Class 9 Maths Chapter 4 NCERT Solutions: Exploring Algebraic Identities

4.1 Introduction

In this chapter we explore algebraic identities — equations that are true for all values of the variables. Identities help simplify calculations and factor algebraic expressions.

Example 1

Consider any three consecutive square numbers. For example, 1, 4, and 9. Add the smallest and the largest squares. Thus, $1 + 9 = 10$ . Then subtract twice the middle square from this sum. This leads to $10 - (2 \times 4) = 10 - 8 = 2$ .

Now try the same process with another set of three consecutive square numbers. Say 9, 16, 25.

$(9 + 25) - (2 \times 16) = 34 - 32 = 2$ .

For example, consider the consecutive squares 25, 36, 49.

Applying the same rule we get $(25 + 49) - (2 \times 36) = 74 - 72 = 2$ .

Repeat this process with other sets of three consecutive square numbers.

Solution

Let's choose the consecutive squares 100, 121, 144 (which are the squares of 10, 11, and 12).

Add the smallest and largest: $100 + 144 = 244$ .

Subtract twice the middle square: $244 - (2 \times 121) = 244 - 242 = 2$ .

The result remains 2.

In general, three consecutive integers can be written as $(𝑛 - 1)$ , $𝑛$ , and $(𝑛 + 1)$ .

Their squares are $(𝑛 - 1) 2$ , $𝑛 2$ , and $(𝑛 + 1) 2$ .

Sum of extremes: $(𝑛 - 1) 2 + (𝑛 + 1) 2 = (𝑛 2 - 2 𝑛 + 1) + (𝑛 2 + 2 𝑛 + 1) = 2 𝑛 2 + 2$ .

Subtract twice the middle square: $(2 𝑛 2 + 2) - 2 𝑛 2 = 2$ .

This proves the result is always $2$ for any three consecutive squares.

Think and Reflect

Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.

Solution

Let's test four consecutive squares: 1, 4, 9, 16 (squares of 1, 2, 3, 4).

Add the extreme squares (first and fourth): $1 + 16 = 17$ .

Add the middle squares (second and third): $4 + 9 = 13$ .

The difference between these sums is $17 - 13 = 4$ .

Let's test another set: 9, 16, 25, 36 (squares of 3, 4, 5, 6).

Sum of extremes: $9 + 36 = 45$ .

Sum of middles: $16 + 25 = 41$ .

The difference is $45 - 41 = 4$ .

The pattern is that for any four consecutive square numbers, the sum of the outermost squares minus the sum of the inner squares always equals 4.

Algebraically, for four consecutive squares $𝑛 2, (𝑛 + 1) 2, (𝑛 + 2) 2, (𝑛 + 3) 2$ :

Sum of extremes: $𝑛 2 + (𝑛 + 3) 2 = 𝑛 2 + (𝑛 2 + 6 𝑛 + 9) = 2 𝑛 2 + 6 𝑛 + 9$ .

Sum of middles: $(𝑛 + 1) 2 + (𝑛 + 2) 2 = (𝑛 2 + 2 𝑛 + 1) + (𝑛 2 + 4 𝑛 + 4) = 2 𝑛 2 + 6 𝑛 + 5$ .

Difference: $(2 𝑛 2 + 6 𝑛 + 9) - (2 𝑛 2 + 6 𝑛 + 5) = 4$ .

So the pattern always equals $4$ .

4.2 Visualising Identities

In this section we revisit algebraic identities using geometrical models — squares and rectangles representing terms.

Consider two line segments of lengths $𝑎$ and $𝑏$ units, and make a longer line segment of length $(𝑎 + 𝑏)$ units as shown in Fig. 4.1.

Fig. 4.1: Line segments of lengths

𝑎

and

𝑏

forming a segment of length

(𝑎 + 𝑏)

We can construct a square of side $(𝑎 + 𝑏)$ units and partition it into smaller squares and rectangles as shown in Fig. 4.2.

Fig. 4.2: Square of side

(𝑎 + 𝑏)

units

The area of the outer square is $(𝑎 + 𝑏) 2$ . The area of the larger inner square is $𝑎 2$ and the smaller square is $𝑏 2$ . The two rectangles each have area $𝑎 𝑏$ . Hence

(𝑎 + 𝑏) 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2 .

An algebraic identity is an equation that is true for all values of the variables, while an equation need not be true for all values.

Example 2

Let $𝑎 = - 2$ and $𝑏 = - 3$ .

Then $(𝑎 + 𝑏) = - 5$ and $(𝑎 + 𝑏) 2 = 25$ .

Also $𝑎 2 = 4$ , $𝑏 2 = 9$ and $2 𝑎 𝑏 = 12$ .

Thus $𝑎 2 + 2 𝑎 𝑏 + 𝑏 2 = 4 + 12 + 9 = 25$ .

Hence, $𝑎 2 + 2 𝑎 𝑏 + 𝑏 2 = (𝑎 + 𝑏) 2$ again!

Now suppose $𝑎$ and $𝑏$ are rational numbers, say $𝑎 = - 2 / 3$ and $𝑏 = 3 / 4$ .

Solution

First, calculate $(𝑎 + 𝑏)$ :

$(𝑎 + 𝑏) = - 2 / 3 + 3 / 4 = - 8 / 12 + 9 / 12 = 1 / 12$ .

Squaring this result:

$(𝑎 + 𝑏) 2 = (1 / 12) 2 = 1 / 144$ .

Next, expand using the formula $𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ :

$𝑎 2 = (- 2 / 3) 2 = 4 / 9$ .

$2 𝑎 𝑏 = 2 \times (- 2 / 3) \times (3 / 4) = - 12 / 12 = - 1$ .

$𝑏 2 = (3 / 4) 2 = 9 / 16$ .

Add them together: $49 - 1 + 916$ .

Common denominator is $144$ :

$49 = 4 \times 16 9 \times 16 = 64144$ , $1 = 144144$ , $916 = 9 \times 9 16 \times 9 = 81144$ .

So $64144 - 144144 + 81144 = 64 - 144 + 81 144 = 145 - 144 144 = 1144$ .

Both methods yield $1 / 144$ , verifying the identity for these rational numbers.

Using the distributive property: $(𝑎 + 𝑏) 2 = (𝑎 + 𝑏) (𝑎 + 𝑏) = 𝑎 2 + 𝑎 𝑏 + 𝑏 𝑎 + 𝑏 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ , which holds for all numbers.

Think and Reflect

What can you say about $𝑎$ and $𝑏$ if $(𝑎 + 𝑏) 2 < 𝑎 2 + 𝑏 2$ ?
What can you say about $𝑎$ and $𝑏$ if $(𝑎 + 𝑏) 2 > 𝑎 2 + 𝑏 2$ ?
When will $(𝑎 + 𝑏) 2$ be equal to $𝑎 2 + 𝑏 2$ ?

Solution

We know that $(𝑎 + 𝑏) 2 = 𝑎 2 + 𝑏 2 + 2 𝑎 𝑏$ . The determining factor is the sign of the $2 𝑎 𝑏$ term.

For $(𝑎 + 𝑏) 2 < 𝑎 2 + 𝑏 2$ , the $2 𝑎 𝑏$ term must be negative. This happens when $𝑎$ and $𝑏$ have opposite signs (one is positive, the other is negative).
For $(𝑎 + 𝑏) 2 > 𝑎 2 + 𝑏 2$ , the $2 𝑎 𝑏$ term must be positive. This happens when $𝑎$ and $𝑏$ have the same sign (both positive or both negative).
For $(𝑎 + 𝑏) 2 = 𝑎 2 + 𝑏 2$ , the $2 𝑎 𝑏$ term must be exactly zero. This happens when either $𝑎 = 0$ , $𝑏 = 0$ , or both are zero.

Example 3

Let us try to expand $(5 𝑥 + 2 𝑦) 2$ .

Solution

Here $𝑎 = 5 𝑥$ and $𝑏 = 2 𝑦$ . Use $(𝑎 + 𝑏) 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ :

$(5 𝑥 + 2 𝑦) 2 = (5 𝑥) 2 + 2 (5 𝑥) (2 𝑦) + (2 𝑦) 2$

$(5 𝑥) 2 = 25 𝑥 2$ , $2 (5 𝑥) (2 𝑦) = 20 𝑥 𝑦$ , $(2 𝑦) 2 = 4 𝑦 2$ .

Therefore $(5 𝑥 + 2 𝑦) 2 = 25 𝑥 2 + 20 𝑥 𝑦 + 4 𝑦 2$ .

Example 4

To calculate $432$ , we can write it as $(40 + 3) 2$ .

Solution

Write $43 = 40 + 3$ . Use $(𝑎 + 𝑏) 2$ with $𝑎 = 40$ , $𝑏 = 3$ :

$(43) 2 = (40 + 3) 2 = 402 + 2 (40) (3) + 32$

$402 = 1600$ , $2 (40) (3) = 240$ , $32 = 9$ .

Therefore $(43) 2 = 1600 + 240 + 9 = 1849$ .

Exercise Set 4.1

Question

1. Using the identity

(𝑎 + 𝑏) 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2

, expand the following:

(i)

(7 𝑥 + 4 𝑦) 2

Solution

Here $𝑎 = 7 𝑥$ and $𝑏 = 4 𝑦$ . Use $(𝑎 + 𝑏) 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ :

$(7 𝑥 + 4 𝑦) 2 = (7 𝑥) 2 + 2 (7 𝑥) (4 𝑦) + (4 𝑦) 2$

$(7 𝑥) 2 = 49 𝑥 2$ , $2 (7 𝑥) (4 𝑦) = 56 𝑥 𝑦$ , $(4 𝑦) 2 = 16 𝑦 2$ .

Therefore $(7 𝑥 + 4 𝑦) 2 = 49 𝑥 2 + 56 𝑥 𝑦 + 16 𝑦 2$ .

(ii)

(75 𝑥 + 32 𝑦) 2

Solution

Here $𝑎 = 75 𝑥$ and $𝑏 = 32 𝑦$ . Using $(𝑎 + 𝑏) 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ :

$(75 𝑥 + 32 𝑦) 2 = (75 𝑥) 2 + 2 (75 𝑥) (32 𝑦) + (32 𝑦) 2$

$= 4925 𝑥 2 + 2 \cdot 75 \cdot 32 𝑥 𝑦 + 94 𝑦 2$

$= 4925 𝑥 2 + 215 𝑥 𝑦 + 94 𝑦 2$ .

(iii)

(2.5 𝑝 + 1.5 𝑞) 2

Solution

Here $𝑎 = 2.5 𝑝$ and $𝑏 = 1.5 𝑞$ . Using $(𝑎 + 𝑏) 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ :

$(2.5 𝑝 + 1.5 𝑞) 2 = (2.5 𝑝) 2 + 2 (2.5 𝑝) (1.5 𝑞) + (1.5 𝑞) 2$

$= 6.25 𝑝 2 + 2 \times 2.5 \times 1.5 𝑝 𝑞 + 2.25 𝑞 2$

$= 6.25 𝑝 2 + 7.5 𝑝 𝑞 + 2.25 𝑞 2$ .

(iv)

(34 𝑠 + 8 𝑡) 2

Solution

Here $𝑎 = 34 𝑠$ and $𝑏 = 8 𝑡$ . Using $(𝑎 + 𝑏) 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ :

$(34 𝑠 + 8 𝑡) 2 = (34 𝑠) 2 + 2 (34 𝑠) (8 𝑡) + (8 𝑡) 2$

$= 916 𝑠 2 + 2 \cdot 34 \cdot 8 𝑠 𝑡 + 64 𝑡 2$

$= 916 𝑠 2 + 12 𝑠 𝑡 + 64 𝑡 2$ .

