . The volume is $3 p \times (s - 1) \times (s - 4)$ .

Possible expressions: Length = $3 p$ units, Breadth = $(s - 1)$ units, Height = $(s - 4)$ units (any permutation of these three factors).

Remember
When simplifying rational expressions, always ensure that the cancelled common factor is not equal to zero. The problem statement usually provides this condition.

Exam-smart
Case/Source-based questions may provide a rational expression representing a real-world scenario (e.g., area of a garden, volume of a box) and ask you to simplify it to find dimensions or other relevant quantities. You will need to factorise the numerator and denominator, identify common factors, and cancel them out, stating any given restrictions on the variables.

Exploring Algebraic Identities — Notes

Read the notes

Introduction

Example 1: Pattern with Consecutive Square Numbers

Visualising Identities

Example 3: Expand (5x+2y)2

Example 4: Calculate 432

Exercise 4.1: Expansion

Factorisation of Algebraic Expressions Using Identities

Example 5: Factorise x2+4x+4

Example 6: Factorise 36x2+12x+1

Example 7: Factorise 50p2+60pq+18q2

Example 8: Calculate 292

Exercise 4.2: Factorisation and Values

More Identities

Identity for the Square of a Sum of Three Numbers

Example 9: Using the identity (a+b+c)2

Summary of Key Identities

Exercise 4.3 Applications

Example: 782

Example: Factorise 16y2−24y+9

Example: Expand (p+3q+7r)2

Exploring the Identity a2−b2=(a+b)(a−b)

Example: Calculating 552 using a2=(a+b)(a−b)+b2