(v)

(𝑥 + 1 2 𝑦) 2

Solution

Here $𝑎 = 𝑥$ and $𝑏 = 1 2 𝑦$ . Using $(𝑎 + 𝑏) 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ :

$(𝑥 + 1 2 𝑦) 2 = 𝑥 2 + 2 (𝑥) (1 2 𝑦) + (1 2 𝑦) 2$

$= 𝑥 2 + 2 𝑥 2 𝑦 + 1 4 𝑦 2$

$= 𝑥 2 + 𝑥 𝑦 + 1 4 𝑦 2$ .

(vi)

(1 𝑥 + 1 𝑦) 2

Solution

Here $𝑎 = 1 𝑥$ and $𝑏 = 1 𝑦$ . Using $(𝑎 + 𝑏) 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ :

$(1 𝑥 + 1 𝑦) 2 = (1 𝑥) 2 + 2 (1 𝑥) (1 𝑦) + (1 𝑦) 2$

$= 1 𝑥 2 + 2 𝑥 𝑦 + 1 𝑦 2$ .

Question

2. Using the same identity, find the values of the following:

(i)

(64) 2

Solution

Rewrite $64 = 60 + 4$ . Use $(𝑎 + 𝑏) 2$ with $𝑎 = 60$ , $𝑏 = 4$ :

$(64) 2 = (60 + 4) 2 = 602 + 2 (60) (4) + 42$

$602 = 3600$ , $2 (60) (4) = 480$ , $42 = 16$ .

Therefore $(64) 2 = 3600 + 480 + 16 = 4096$ .

(ii)

(105) 2

Solution

Rewrite $105 = 100 + 5$ . Use $(𝑎 + 𝑏) 2$ with $𝑎 = 100$ , $𝑏 = 5$ :

$(105) 2 = (100 + 5) 2 = 1002 + 2 (100) (5) + 52$

$1002 = 10000$ , $2 (100) (5) = 1000$ , $52 = 25$ .

Therefore $(105) 2 = 10000 + 1000 + 25 = 11025$ .

(iii)

(205) 2

Solution

Rewrite $205 = 200 + 5$ . Use $(𝑎 + 𝑏) 2$ with $𝑎 = 200$ , $𝑏 = 5$ :

$(205) 2 = (200 + 5) 2 = 2002 + 2 (200) (5) + 52$

$2002 = 40000$ , $2 (200) (5) = 2000$ , $52 = 25$ .

Therefore $(205) 2 = 40000 + 2000 + 25 = 42025$ .

4.3 Factorisation of Algebraic Expressions Using Identities

Example 5

Consider the algebraic expression $𝑥 2 + 4 𝑥 + 4$ .

Solution

We observe that $𝑥 2 = (𝑥) 2$ , so $𝑎 = 𝑥$ .

Also $4 = 22$ , so $𝑏 = 2$ .

Check middle term: $2 𝑎 𝑏 = 2 (𝑥) (2) = 4 𝑥$ , which matches the given middle term $4 𝑥$ .

Therefore $𝑥 2 + 4 𝑥 + 4 = (𝑥 + 2) 2$ . The factor is $(𝑥 + 2)$ .

Example 6

Let us try to find factors of another algebraic expression: $36 𝑥 2 + 12 𝑥 + 1$ .

Solution

Writing this in the form $𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ :

$36 𝑥 2 = (6 𝑥) 2$ , so take $𝑎 = 6 𝑥$ .

$1 = (1) 2$ , so take $𝑏 = 1$ .

Check middle term: $2 𝑎 𝑏 = 2 (6 𝑥) (1) = 12 𝑥$ , which matches the given middle term.

Therefore $36 𝑥 2 + 12 𝑥 + 1 = (6 𝑥 + 1) 2$ . The factor is $(6 𝑥 + 1)$ .

Example 7

Let us try to factor $50 𝑝 2 + 60 𝑝 𝑞 + 18 𝑞 2$ .

Solution

First, factor out the common multiplier, 2:

$2 (25 𝑝 2 + 30 𝑝 𝑞 + 9 𝑞 2)$ .

Focus on the expression inside the parentheses. We can rewrite the squared terms:

$25 𝑝 2 = (5 𝑝) 2$ and $9 𝑞 2 = (3 𝑞) 2$ .

Check the middle term: $2 (5 𝑝) (3 𝑞) = 30 𝑝 𝑞$ . This fits perfectly.

Therefore, the factored form is $2 (5 𝑝 + 3 𝑞) 2$ .

Think and Reflect

What if we replace $𝑏$ by $- 𝑏$ in $(𝑎 + 𝑏) 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ ?

Solution

Substituting $- 𝑏$ into the identity gives us $(𝑎 + (- 𝑏)) 2 = 𝑎 2 + 2 (𝑎) (- 𝑏) + (- 𝑏) 2$ .

This simplifies to a new identity: $(𝑎 - 𝑏) 2 = 𝑎 2 - 2 𝑎 𝑏 + 𝑏 2$ .

To visualise $(𝑎 - 𝑏) 2 = 𝑎 2 - 2 𝑎 𝑏 + 𝑏 2$ , draw a square of side $𝑎$ units split into parts of length $(𝑎 - 𝑏)$ and $𝑏$ , as in Fig. 4.3.

Fig. 4.3: A square of side

𝑎

units

The big square has area $𝑎 2$ . The small square has area $(𝑎 - 𝑏) 2$ . The larger rectangle has area $𝑎 𝑏$ and the smaller rectangle has area $𝑏 (𝑎 - 𝑏)$ . Subtracting the rectangles from the big square:

(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

Suppose we have to calculate $292$ . We can express this as $(30 - 1) 2$ .

Solution

Write $29 = 30 - 1$ . Use $(𝑎 - 𝑏) 2$ with $𝑎 = 30$ , $𝑏 = 1$ :

$(29) 2 = (30 - 1) 2 = 302 - 2 (30) (1) + 12$

$302 = 900$ , $2 (30) (1) = 60$ , $12 = 1$ .

Therefore $(29) 2 = 900 - 60 + 1 = 841$ .

Exercise Set 4.2

Question

1. Factor completely:

(i)

9 𝑥 2 + 24 𝑥 𝑦 + 16 𝑦 2

Solution

Compare with $𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ .

Here $9 𝑥 2 = (3 𝑥) 2$ , so $𝑎 = 3 𝑥$ ; $16 𝑦 2 = (4 𝑦) 2$ , so $𝑏 = 4 𝑦$ .

Check middle term: $2 𝑎 𝑏 = 2 (3 𝑥) (4 𝑦) = 24 𝑥 𝑦$ , which matches the given middle term.

Therefore $9 𝑥 2 + 24 𝑥 𝑦 + 16 𝑦 2 = (3 𝑥 + 4 𝑦) 2$ .

(ii)

4 𝑠 2 + 20 𝑠 𝑡 + 25 𝑡 2

Solution

Compare with $𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ .

Here $4 𝑠 2 = (2 𝑠) 2$ , so $𝑎 = 2 𝑠$ ; $25 𝑡 2 = (5 𝑡) 2$ , so $𝑏 = 5 𝑡$ .

Check middle term: $2 𝑎 𝑏 = 2 (2 𝑠) (5 𝑡) = 20 𝑠 𝑡$ , which matches.

Therefore $4 𝑠 2 + 20 𝑠 𝑡 + 25 𝑡 2 = (2 𝑠 + 5 𝑡) 2$ .

(iii)

49 𝑥 2 + 28 𝑥 𝑦 + 4 𝑦 2

Solution

Compare with $𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ .

Here $49 𝑥 2 = (7 𝑥) 2$ , so $𝑎 = 7 𝑥$ ; $4 𝑦 2 = (2 𝑦) 2$ , so $𝑏 = 2 𝑦$ .

Check middle term: $2 𝑎 𝑏 = 2 (7 𝑥) (2 𝑦) = 28 𝑥 𝑦$ , which matches.

Therefore $49 𝑥 2 + 28 𝑥 𝑦 + 4 𝑦 2 = (7 𝑥 + 2 𝑦) 2$ .

(iv)

64 𝑝 2 + 323 𝑝 𝑞 + 49 𝑞 2

Solution

Compare with $𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ .

Here $64 𝑝 2 = (8 𝑝) 2$ , so $𝑎 = 8 𝑝$ ; $49 𝑞 2 = (23 𝑞) 2$ , so $𝑏 = 23 𝑞$ .

Check middle term: $2 𝑎 𝑏 = 2 (8 𝑝) (23 𝑞) = 323 𝑝 𝑞$ , which matches.

Therefore $64 𝑝 2 + 323 𝑝 𝑞 + 49 𝑞 2 = (8 𝑝 + 23 𝑞) 2$ .

*(v)

3 𝑎 2 + 4 𝑎 𝑏 + 43 𝑏 2

Solution

Factor out the common factor $3$ :

$3 𝑎 2 + 4 𝑎 𝑏 + 43 𝑏 2 = 3 (𝑎 2 + 43 𝑎 𝑏 + 49 𝑏 2)$ .

Inside the brackets, take $𝑎$ as the first term and $𝑏 i n = 23 𝑏$ so that $(23 𝑏) 2 = 49 𝑏 2$ .

Check middle term: $2 \cdot 𝑎 \cdot 23 𝑏 = 43 𝑎 𝑏$ , which matches.

Therefore $3 𝑎 2 + 4 𝑎 𝑏 + 43 𝑏 2 = 3 (𝑎 + 23 𝑏) 2$ .

*(vi)

95 𝑠 2 + 6 𝑠 𝑣 + 5 𝑣 2

Solution

Factor out $15$ :

$95 𝑠 2 + 6 𝑠 𝑣 + 5 𝑣 2 = 15 (9 𝑠 2 + 30 𝑠 𝑣 + 25 𝑣 2)$ .

Inside: $9 𝑠 2 = (3 𝑠) 2$ and $25 𝑣 2 = (5 𝑣) 2$ .

Check middle term: $2 (3 𝑠) (5 𝑣) = 30 𝑠 𝑣$ , which matches.

Therefore $95 𝑠 2 + 6 𝑠 𝑣 + 5 𝑣 2 = 15 (3 𝑠 + 5 𝑣) 2$ .

Question

2. Find the values of the following using the identity

(𝑎 - 𝑏) 2 = 𝑎 2 - 2 𝑎 𝑏 + 𝑏 2

(i)

(79) 2

Solution

Rewrite $79 = 80 - 1$ . Use $(𝑎 - 𝑏) 2$ with $𝑎 = 80$ , $𝑏 = 1$ :

$(79) 2 = (80 - 1) 2 = 802 - 2 (80) (1) + 12$

$802 = 6400$ , $2 (80) (1) = 160$ , $12 = 1$ .

Therefore $(79) 2 = 6400 - 160 + 1 = 6241$ .

(ii)

(193) 2

Solution

Rewrite $193 = 200 - 7$ . Use $(𝑎 - 𝑏) 2$ with $𝑎 = 200$ , $𝑏 = 7$ :

$(193) 2 = (200 - 7) 2 = 2002 - 2 (200) (7) + 72$

$2002 = 40000$ , $2 (200) (7) = 2800$ , $72 = 49$ .

Therefore $(193) 2 = 40000 - 2800 + 49 = 37249$ .

(iii)

(299) 2

Solution

Rewrite $299 = 300 - 1$ . Use $(𝑎 - 𝑏) 2$ with $𝑎 = 300$ , $𝑏 = 1$ :

$(299) 2 = (300 - 1) 2 = 3002 - 2 (300) (1) + 12$

$3002 = 90000$ , $2 (300) (1) = 600$ , $12 = 1$ .

Therefore $(299) 2 = 90000 - 600 + 1 = 89401$ .

4.4 More Identities

For the square of the sum of three numbers:

(𝑎 + 𝑏 + 𝑐) 2 = 𝑎 2 + 𝑏 2 + 𝑐 2 + 2 𝑎 𝑏 + 2 𝑏 𝑐 + 2 𝑐 𝑎 .

Geometrically, this is a square of side $𝑎 + 𝑏 + 𝑐$ partitioned as in Fig. 4.4.

Fig. 4.4: A geometrical model representing the identity

(𝑎 + 𝑏 + 𝑐) 2 = 𝑎 2 + 𝑏 2 + 𝑐 2 + 2 𝑎 𝑏 + 2 𝑏 𝑐 + 2 𝑐 𝑎

Think and Reflect

Label the squares and rectangles in Fig. 4.4 so that it represents the identity $(𝑎 + 𝑏 + 𝑐) 2 = 𝑎 2 + 𝑏 2 + 𝑐 2 + 2 𝑎 𝑏 + 2 𝑏 𝑐 + 2 𝑐 𝑎$ .

Solution

In Fig. 4.4, the largest square area is $(𝑎 + 𝑏 + 𝑐) 2$ . It is subdivided into:

Three squares along the diagonal representing areas $𝑎 2$ , $𝑏 2$ , and $𝑐 2$ .
Two rectangles of area $𝑎 𝑏$ (one is $𝑎$ tall and $𝑏$ wide, the other $𝑏$ tall and $𝑎$ wide).
Two rectangles of area $𝑏 𝑐$ (one is $𝑏$ tall and $𝑐$ wide, the other $𝑐$ tall and $𝑏$ wide).
Two rectangles of area $𝑎 𝑐$ (one is $𝑎$ tall and $𝑐$ wide, the other $𝑐$ tall and $𝑎$ wide).

Adding these sub-areas together gives the expanded identity. (The figure above already shows these labels.)

Example 9

Let us use this identity to find the square of a number, say 119:

$1192 = (100 + 10 + 9) 2$

Solution

Write $119 = 100 + 10 + 9$ . Using $(𝑎 + 𝑏 + 𝑐) 2 = 𝑎 2 + 𝑏 2 + 𝑐 2 + 2 𝑎 𝑏 + 2 𝑏 𝑐 + 2 𝑐 𝑎$ with $𝑎 = 100$ , $𝑏 = 10$ , $𝑐 = 9$ :

$1192 = (100 + 10 + 9) 2$

$= 1002 + 102 + 92 + 2 (100) (10) + 2 (10) (9) + 2 (9) (100)$

$= 10000 + 100 + 81 + 2000 + 180 + 1800$

$= 10000 + 100 = 10100$ ; $10100 + 81 = 10181$ ; $10181 + 2000 = 12181$ ; $12181 + 180 = 12361$ ; $12361 + 1800 = 14161$ .

Exercise Set 4.3

Question

1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

(i)

1172

Solution

Using $(𝑎 - 𝑏) 2$ is easier. Write $117 = 120 - 3$ with $𝑎 = 120$ , $𝑏 = 3$ :

$1172 = (120 - 3) 2 = 1202 - 2 (120) (3) + 32$

$1202 = 14400$ , $2 (120) (3) = 720$ , $32 = 9$ .

Therefore $1172 = 14400 - 720 + 9 = 13689$ .

(ii)

782

Solution

Using $(𝑎 - 𝑏) 2$ . Write $78 = 80 - 2$ with $𝑎 = 80$ , $𝑏 = 2$ :

$782 = (80 - 2) 2 = 802 - 2 (80) (2) + 22$

$802 = 6400$ , $2 (80) (2) = 320$ , $22 = 4$ .

Therefore $782 = 6400 - 320 + 4 = 6084$ .

(iii)

1982

Solution

Using $(𝑎 - 𝑏) 2$ . Write $198 = 200 - 2$ with $𝑎 = 200$ , $𝑏 = 2$ :

$1982 = (200 - 2) 2 = 2002 - 2 (200) (2) + 22$

$2002 = 40000$ , $2 (200) (2) = 800$ , $22 = 4$ .

Therefore $1982 = 40000 - 800 + 4 = 39204$ .

(iv)

2142

Solution

Using $(𝑎 + 𝑏) 2$ . Write $214 = 200 + 14$ with $𝑎 = 200$ , $𝑏 = 14$ :

$2142 = (200 + 14) 2 = 2002 + 2 (200) (14) + 142$

$2002 = 40000$ , $2 (200) (14) = 5600$ , $142 = 196$ .

Therefore $2142 = 40000 + 5600 + 196 = 45796$ .

(v)

11042

Solution

Using $(𝑎 + 𝑏) 2$ . Write $1104 = 1100 + 4$ with $𝑎 = 1100$ , $𝑏 = 4$ :

$11042 = (1100 + 4) 2 = 11002 + 2 (1100) (4) + 42$

$11002 = 1210000$ , $2 (1100) (4) = 8800$ , $42 = 16$ .

Therefore $11042 = 1210000 + 8800 + 16 = 1218816$ .

(vi)

11202

Solution

Using $(𝑎 + 𝑏) 2$ . Write $1120 = 1100 + 20$ with $𝑎 = 1100$ , $𝑏 = 20$ :

$11202 = (1100 + 20) 2 = 11002 + 2 (1100) (20) + 202$

$11002 = 1210000$ , $2 (1100) (20) = 44000$ , $202 = 400$ .

Therefore $11202 = 1210000 + 44000 + 400 = 1254400$ .

Question

2. Factor using suitable identities:

(i)

16 𝑦 2 - 24 𝑦 + 9

Solution

Compare with $𝑎 2 - 2 𝑎 𝑏 + 𝑏 2$ .

Here $16 𝑦 2 = (4 𝑦) 2$ , so $𝑎 = 4 𝑦$ ; $9 = 32$ , so $𝑏 = 3$ .

Check middle term: $- 2 𝑎 𝑏 = - 2 (4 𝑦) (3) = - 24 𝑦$ , which matches.

Therefore $16 𝑦 2 - 24 𝑦 + 9 = (4 𝑦 - 3) 2$ .

(ii)

94 𝑠 2 + 6 𝑠 𝑡 + 4 𝑡 2

Solution

Compare with $𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ .

Here $94 𝑠 2 = (32 𝑠) 2$ , so $𝑎 = 32 𝑠$ ; $4 𝑡 2 = (2 𝑡) 2$ , so $𝑏 = 2 𝑡$ .

Check middle term: $2 𝑎 𝑏 = 2 (32 𝑠) (2 𝑡) = 6 𝑠 𝑡$ , which matches.

Therefore $94 𝑠 2 + 6 𝑠 𝑡 + 4 𝑡 2 = (32 𝑠 + 2 𝑡) 2$ .

(iii)

𝑚 2 9 + 𝑚 𝑘 3 + 𝑘 2 4 + 3 𝑛 𝑘 + 2 𝑚 𝑛 + 9 𝑛 2

Solution

Group the perfect-square and cross terms:

$𝑚 2 9 + 𝑘 2 4 + 9 𝑛 2 + 𝑚 𝑘 3 + 3 𝑛 𝑘 + 2 𝑚 𝑛$ .

Take $𝑎 = 𝑚 3$ , $𝑏 = 𝑘 2$ , $𝑐 = 3 𝑛$ . Then:

$𝑎 2 = 𝑚 2 9$ , $𝑏 2 = 𝑘 2 4$ , $𝑐 2 = 9 𝑛 2$ ,

$2 𝑎 𝑏 = 2 \cdot 𝑚 3 \cdot 𝑘 2 = 𝑚 𝑘 3$ ,

$2 𝑏 𝑐 = 2 \cdot 𝑘 2 \cdot 3 𝑛 = 3 𝑛 𝑘$ ,

$2 𝑐 𝑎 = 2 \cdot 3 𝑛 \cdot 𝑚 3 = 2 𝑚 𝑛$ .

All terms match. Therefore the expression equals $(𝑚 3 + 𝑘 2 + 3 𝑛) 2$ .

(iv)

𝑝 2 16 - 2 + 16 𝑝 2

Solution

Compare with $𝑎 2 - 2 𝑎 𝑏 + 𝑏 2$ .

Take $𝑎 = 𝑝 4$ and $𝑏 = 4 𝑝$ . Then $𝑎 2 = 𝑝 2 16$ , $𝑏 2 = 16 𝑝 2$ .

$2 𝑎 𝑏 = 2 (𝑝 4) (4 𝑝) = 2$ , so $- 2 𝑎 𝑏 = - 2$ .

All terms match. Therefore $𝑝 2 16 - 2 + 16 𝑝 2 = (𝑝 4 - 4 𝑝) 2$ .

(v)

9 𝑎 2 + 4 𝑏 2 + 𝑐 2 - 12 𝑎 𝑏 + 6 𝑎 𝑐 - 4 𝑏 𝑐

Solution

Compare with $(𝑥 + 𝑦 + 𝑧) 2 = 𝑥 2 + 𝑦 2 + 𝑧 2 + 2 𝑥 𝑦 + 2 𝑦 𝑧 + 2 𝑧 𝑥$ .

The square terms suggest $𝑥 = 3 𝑎$ (from $9 𝑎 2$ ), $𝑦 = \pm 2 𝑏$ (from $4 𝑏 2$ ), $𝑧 = 𝑐$ (from $𝑐 2$ ).

The cross terms $- 12 𝑎 𝑏$ and $- 4 𝑏 𝑐$ are negative in $𝑏$ , while $6 𝑎 𝑐$ is positive.

So take $𝑥 = 3 𝑎$ , $𝑦 = - 2 𝑏$ , $𝑧 = 𝑐$ .

Check: $2 𝑥 𝑦 = 2 (3 𝑎) (- 2 𝑏) = - 12 𝑎 𝑏$ , $2 𝑦 𝑧 = 2 (- 2 𝑏) (𝑐) = - 4 𝑏 𝑐$ , $2 𝑧 𝑥 = 2 (𝑐) (3 𝑎) = 6 𝑎 𝑐$ . All match.

Therefore $9 𝑎 2 + 4 𝑏 2 + 𝑐 2 - 12 𝑎 𝑏 + 6 𝑎 𝑐 - 4 𝑏 𝑐 = (3 𝑎 - 2 𝑏 + 𝑐) 2$ .

Question

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

(i)

(𝑝 + 3 𝑞 + 7 𝑟) 2

Solution

Using $(𝑎 + 𝑏 + 𝑐) 2 = 𝑎 2 + 𝑏 2 + 𝑐 2 + 2 𝑎 𝑏 + 2 𝑏 𝑐 + 2 𝑐 𝑎$ with $𝑎 = 𝑝$ , $𝑏 = 3 𝑞$ , $𝑐 = 7 𝑟$ :

$(𝑝 + 3 𝑞 + 7 𝑟) 2 = 𝑝 2 + (3 𝑞) 2 + (7 𝑟) 2 + 2 (𝑝) (3 𝑞) + 2 (3 𝑞) (7 𝑟) + 2 (7 𝑟) (𝑝)$

$= 𝑝 2 + 9 𝑞 2 + 49 𝑟 2 + 6 𝑝 𝑞 + 42 𝑞 𝑟 + 14 𝑝 𝑟$ .

(ii)

(3 𝑥 - 2 𝑦 + 4 𝑧) 2

Solution

Using $(𝑎 + 𝑏 + 𝑐) 2$ with $𝑎 = 3 𝑥$ , $𝑏 = - 2 𝑦$ , $𝑐 = 4 𝑧$ :

$(3 𝑥 - 2 𝑦 + 4 𝑧) 2 = (3 𝑥) 2 + (- 2 𝑦) 2 + (4 𝑧) 2 + 2 (3 𝑥) (- 2 𝑦) + 2 (- 2 𝑦) (4 𝑧) + 2 (4 𝑧) (3 𝑥)$

$= 9 𝑥 2 + 4 𝑦 2 + 16 𝑧 2 - 12 𝑥 𝑦 - 16 𝑦 𝑧 + 24 𝑥 𝑧$ .

Question

4. Is this an identity?

(𝑎 + 𝑏 - 𝑐) 2 + (𝑎 - 𝑏 + 𝑐) 2 + (𝑎 - 𝑏 - 𝑐) 2 = 2 𝑎 2 + 2 𝑏 2 + 2 𝑐 2

Solution

Let's expand the left-hand side (LHS) by squaring each term using the $(𝑥 + 𝑦 + 𝑧) 2$ expansion:

$(𝑎 + 𝑏 - 𝑐) 2 = 𝑎 2 + 𝑏 2 + 𝑐 2 + 2 𝑎 𝑏 - 2 𝑎 𝑐 - 2 𝑏 𝑐$

$(𝑎 - 𝑏 + 𝑐) 2 = 𝑎 2 + 𝑏 2 + 𝑐 2 - 2 𝑎 𝑏 + 2 𝑎 𝑐 - 2 𝑏 𝑐$

$(𝑎 - 𝑏 - 𝑐) 2 = 𝑎 2 + 𝑏 2 + 𝑐 2 - 2 𝑎 𝑏 - 2 𝑎 𝑐 + 2 𝑏 𝑐$

Adding all three equations:

LHS $= 3 𝑎 2 + 3 𝑏 2 + 3 𝑐 2 + (2 𝑎 𝑏 - 2 𝑎 𝑏 - 2 𝑎 𝑏) + (- 2 𝑎 𝑐 + 2 𝑎 𝑐 - 2 𝑎 𝑐) + (- 2 𝑏 𝑐 - 2 𝑏 𝑐 + 2 𝑏 𝑐)$

LHS $= 3 𝑎 2 + 3 𝑏 2 + 3 𝑐 2 - 2 𝑎 𝑏 - 2 𝑎 𝑐 - 2 𝑏 𝑐$ .

This is not equal to $2 𝑎 2 + 2 𝑏 2 + 2 𝑐 2$ . Therefore, No, it is not an identity.

In Grade 8, you were introduced to $𝑎 2 - 𝑏 2 = (𝑎 + 𝑏) (𝑎 - 𝑏)$ , which can be rewritten as $𝑎 2 = (𝑎 + 𝑏) (𝑎 - 𝑏) + 𝑏 2$ .

Fig. 4.5: Geometric justification of

𝑎 2 = (𝑎 + 𝑏) (𝑎 - 𝑏) + 𝑏 2

(Śhrīdharāchārya, 750 CE)

Think and Reflect

Look at the following figure (Fig. 4.5). Justify the identity $𝑎 2 = (𝑎 + 𝑏) (𝑎 - 𝑏) + 𝑏 2$ for yourself.

Solution

Fig. 4.5 starts with a large green square of side $𝑎$ , giving area $𝑎 2$ . A vertical rectangular strip of width $𝑏$ and height $𝑎 - 𝑏$ is removed from the right side and rotated to sit horizontally at the bottom.

The remaining main shape is a rectangle with length $(𝑎 + 𝑏)$ and width $(𝑎 - 𝑏)$ , giving an area of $(𝑎 + 𝑏) (𝑎 - 𝑏)$ . However, moving the strip leaves an empty square corner of dimensions $𝑏 \times 𝑏$ , so area $𝑏 2$ is missing from the original total area $𝑎 2$ .

Thus, the total original area $𝑎 2$ equals the new rectangle's area plus the missing corner: $𝑎 2 = (𝑎 + 𝑏) (𝑎 - 𝑏) + 𝑏 2$ .

Think and Reflect

1. Try to evaluate the following using a suitable identity:

(i) $352$ (ii) $652$ (iii) $852$ (iv) $1052$

Do you observe any interesting pattern?

Solution

We can use the identity $𝑎 2 = (𝑎 + 𝑏) (𝑎 - 𝑏) + 𝑏 2$ where $𝑏 = 5$ .

(i) $352 = (35 - 5) (35 + 5) + 52 = (30) (40) + 25 = 1200 + 25 = 1225$ .

(ii) $652 = (60) (70) + 25 = 4200 + 25 = 4225$ .

(iii) $852 = (80) (90) + 25 = 7200 + 25 = 7225$ .

(iv) $1052 = (100) (110) + 25 = 11000 + 25 = 11025$ .

Pattern observed: To square a number ending in 5, multiply the part of the number before the 5 (let's call it $𝑛$ ) by the next integer ( $𝑛 + 1$ ), and append 25 to the end. For example, for $352$ , $𝑛 = 3$ , so $3 \times 4 = 12$ , resulting in 1225.

Think and Reflect

2. Observe the two rows of figures below (Fig. 4.6). They represent an algebraic identity. Try to identify it.

Fig. 4.6: Visual rearrangement illustrating a four-square identity

Solution

The top row shows four squares with sides $𝑎 + 𝑏 + 𝑐$ , $𝑎 + 𝑏 - 𝑐$ , $𝑎 - 𝑏 + 𝑐$ , and $𝑎 - 𝑏 - 𝑐$ . The bottom row rearranges the same total area into three squares with sides $2 𝑎$ , $2 𝑏$ , and $2 𝑐$ .

Expanding confirms the identity:

(𝑎 + 𝑏 + 𝑐) 2 + (𝑎 + 𝑏 - 𝑐) 2 + (𝑎 - 𝑏 + 𝑐) 2 + (𝑎 - 𝑏 - 𝑐) 2 = 4 (𝑎 2 + 𝑏 2 + 𝑐 2) = (2 𝑎) 2 + (2 𝑏) 2 + (2 𝑐) 2 .

(Each cross term cancels when the four expansions are added, leaving four copies of each square term.)

4.5 Factorisation Using Algebra Tiles

Consider a rectangle with sides $𝑥 + 3$ and $𝑥 + 4$ . Its area is $(𝑥 + 3) (𝑥 + 4) = 𝑥 2 + 7 𝑥 + 12$ . Fig. 4.7 visualises this product with algebra tiles.

Fig. 4.7: Factorisation of

𝑥 2 + 7 𝑥 + 12

The $7 𝑥$ is split as $3 𝑥 + 4 𝑥$ (three $𝑥$ -tiles to the right of $𝑥 2$ and four below). The 12 unit tiles form a $3 \times 4$ array. The rectangle dimensions are $𝑥 + 3$ and $𝑥 + 4$ .

Think and Reflect

Suppose $7 𝑥$ is split as $2 𝑥 + 5 𝑥$ ; can a similar rectangular arrangement be formed? Consider other possibilities and check.

Solution

No, a solid rectangle cannot be formed if $7 𝑥$ is split as $2 𝑥 + 5 𝑥$ for the expression $𝑥 2 + 7 𝑥 + 12$ . To form a rectangle with 12 unit tiles, the lengths of the $𝑥$ -tiles strips must multiply to 12. Since $2 \times 5 = 10 \neq 12$ , you would have gaps. The only way to form a perfect rectangle with 12 units is by splitting $7 𝑥$ into $3 𝑥$ and $4 𝑥$ (since $3 \times 4 = 12$ ).

Think and Reflect

1. Figure out the product of $𝑥 + 2$ and $𝑥 + 3$ using algebra tiles.

2. Lay out algebra tiles for $𝑥 2 + 11 𝑥 + 30$ in such a way that you will see its factors.

Solution

1. A rectangle with sides $(𝑥 + 2)$ and $(𝑥 + 3)$ will consist of one $𝑥 2$ -tile, two $𝑥$ -tiles on one side, three $𝑥$ -tiles on the adjacent side, and a 2 by 3 grid of unit tiles. Adding them gives $𝑥 2 + 5 𝑥 + 6$ .

2. For $𝑥 2 + 11 𝑥 + 30$ , we need two numbers that multiply to 30 and add to 11. These are 5 and 6. The tile layout will be one $𝑥 2$ -tile, a column of 5 $𝑥$ -tiles, a row of 6 $𝑥$ -tiles, and a 5 by 6 grid of 30 unit tiles. The dimensions (factors) are $(𝑥 + 5)$ and $(𝑥 + 6)$ .

Think and Reflect

We have seen that $(𝑥 + 3) (𝑥 + 4) = 𝑥 2 + 7 𝑥 + 12$ .

Also $(𝑥 + 6) (𝑥 + 7) = 𝑥 2 + 13 𝑥 + 42$ .

Generalise the pattern to get an expression for $(𝑥 + 𝑎) (𝑥 + 𝑏)$ .

Solution

Looking at the pattern, the middle term coefficient is the sum of the constants, and the final term is their product.

Generalizing: $(𝑥 + 𝑎) (𝑥 + 𝑏) = 𝑥 2 + (𝑎 + 𝑏) 𝑥 + 𝑎 𝑏$ .

Now consider a rectangle of side-lengths $2 𝑥 + 3$ and $3 𝑥 + 1$ , as in Fig. 4.8.

Fig. 4.8: Using algebra tiles to represent

(2 𝑥 + 3) \times (3 𝑥 + 1)

The area is $(2 𝑥 + 3) (3 𝑥 + 1) = 6 𝑥 2 + 11 𝑥 + 3$ (six $𝑥 2$ -tiles, eleven $𝑥$ -tiles, three unit tiles).

Question

Task: Fill in the blanks:

(𝑝 𝑥 + 𝑎) (𝑞 𝑥 + 𝑏) = (__) 𝑥 2 + (__) 𝑥 +__

. Also verify using the distributive property.

Solution

Expand using distributivity step by step:

$(𝑝 𝑥 + 𝑎) (𝑞 𝑥 + 𝑏) = 𝑝 𝑥 (𝑞 𝑥 + 𝑏) + 𝑎 (𝑞 𝑥 + 𝑏)$

$= 𝑝 𝑥 \cdot 𝑞 𝑥 + 𝑝 𝑥 \cdot 𝑏 + 𝑎 \cdot 𝑞 𝑥 + 𝑎 \cdot 𝑏$

$= 𝑝 𝑞 𝑥 2 + 𝑝 𝑏 𝑥 + 𝑞 𝑎 𝑥 + 𝑎 𝑏$

$= 𝑝 𝑞 𝑥 2 + (𝑝 𝑏 + 𝑞 𝑎) 𝑥 + 𝑎 𝑏$ .

So the blanks are: $𝑝 𝑞$ , $𝑝 𝑏 + 𝑞 𝑎$ , and $𝑎 𝑏$ .

4.6 Factorisation Without Using Algebra Tiles

Example 10

Let us begin with $𝑥 2 + 7 𝑥 + 12 = 𝑥 2 + (𝑎 + 𝑏) 𝑥 + 𝑎 𝑏$ .

Solution

Compare $𝑥 2 + 7 𝑥 + 12$ with $𝑥 2 + (𝑎 + 𝑏) 𝑥 + 𝑎 𝑏$ .

We need $𝑎 + 𝑏 = 7$ and $𝑎 𝑏 = 12$ . The pair $𝑎 = 3$ , $𝑏 = 4$ works.

Split the middle term: $𝑥 2 + 7 𝑥 + 12 = 𝑥 2 + 3 𝑥 + 4 𝑥 + 12$

$= 𝑥 (𝑥 + 3) + 4 (𝑥 + 3) = (𝑥 + 3) (𝑥 + 4)$ .

Example 11

Let us try to factor $𝑥 2 + 11 𝑥 + 30$ in a similar manner.

Solution

Compare $𝑥 2 + 11 𝑥 + 30$ with $𝑥 2 + (𝑎 + 𝑏) 𝑥 + 𝑎 𝑏$ .

We need $𝑎 + 𝑏 = 11$ and $𝑎 𝑏 = 30$ . The pair $𝑎 = 5$ , $𝑏 = 6$ works.

Split the middle term: $𝑥 2 + 11 𝑥 + 30 = 𝑥 2 + 5 𝑥 + 6 𝑥 + 30$

$= 𝑥 (𝑥 + 5) + 6 (𝑥 + 5) = (𝑥 + 5) (𝑥 + 6)$ .

Example 12

In order to factor $𝑥 2 - 5 𝑥 + 6$ we first note that the coefficient of $𝑥$ is negative.

Solution

Compare $𝑥 2 - 5 𝑥 + 6$ with $𝑥 2 + (𝑎 + 𝑏) 𝑥 + 𝑎 𝑏$ .

We need $𝑎 + 𝑏 = - 5$ and $𝑎 𝑏 = 6$ . The pair $𝑎 = - 2$ , $𝑏 = - 3$ works.

Split the middle term: $𝑥 2 - 5 𝑥 + 6 = 𝑥 2 - 2 𝑥 - 3 𝑥 + 6$

$= 𝑥 (𝑥 - 2) - 3 (𝑥 - 2) = (𝑥 - 2) (𝑥 - 3)$ .

Check: $(𝑥 - 2) (𝑥 - 3) = 𝑥 2 - 5 𝑥 + 6$ .

Exercise Set 4.4

Question

1. Fill in the blanks to complete the following identities:

(i)

𝑠 2 - 11 𝑠 + 24 = (__) (__)

Solution

We need two numbers whose product is $24$ and whose sum is $- 11$ .

Possible factor pairs of $24$ include $(\pm 3, \pm 8)$ , $(\pm 4, \pm 6)$ , etc.

The pair $- 8$ and $- 3$ works: $(- 8) \times (- 3) = 24$ and $(- 8) + (- 3) = - 11$ .

Therefore $𝑠 2 - 11 𝑠 + 24 = (𝑠 - 8) (𝑠 - 3)$ .

Check: $(𝑠 - 8) (𝑠 - 3) = 𝑠 2 - 3 𝑠 - 8 𝑠 + 24 = 𝑠 2 - 11 𝑠 + 24$ .

(ii)

(__) (𝑥 + 1) = 3 𝑥 2 - 4 𝑥 - 7

Solution

Let the missing factor be $(𝐴 𝑥 + 𝐵)$ . Then $(𝐴 𝑥 + 𝐵) (𝑥 + 1) = 𝐴 𝑥 2 + (𝐴 + 𝐵) 𝑥 + 𝐵$ .

Comparing with $3 𝑥 2 - 4 𝑥 - 7$ : $𝐴 = 3$ and $𝐵 = - 7$ .

Check middle coefficient: $𝐴 + 𝐵 = 3 + (- 7) = - 4$ , which matches.

Therefore the missing factor is $(3 𝑥 - 7)$ .

Verify: $(3 𝑥 - 7) (𝑥 + 1) = 3 𝑥 2 + 3 𝑥 - 7 𝑥 - 7 = 3 𝑥 2 - 4 𝑥 - 7$ .

(iii)

10 𝑥 2 - 11 𝑥 - 6 = (2 𝑥 -__) (__+ 2)

Solution

We need $(2 𝑥 - ――) (―― + 2) = 10 𝑥 2 - 11 𝑥 - 6$ .

For the $𝑥 2$ term: $2 𝑥$ times the first term of the second factor must give $10 𝑥 2$ , so that first term is $5 𝑥$ .

For the constant: $(- b l a n k) \times 2 = - 6$ implies the blank is $3$ .

Thus try $(2 𝑥 - 3) (5 𝑥 + 2)$ :

$(2 𝑥 - 3) (5 𝑥 + 2) = 10 𝑥 2 + 4 𝑥 - 15 𝑥 - 6 = 10 𝑥 2 - 11 𝑥 - 6$ .

This matches. So $10 𝑥 2 - 11 𝑥 - 6 = (2 𝑥 - 3) (5 𝑥 + 2)$ .

(iv)

6 𝑥 2 + 7 𝑥 + 2 = (__) (__)

Solution

For $6 𝑥 2 + 7 𝑥 + 2$ , we have coefficient product $𝑎 𝑐 = 6 \times 2 = 12$ .

Find two numbers whose product is $12$ and sum is $7$ : these are $4$ and $3$ .

Split the middle term: $6 𝑥 2 + 7 𝑥 + 2 = 6 𝑥 2 + 4 𝑥 + 3 𝑥 + 2$

$= 2 𝑥 (3 𝑥 + 2) + 1 (3 𝑥 + 2)$

$= (2 𝑥 + 1) (3 𝑥 + 2)$ .

Check: $(2 𝑥 + 1) (3 𝑥 + 2) = 6 𝑥 2 + 4 𝑥 + 3 𝑥 + 2 = 6 𝑥 2 + 7 𝑥 + 2$ .

Question

2. Select and use the identity that will help you to find the following products without multiplying directly:

(i)

(41) 2

Solution

Use $(𝑎 + 𝑏) 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ with $𝑎 = 40$ , $𝑏 = 1$ :

$(41) 2 = (40 + 1) 2 = 402 + 2 (40) (1) + 12$

$402 = 1600$ , $2 (40) (1) = 80$ , $12 = 1$ .

Therefore $(41) 2 = 1600 + 80 + 1 = 1681$ .

(ii)

(27) 2

Solution

Use $(𝑎 - 𝑏) 2 = 𝑎 2 - 2 𝑎 𝑏 + 𝑏 2$ with $𝑎 = 30$ , $𝑏 = 3$ :

$(27) 2 = (30 - 3) 2 = 302 - 2 (30) (3) + 32$

$302 = 900$ , $2 (30) (3) = 180$ , $32 = 9$ .

Therefore $(27) 2 = 900 - 180 + 9 = 729$ .

(iii)

(23 \times 17)

Solution

Note that $23 = 20 + 3$ and $17 = 20 - 3$ .

Use $(𝑎 + 𝑏) (𝑎 - 𝑏) = 𝑎 2 - 𝑏 2$ with $𝑎 = 20$ , $𝑏 = 3$ :

$23 \times 17 = (20 + 3) (20 - 3) = 202 - 32$

$202 = 400$ , $32 = 9$ .

Therefore $23 \times 17 = 400 - 9 = 391$ .

(iv)

(135) 2

Solution

Use $𝑎 2 = (𝑎 + 5) (𝑎 - 5) + 52$ with $𝑎 = 135$ :

$1352 = (135 + 5) (135 - 5) + 25 = 140 \times 130 + 25$ .

$140 \times 130 = 18200$ , so $1352 = 18200 + 25 = 18225$ .

(Shortcut: for a number ending in 5 with front part $𝑛 = 13$ , compute $𝑛 (𝑛 + 1) = 13 \times 14 = 182$ and append $25$ .)

(v)

(97) 2

Solution

Use $(𝑎 - 𝑏) 2 = 𝑎 2 - 2 𝑎 𝑏 + 𝑏 2$ with $𝑎 = 100$ , $𝑏 = 3$ :

$(97) 2 = (100 - 3) 2 = 1002 - 2 (100) (3) + 32$

$1002 = 10000$ , $2 (100) (3) = 600$ , $32 = 9$ .

Therefore $(97) 2 = 10000 - 600 + 9 = 9409$ .

(vi)

(18 \times 29)

Solution

Write $18 = 20 + (- 2)$ and $29 = 20 + 9$ . Use $(𝑥 + 𝑎) (𝑥 + 𝑏) = 𝑥 2 + (𝑎 + 𝑏) 𝑥 + 𝑎 𝑏$ with $𝑥 = 20$ , $𝑎 = - 2$ , $𝑏 = 9$ :

$18 \times 29 = (20 - 2) (20 + 9) = 202 + (- 2 + 9) (20) + (- 2) (9)$

$= 400 + 7 \times 20 - 18 = 400 + 140 - 18 = 522$ .

(vii)

(34 \times 43)

Solution

Write $34 = 30 + 4$ and $43 = 40 + 3$ . Expand using distributivity:

$(30 + 4) (40 + 3) = 30 \cdot 40 + 30 \cdot 3 + 4 \cdot 40 + 4 \cdot 3$

$= 1200 + 90 + 160 + 12 = 1462$ .

(viii)

(205) 2

Solution

Use $𝑎 2 = (𝑎 + 5) (𝑎 - 5) + 52$ with $𝑎 = 205$ :

$2052 = (205 + 5) (205 - 5) + 25 = 210 \times 200 + 25 = 42000 + 25 = 42025$ .

(Shortcut: front part $𝑛 = 20$ , so $𝑛 (𝑛 + 1) = 20 \times 21 = 420$ , append $25$ .)

Question

3. Factor the following:

(i)

9 𝑎 2 + 𝑏 2 + 4 𝑐 2 - 6 𝑎 𝑏 + 12 𝑎 𝑐 - 4 𝑏 𝑐

Solution

Compare with $(𝑥 + 𝑦 + 𝑧) 2$ . Square terms: $9 𝑎 2 = (3 𝑎) 2$ , $𝑏 2 = (\pm 𝑏) 2$ , $4 𝑐 2 = (2 𝑐) 2$ .

Cross terms: $- 6 𝑎 𝑏 = 2 (3 𝑎) (- 𝑏)$ , $12 𝑎 𝑐 = 2 (3 𝑎) (2 𝑐)$ , $- 4 𝑏 𝑐 = 2 (- 𝑏) (2 𝑐)$ .

So take $𝑥 = 3 𝑎$ , $𝑦 = - 𝑏$ , $𝑧 = 2 𝑐$ .

Therefore $9 𝑎 2 + 𝑏 2 + 4 𝑐 2 - 6 𝑎 𝑏 + 12 𝑎 𝑐 - 4 𝑏 𝑐 = (3 𝑎 - 𝑏 + 2 𝑐) 2$ .

Check: $(3 𝑎 - 𝑏 + 2 𝑐) 2 = 9 𝑎 2 + 𝑏 2 + 4 𝑐 2 - 6 𝑎 𝑏 - 4 𝑏 𝑐 + 12 𝑎 𝑐$ .

(ii)

16 𝑠 2 + 25 𝑡 2 - 40 𝑠 𝑡

Solution

Compare with $𝑎 2 - 2 𝑎 𝑏 + 𝑏 2$ .

$16 𝑠 2 = (4 𝑠) 2$ so $𝑎 = 4 𝑠$ ; $25 𝑡 2 = (5 𝑡) 2$ so $𝑏 = 5 𝑡$ .

Middle term: $- 2 𝑎 𝑏 = - 2 (4 𝑠) (5 𝑡) = - 40 𝑠 𝑡$ , which matches.

Therefore $16 𝑠 2 + 25 𝑡 2 - 40 𝑠 𝑡 = (4 𝑠 - 5 𝑡) 2$ .

(iii)

𝑟 2 - 𝑟 - 42

Solution

We need two numbers whose product is $- 42$ and whose sum is $- 1$ .

The pair $- 7$ and $6$ works: $(- 7) \times 6 = - 42$ and $- 7 + 6 = - 1$ .

Therefore $𝑟 2 - 𝑟 - 42 = (𝑟 - 7) (𝑟 + 6)$ .

Check: $(𝑟 - 7) (𝑟 + 6) = 𝑟 2 + 6 𝑟 - 7 𝑟 - 42 = 𝑟 2 - 𝑟 - 42$ .

(iv)

49 𝑔 2 + 14 𝑔 ℎ + ℎ 2

Solution

Compare with $𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ .

$49 𝑔 2 = (7 𝑔) 2$ so $𝑎 = 7 𝑔$ ; $ℎ 2 = ℎ 2$ so $𝑏 = ℎ$ .

Middle term: $2 𝑎 𝑏 = 2 (7 𝑔) (ℎ) = 14 𝑔 ℎ$ , which matches.

Therefore $49 𝑔 2 + 14 𝑔 ℎ + ℎ 2 = (7 𝑔 + ℎ) 2$ .

(v)

64 𝑢 2 + 121 𝑣 2 + 4 𝑤 2 - 176 𝑢 𝑣 - 32 𝑢 𝑤 + 44 𝑣 𝑤

Solution

Square terms: $64 𝑢 2 = (8 𝑢) 2$ , $121 𝑣 2 = (11 𝑣) 2$ , $4 𝑤 2 = (2 𝑤) 2$ .

Cross terms: $- 176 𝑢 𝑣 = 2 (- 8 𝑢) (11 𝑣)$ ; $- 32 𝑢 𝑤 = 2 (- 8 𝑢) (2 𝑤)$ ; $44 𝑣 𝑤 = 2 (11 𝑣) (2 𝑤)$ .

Taking $𝑥 = - 8 𝑢$ , $𝑦 = 11 𝑣$ , $𝑧 = 2 𝑤$ all cross terms match.

Therefore $64 𝑢 2 + 121 𝑣 2 + 4 𝑤 2 - 176 𝑢 𝑣 - 32 𝑢 𝑤 + 44 𝑣 𝑤 = (- 8 𝑢 + 11 𝑣 + 2 𝑤) 2$ .

Equivalently (overall sign flip): $(8 𝑢 - 11 𝑣 - 2 𝑤) 2$ .

Think and Reflect

James and Reshma were talking about algebraic identities... According to you, who is correct and why?

James expands $(𝑎 - 𝑏) 2 (𝑎 + 𝑏)$ as $(𝑎 2 - 2 𝑎 𝑏 + 𝑏 2) (𝑎 + 𝑏)$ . Reshma rewrites it as $(𝑎 - 𝑏) [(𝑎 - 𝑏) (𝑎 + 𝑏)] = (𝑎 - 𝑏) (𝑎 2 - 𝑏 2)$ .

Solution

James: $(𝑎 - 𝑏) 2 (𝑎 + 𝑏) = (𝑎 2 - 2 𝑎 𝑏 + 𝑏 2) (𝑎 + 𝑏)$ — This is correct because he properly expanded $(𝑎 - 𝑏) 2$ first before multiplying.

Reshma: $(𝑎 - 𝑏) 2 (𝑎 + 𝑏) = (𝑎 - 𝑏) [(𝑎 - 𝑏) (𝑎 + 𝑏)] = (𝑎 - 𝑏) (𝑎 2 - 𝑏 2)$ — This is also correct. She regrouped the factors intelligently to use the $(𝑎 - 𝑏) (𝑎 + 𝑏) = 𝑎 2 - 𝑏 2$ identity. Both are mathematically correct, though Reshma's method is slightly more elegant and faster for manual calculation.

4.7 Finding New Identities

Question

Task: Try to multiply the following using the distributive property.
1.

(𝑥 - 𝑦) (𝑥 2 + 𝑥 𝑦 + 𝑦 2)

(𝑥 + 𝑦) (𝑥 2 - 𝑥 𝑦 + 𝑦 2)

Solution

1. Expand

(𝑥 - 𝑦) (𝑥 2 + 𝑥 𝑦 + 𝑦 2)

by distributivity:

$= 𝑥 (𝑥 2 + 𝑥 𝑦 + 𝑦 2) - 𝑦 (𝑥 2 + 𝑥 𝑦 + 𝑦 2)$

$= 𝑥 3 + 𝑥 2 𝑦 + 𝑥 𝑦 2 - 𝑥 2 𝑦 - 𝑥 𝑦 2 - 𝑦 3 = 𝑥 3 - 𝑦 3$ .

2. Expand

(𝑥 + 𝑦) (𝑥 2 - 𝑥 𝑦 + 𝑦 2)

$= 𝑥 (𝑥 2 - 𝑥 𝑦 + 𝑦 2) + 𝑦 (𝑥 2 - 𝑥 𝑦 + 𝑦 2)$

$= 𝑥 3 - 𝑥 2 𝑦 + 𝑥 𝑦 2 + 𝑥 2 𝑦 - 𝑥 𝑦 2 + 𝑦 3 = 𝑥 3 + 𝑦 3$ .

Thus $𝑥 3 - 𝑦 3 = (𝑥 - 𝑦) (𝑥 2 + 𝑥 𝑦 + 𝑦 2)$ and $𝑥 3 + 𝑦 3 = (𝑥 + 𝑦) (𝑥 2 - 𝑥 𝑦 + 𝑦 2)$ .

Example 13

What is the side of the cube whose volume is $𝑝 3 + 6 𝑝 2 𝑞 + 12 𝑝 𝑞 2 + 8 𝑞 3$ cubic units?

Solution

Compare $𝑝 3 + 6 𝑝 2 𝑞 + 12 𝑝 𝑞 2 + 8 𝑞 3$ with $(𝑎 + 𝑏) 3 = 𝑎 3 + 3 𝑎 2 𝑏 + 3 𝑎 𝑏 2 + 𝑏 3$ .

The first term $𝑝 3$ suggests $𝑎 = 𝑝$ . The last term $8 𝑞 3 = (2 𝑞) 3$ suggests $𝑏 = 2 𝑞$ .

Check: $3 𝑎 2 𝑏 = 3 𝑝 2 (2 𝑞) = 6 𝑝 2 𝑞$ , and $3 𝑎 𝑏 2 = 3 𝑝 (4 𝑞 2) = 12 𝑝 𝑞 2$ . Both match.

Therefore the volume is $(𝑝 + 2 𝑞) 3$ , so the side is $𝑝 + 2 𝑞$ .

Example 14

Now consider the expression $8 𝑛 3 - 60 𝑛 2 𝑚 + 150 𝑛 𝑚 2 - 125 𝑚 3$ . If you write it in the form $(𝑎 - 𝑏) 3$ what will be $𝑎$ and $𝑏$ ?

Solution

Compare $8 𝑛 3 - 60 𝑛 2 𝑚 + 150 𝑛 𝑚 2 - 125 𝑚 3$ with $(𝑎 - 𝑏) 3 = 𝑎 3 - 3 𝑎 2 𝑏 + 3 𝑎 𝑏 2 - 𝑏 3$ .

First term $8 𝑛 3 = (2 𝑛) 3$ suggests $𝑎 = 2 𝑛$ . Last term $125 𝑚 3 = (5 𝑚) 3$ suggests $𝑏 = 5 𝑚$ .

Check: $- 3 𝑎 2 𝑏 = - 3 (4 𝑛 2) (5 𝑚) = - 60 𝑛 2 𝑚$ , and $3 𝑎 𝑏 2 = 3 (2 𝑛) (25 𝑚 2) = 150 𝑛 𝑚 2$ . Both match.

Therefore the expression is $(2 𝑛 - 5 𝑚) 3$ .

Think and Reflect

Do you think $𝑥 - 𝑦$ is also a factor of $𝑥 4 - 𝑦 4$ ? Can you see how $𝑥 - 𝑦$ is a factor of $𝑥 4 - 𝑦 4$ ? How about $𝑥 5 - 𝑦 5$ ?

Solution

Yes, $𝑥 - 𝑦$ is a factor of $𝑥 4 - 𝑦 4$ . Since $𝑥 4 - 𝑦 4 = (𝑥 2 - 𝑦 2) (𝑥 2 + 𝑦 2) = (𝑥 - 𝑦) (𝑥 + 𝑦) (𝑥 2 + 𝑦 2)$ , $𝑥 - 𝑦$ is clearly a factor.

For $𝑥 5 - 𝑦 5$ , yes, $𝑥 - 𝑦$ is also a factor. In general, $𝑥 - 𝑦$ is a factor of $𝑥 𝑛 - 𝑦 𝑛$ for any positive integer $𝑛$ , because setting $𝑥 = 𝑦$ makes the expression equal to zero (Factor Theorem). The expansion is $𝑥 5 - 𝑦 5 = (𝑥 - 𝑦) (𝑥 4 + 𝑥 3 𝑦 + 𝑥 2 𝑦 2 + 𝑥 𝑦 3 + 𝑦 4)$ .

Example 15

The sum of three numbers is 10 and their product is 25. The sum of their squares is 38. Try to use the previous identity to find the sum of the cubes of these three numbers.

Solution

Given: $𝑥 + 𝑦 + 𝑧 = 10$ , $𝑥 𝑦 𝑧 = 25$ , $𝑥 2 + 𝑦 2 + 𝑧 2 = 38$ . Find $𝑥 3 + 𝑦 3 + 𝑧 3$ .

First, find $(𝑥 𝑦 + 𝑥 𝑧 + 𝑦 𝑧)$ using $(𝑥 + 𝑦 + 𝑧) 2 = 𝑥 2 + 𝑦 2 + 𝑧 2 + 2 (𝑥 𝑦 + 𝑥 𝑧 + 𝑦 𝑧)$ .

$102 = 38 + 2 (𝑥 𝑦 + 𝑥 𝑧 + 𝑦 𝑧) ⟹ 100 - 38 = 2 (𝑥 𝑦 + 𝑥 𝑧 + 𝑦 𝑧) ⟹ 62 = 2 (𝑥 𝑦 + 𝑥 𝑧 + 𝑦 𝑧) ⟹ 𝑥 𝑦 + 𝑥 𝑧 + 𝑦 𝑧 = 31$ .

Next, use the identity: $(𝑥 + 𝑦 + 𝑧) (𝑥 2 + 𝑦 2 + 𝑧 2 - 𝑥 𝑦 - 𝑥 𝑧 - 𝑦 𝑧) = 𝑥 3 + 𝑦 3 + 𝑧 3 - 3 𝑥 𝑦 𝑧$ .

$10 (38 - 31) = 𝑥 3 + 𝑦 3 + 𝑧 3 - 3 (25)$

$10 (7) = 𝑥 3 + 𝑦 3 + 𝑧 3 - 75$

$70 + 75 = 𝑥 3 + 𝑦 3 + 𝑧 3$

$𝑥 3 + 𝑦 3 + 𝑧 3 = 145$ .

4.8 Simplifying Rational Expressions

Example 16

Simplify the rational expression $𝑥 2 - 7 𝑥 + 12 5 𝑥 2 + 5 𝑥 - 100$ assuming that $5 𝑥 2 + 5 𝑥 - 100 \neq 0$ .

Solution

Numerator: $𝑥 2 - 7 𝑥 + 12$ . Numbers with product $12$ and sum $- 7$ are $- 3$ and $- 4$ .

So $𝑥 2 - 7 𝑥 + 12 = (𝑥 - 3) (𝑥 - 4)$ .

Denominator: $5 𝑥 2 + 5 𝑥 - 100 = 5 (𝑥 2 + 𝑥 - 20)$ .

For $𝑥 2 + 𝑥 - 20$ : numbers with product $- 20$ and sum $1$ are $5$ and $- 4$ .

So $𝑥 2 + 𝑥 - 20 = (𝑥 + 5) (𝑥 - 4)$ , and denominator $= 5 (𝑥 + 5) (𝑥 - 4)$ .

Expression: $(𝑥 - 3) (𝑥 - 4) 5 (𝑥 + 5) (𝑥 - 4)$ . Cancel $(𝑥 - 4)$ : $𝑥 - 3 5 (𝑥 + 5)$ .

Think and Reflect

Try to simplify the following rational expression: $36 𝑠 2 - 12 𝑠 𝑡 + 𝑡 2 𝑡 2 + 2 𝑡 𝑠 - 48 𝑠 2 = (6 𝑠 - 𝑡) 2 (__+__) (__+__)$ .

Solution

Numerator: $36 𝑠 2 - 12 𝑠 𝑡 + 𝑡 2 = (6 𝑠 - 𝑡) 2$ (since $2 (6 𝑠) (𝑡) = 12 𝑠 𝑡$ ).

Also $(6 𝑠 - 𝑡) 2 = (𝑡 - 6 𝑠) 2$ .

Denominator: $𝑡 2 + 2 𝑡 𝑠 - 48 𝑠 2$ . Numbers with product $- 48$ and sum $2$ are $8$ and $- 6$ .

So $𝑡 2 + 2 𝑡 𝑠 - 48 𝑠 2 = 𝑡 2 + 8 𝑡 𝑠 - 6 𝑡 𝑠 - 48 𝑠 2 = 𝑡 (𝑡 + 8 𝑠) - 6 𝑠 (𝑡 + 8 𝑠) = (𝑡 + 8 𝑠) (𝑡 - 6 𝑠)$ .

Expression: $(𝑡 - 6 𝑠) 2 (𝑡 + 8 𝑠) (𝑡 - 6 𝑠) = 𝑡 - 6 𝑠 𝑡 + 8 𝑠$ after cancelling $(𝑡 - 6 𝑠)$ .

Filled blanks: $(6 𝑠 - 𝑡) 2 (𝑡 + 8 𝑠) (𝑡 - 6 𝑠)$ .

Exercise Set 4.5

Question

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

(i)

3 𝑝 2 - 3 𝑝 𝑞 - 18 𝑞 2 𝑝 2 + 3 𝑝 𝑞 - 10 𝑞 2

Solution

Numerator: $3 𝑝 2 - 3 𝑝 𝑞 - 18 𝑞 2 = 3 (𝑝 2 - 𝑝 𝑞 - 6 𝑞 2)$ .

Factor $𝑝 2 - 𝑝 𝑞 - 6 𝑞 2$ : numbers with product $- 6$ and sum $- 1$ are $- 3$ and $2$ .

So $𝑝 2 - 𝑝 𝑞 - 6 𝑞 2 = (𝑝 - 3 𝑞) (𝑝 + 2 𝑞)$ . Numerator $= 3 (𝑝 - 3 𝑞) (𝑝 + 2 𝑞)$ .

Denominator: $𝑝 2 + 3 𝑝 𝑞 - 10 𝑞 2$ . Numbers with product $- 10$ and sum $3$ are $5$ and $- 2$ .

So $𝑝 2 + 3 𝑝 𝑞 - 10 𝑞 2 = (𝑝 + 5 𝑞) (𝑝 - 2 𝑞)$ .

There is no common factor. The simplified form is $3 (𝑝 - 3 𝑞) (𝑝 + 2 𝑞) (𝑝 + 5 𝑞) (𝑝 - 2 𝑞)$ .

(ii)

𝑛 3 - 3 𝑛 2 𝑚 + 3 𝑛 𝑚 2 - 𝑚 3 5 𝑚 2 - 10 𝑚 𝑛 + 5 𝑛 2

Solution

Numerator: $𝑛 3 - 3 𝑛 2 𝑚 + 3 𝑛 𝑚 2 - 𝑚 3 = (𝑛 - 𝑚) 3$ .

Denominator: $5 𝑚 2 - 10 𝑚 𝑛 + 5 𝑛 2 = 5 (𝑚 2 - 2 𝑚 𝑛 + 𝑛 2) = 5 (𝑚 - 𝑛) 2 = 5 (𝑛 - 𝑚) 2$ .

Therefore $(𝑛 - 𝑚) 3 5 (𝑛 - 𝑚) 2 = 𝑛 - 𝑚 5$ (cancelling $(𝑛 - 𝑚) 2$ , assuming $𝑛 \neq 𝑚$ ).

(iii)

𝑤 3 - 𝑣 3 + 𝑥 3 + 3 𝑤 𝑣 𝑥 𝑤 2 + 𝑣 2 + 𝑥 2 - 2 𝑤 𝑣 - 2 𝑣 𝑥 + 2 𝑤 𝑥

Solution

Numerator $𝑤 3 - 𝑣 3 + 𝑥 3 + 3 𝑤 𝑣 𝑥$ matches $𝑎 3 + 𝑏 3 + 𝑐 3 - 3 𝑎 𝑏 𝑐$ with $𝑎 = 𝑤$ , $𝑏 = - 𝑣$ , $𝑐 = 𝑥$ :

because $𝑏 3 = - 𝑣 3$ and $- 3 𝑎 𝑏 𝑐 = - 3 (𝑤) (- 𝑣) (𝑥) = 3 𝑤 𝑣 𝑥$ .

Using the identity: $𝑎 + 𝑏 + 𝑐 = 𝑤 - 𝑣 + 𝑥$ , and

$𝑎 2 + 𝑏 2 + 𝑐 2 - 𝑎 𝑏 - 𝑏 𝑐 - 𝑐 𝑎 = 𝑤 2 + 𝑣 2 + 𝑥 2 + 𝑤 𝑣 + 𝑣 𝑥 - 𝑤 𝑥$ .

Denominator: $𝑤 2 + 𝑣 2 + 𝑥 2 - 2 𝑤 𝑣 - 2 𝑣 𝑥 + 2 𝑤 𝑥 = (𝑤 - 𝑣 + 𝑥) 2$ .

So the fraction simplifies to $𝑤 2 + 𝑣 2 + 𝑥 2 + 𝑤 𝑣 + 𝑣 𝑥 - 𝑤 𝑥 𝑤 - 𝑣 + 𝑥$ .

(iv)

4 𝑦 2 - 20 𝑦 𝑧 + 25 𝑧 2 25 𝑧 2 - 4 𝑦 2

Solution

Numerator: $4 𝑦 2 - 20 𝑦 𝑧 + 25 𝑧 2 = (2 𝑦 - 5 𝑧) 2$ .

Denominator: $25 𝑧 2 - 4 𝑦 2 = (5 𝑧) 2 - (2 𝑦) 2 = (5 𝑧 - 2 𝑦) (5 𝑧 + 2 𝑦)$ .

Note $(2 𝑦 - 5 𝑧) 2 = (5 𝑧 - 2 𝑦) 2$ .

So $(5 𝑧 - 2 𝑦) 2 (5 𝑧 - 2 𝑦) (5 𝑧 + 2 𝑦) = 5 𝑧 - 2 𝑦 5 𝑧 + 2 𝑦$ .

(v)

(𝑥 2 + 𝑥 - 6) (𝑥 2 - 7 𝑥 + 12) (𝑥 2 - 6 𝑥 + 8) (𝑥 2 - 9)

Solution

Factor each quadratic:

$𝑥 2 + 𝑥 - 6 = (𝑥 + 3) (𝑥 - 2)$ (product $- 6$ , sum $1$ ),

$𝑥 2 - 7 𝑥 + 12 = (𝑥 - 3) (𝑥 - 4)$ (product $12$ , sum $- 7$ ),

$𝑥 2 - 6 𝑥 + 8 = (𝑥 - 2) (𝑥 - 4)$ (product $8$ , sum $- 6$ ),

$𝑥 2 - 9 = (𝑥 + 3) (𝑥 - 3)$ (difference of squares).

So $(𝑥 + 3) (𝑥 - 2) (𝑥 - 3) (𝑥 - 4) (𝑥 - 2) (𝑥 - 4) (𝑥 + 3) (𝑥 - 3)$ .

Cancel common factors (where defined). The result is $1$ .

(vi)

𝑝 4 - 16 𝑝 2 - 4 𝑝 + 4

Solution

Numerator: $𝑝 4 - 16 = (𝑝 2) 2 - 42 = (𝑝 2 - 4) (𝑝 2 + 4)$ .

Further, $𝑝 2 - 4 = (𝑝 - 2) (𝑝 + 2)$ , so numerator $= (𝑝 - 2) (𝑝 + 2) (𝑝 2 + 4)$ .

Denominator: $𝑝 2 - 4 𝑝 + 4 = (𝑝 - 2) 2$ .

Therefore $(𝑝 - 2) (𝑝 + 2) (𝑝 2 + 4) (𝑝 - 2) 2 = (𝑝 + 2) (𝑝 2 + 4) 𝑝 - 2$ .

Example 17

Saira has arranged a square of side $𝑥$ units, 8 rectangular strips of sides $𝑥$ units and width 1 unit, and 15 squares of side 1 unit to form a bigger rectangle. Find the length and breadth of the rectangle in terms of $𝑥$ .

Solution

Area of the $𝑥 \times 𝑥$ square: $𝑥 2$ .

Area of 8 strips of size $𝑥 \times 1$ : $8 𝑥$ .

Area of 15 unit squares: $15$ .

Total area $= 𝑥 2 + 8 𝑥 + 15$ .

Factor: numbers with product $15$ and sum $8$ are $3$ and $5$ .

$𝑥 2 + 8 𝑥 + 15 = 𝑥 2 + 3 𝑥 + 5 𝑥 + 15 = 𝑥 (𝑥 + 3) + 5 (𝑥 + 3) = (𝑥 + 3) (𝑥 + 5)$ .

So length $= 𝑥 + 5$ units and breadth $= 𝑥 + 3$ units (or vice versa).

Example 18

A rectangular pool is such that its breadth is 4 metres less than its length and its area is 96 sq. metres. Find the length and breadth of the pool.

Solution

Let length $= 𝑥$ metres. Then breadth $= 𝑥 - 4$ metres.

Area: $𝑥 (𝑥 - 4) = 96 \Rightarrow 𝑥 2 - 4 𝑥 - 96 = 0$ .

Factor: numbers with product $- 96$ and sum $- 4$ are $- 12$ and $8$ .

$𝑥 2 - 4 𝑥 - 96 = 𝑥 2 - 12 𝑥 + 8 𝑥 - 96 = 𝑥 (𝑥 - 12) + 8 (𝑥 - 12) = (𝑥 - 12) (𝑥 + 8)$ .

So $(𝑥 - 12) (𝑥 + 8) = 0 \Rightarrow 𝑥 = 12$ or $𝑥 = - 8$ .

Length cannot be negative, so length $= 12$ m and breadth $= 12 - 4 = 8$ m.

End-of-Chapter Exercises

Question

1. Use suitable identities to find the following products:

(i)

(- 3 𝑥 + 4) 2

Solution

Use $(𝑎 + 𝑏) 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ with $𝑎 = - 3 𝑥$ , $𝑏 = 4$ :

$(- 3 𝑥 + 4) 2 = (- 3 𝑥) 2 + 2 (- 3 𝑥) (4) + 42$

$= (- 3 𝑥) 2 = 9 𝑥 2$ , $2 (- 3 𝑥) (4) = - 24 𝑥$ , $42 = 16$ .

Therefore $(- 3 𝑥 + 4) 2 = 9 𝑥 2 - 24 𝑥 + 16$ .

(ii)

(2 𝑠 + 7) (2 𝑠 - 7)

Solution

Use $(𝑎 + 𝑏) (𝑎 - 𝑏) = 𝑎 2 - 𝑏 2$ with $𝑎 = 2 𝑠$ , $𝑏 = 7$ :

$(2 𝑠 + 7) (2 𝑠 - 7) = (2 𝑠) 2 - 72$

$(2 𝑠) 2 = 4 𝑠 2$ , $72 = 49$ .

Therefore $(2 𝑠 + 7) (2 𝑠 - 7) = 4 𝑠 2 - 49$ .

(iii)

(𝑝 2 + 12) (𝑝 2 - 12)

Solution

Use $(𝑎 + 𝑏) (𝑎 - 𝑏) = 𝑎 2 - 𝑏 2$ with $𝑎 = 𝑝 2$ , $𝑏 = 12$ :

$(𝑝 2 + 12) (𝑝 2 - 12) = (𝑝 2) 2 - (12) 2$

$(𝑝 2) 2 = 𝑝 4$ , $(12) 2 = 14$ .

Therefore the product equals $𝑝 4 - 14$ .

(iv)

(2 𝑛 + 7) (2 𝑛 - 7)

Solution

Use $(𝑎 + 𝑏) (𝑎 - 𝑏) = 𝑎 2 - 𝑏 2$ with $𝑎 = 2 𝑛$ , $𝑏 = 7$ :

$(2 𝑛 + 7) (2 𝑛 - 7) = (2 𝑛) 2 - 72$

$(2 𝑛) 2 = 4 𝑛 2$ , $72 = 49$ .

Therefore $(2 𝑛 + 7) (2 𝑛 - 7) = 4 𝑛 2 - 49$ .

(v)

(𝑠 - 2 𝑡) (𝑠 2 + 2 𝑠 𝑡 + 4 𝑡 2)

Solution

This matches $𝑎 3 - 𝑏 3 = (𝑎 - 𝑏) (𝑎 2 + 𝑎 𝑏 + 𝑏 2)$ with $𝑎 = 𝑠$ and $𝑏 = 2 𝑡$ .

Check the second factor: $𝑎 2 + 𝑎 𝑏 + 𝑏 2 = 𝑠 2 + 𝑠 (2 𝑡) + (2 𝑡) 2 = 𝑠 2 + 2 𝑠 𝑡 + 4 𝑡 2$ , which matches.

Therefore $(𝑠 - 2 𝑡) (𝑠 2 + 2 𝑠 𝑡 + 4 𝑡 2) = 𝑠 3 - (2 𝑡) 3 = 𝑠 3 - 8 𝑡 3$ .

(vi)

(1 2 𝑟 - 4 𝑟) 2

Solution

Using $(𝑎 - 𝑏) 2 = 𝑎 2 - 2 𝑎 𝑏 + 𝑏 2$ with $𝑎 = 1 2 𝑟$ , $𝑏 = 4 𝑟$ :

$(1 2 𝑟 - 4 𝑟) 2 = (1 2 𝑟) 2 - 2 (1 2 𝑟) (4 𝑟) + (4 𝑟) 2$

$= 1 4 𝑟 2 - 2 \cdot 1 2 𝑟 \cdot 4 𝑟 + 16 𝑟 2$

$= 1 4 𝑟 2 - 4 + 16 𝑟 2$ .

(vii)

(- 3 𝑚 + 4 𝑘 - 𝑙) 2

Solution

Using $(𝑎 + 𝑏 + 𝑐) 2$ with $𝑎 = - 3 𝑚$ , $𝑏 = 4 𝑘$ , $𝑐 = - 𝑙$ :

$(- 3 𝑚 + 4 𝑘 - 𝑙) 2 = (- 3 𝑚) 2 + (4 𝑘) 2 + (- 𝑙) 2 + 2 (- 3 𝑚) (4 𝑘) + 2 (4 𝑘) (- 𝑙) + 2 (- 𝑙) (- 3 𝑚)$

$= 9 𝑚 2 + 16 𝑘 2 + 𝑙 2 - 24 𝑚 𝑘 - 8 𝑘 𝑙 + 6 𝑚 𝑙$ .

(viii)

(𝑥 - 13 𝑦) 3

Solution

Using $(𝑎 - 𝑏) 3 = 𝑎 3 - 3 𝑎 2 𝑏 + 3 𝑎 𝑏 2 - 𝑏 3$ with $𝑎 = 𝑥$ , $𝑏 = 13 𝑦$ :

$(𝑥 - 13 𝑦) 3 = 𝑥 3 - 3 𝑥 2 (𝑦 3) + 3 𝑥 (𝑦 3) 2 - (𝑦 3) 3$

$= 𝑥 3 - 𝑥 2 𝑦 + 3 𝑥 \cdot 𝑦 2 9 - 𝑦 3 27$

$= 𝑥 3 - 𝑥 2 𝑦 + 13 𝑥 𝑦 2 - 127 𝑦 3$ .

(ix)

(72 𝑘 - 23 𝑚) 3

Solution

Using $(𝑎 - 𝑏) 3$ with $𝑎 = 72 𝑘$ , $𝑏 = 23 𝑚$ :

$(72 𝑘 - 23 𝑚) 3 = (72 𝑘) 3 - 3 (72 𝑘) 2 (23 𝑚) + 3 (72 𝑘) (23 𝑚) 2 - (23 𝑚) 3$ .

First term: $(72 𝑘) 3 = 3438 𝑘 3$ .

Second term: $3 \cdot 494 𝑘 2 \cdot 23 𝑚 = 492 𝑘 2 𝑚$ , so with the minus sign: $- 492 𝑘 2 𝑚$ .

Third term: $3 \cdot 72 𝑘 \cdot 49 𝑚 2 = 143 𝑘 𝑚 2$ .

Fourth term: $(23 𝑚) 3 = 827 𝑚 3$ .

Therefore: $3438 𝑘 3 - 492 𝑘 2 𝑚 + 143 𝑘 𝑚 2 - 827 𝑚 3$ .

Question

2. Find the values using suitable identities:

(i)

17 \times 21

Solution

Note that $17 = 19 - 2$ and $21 = 19 + 2$ .

Use $(𝑎 - 𝑏) (𝑎 + 𝑏) = 𝑎 2 - 𝑏 2$ with $𝑎 = 19$ , $𝑏 = 2$ :

$17 \times 21 = (19 - 2) (19 + 2) = 192 - 22$

$192 = 361$ , $22 = 4$ .

Therefore $17 \times 21 = 361 - 4 = 357$ .

(ii)

104 \times 96

Solution

Note that $104 = 100 + 4$ and $96 = 100 - 4$ .

Use $(𝑎 + 𝑏) (𝑎 - 𝑏) = 𝑎 2 - 𝑏 2$ with $𝑎 = 100$ , $𝑏 = 4$ :

$104 \times 96 = (100 + 4) (100 - 4) = 1002 - 42$

$1002 = 10000$ , $42 = 16$ .

Therefore $104 \times 96 = 10000 - 16 = 9984$ .

(iii)

24 \times 16

Solution

Note that $24 = 20 + 4$ and $16 = 20 - 4$ .

Use $(𝑎 + 𝑏) (𝑎 - 𝑏) = 𝑎 2 - 𝑏 2$ with $𝑎 = 20$ , $𝑏 = 4$ :

$24 \times 16 = (20 + 4) (20 - 4) = 202 - 42$

$202 = 400$ , $42 = 16$ .

Therefore $24 \times 16 = 400 - 16 = 384$ .

(iv)

1473

Solution

Write $147 = 150 - 3$ . Using $(𝑎 - 𝑏) 3 = 𝑎 3 - 3 𝑎 2 𝑏 + 3 𝑎 𝑏 2 - 𝑏 3$ with $𝑎 = 150$ , $𝑏 = 3$ :

$1473 = (150 - 3) 3 = 1503 - 3 (150) 2 (3) + 3 (150) (3) 2 - 33$

$= 3375000 - 3 \cdot 22500 \cdot 3 + 450 \cdot 9 - 27$

$= 3375000 - 202500 + 4050 - 27$ .

Compute step by step: $3375000 - 202500 = 3172500$ ; $3172500 + 4050 = 3176550$ ; $3176550 - 27 = 3176523$ .

(v)

1993

Solution

Write $199 = 200 - 1$ . Using $(𝑎 - 𝑏) 3$ with $𝑎 = 200$ , $𝑏 = 1$ :

$1993 = (200 - 1) 3 = 2003 - 3 (200) 2 (1) + 3 (200) (1) 2 - 1$

$= 8000000 - 3 \cdot 40000 + 600 - 1$

$= 8000000 - 120000 + 600 - 1 = 7880599$ .

(vi)

1273

Solution

Write $127 = 125 + 2$ . Using $(𝑎 + 𝑏) 3 = 𝑎 3 + 3 𝑎 2 𝑏 + 3 𝑎 𝑏 2 + 𝑏 3$ with $𝑎 = 125$ , $𝑏 = 2$ :

$1273 = (125 + 2) 3 = 1253 + 3 (125) 2 (2) + 3 (125) (2) 2 + 8$

$= 1953125 + 3 \cdot 15625 \cdot 2 + 375 \cdot 4 + 8$

$= 1953125 + 93750 + 1500 + 8$ .

Step by step: $1953125 + 93750 = 2046875$ ; $2046875 + 1500 = 2048375$ ; $2048375 + 8 = 2048383$ .

(vii)

(- 107) 3

Solution

Note $(- 107) 3 = - (107) 3$ . Write $107 = 100 + 7$ . Using $(𝑎 + 𝑏) 3$ with $𝑎 = 100$ , $𝑏 = 7$ :

$1073 = 1003 + 3 (100) 2 (7) + 3 (100) (7) 2 + 73$

$= 1000000 + 210000 + 14700 + 343 = 1225043$ .

Therefore $(- 107) 3 = - 1225043$ .

(viii)

(- 299) 3

Solution

Note $(- 299) 3 = - (299) 3$ . Write $299 = 300 - 1$ . Using $(𝑎 - 𝑏) 3$ with $𝑎 = 300$ , $𝑏 = 1$ :

$2993 = 3003 - 3 (300) 2 (1) + 3 (300) (1) - 1$

$= 27000000 - 270000 + 900 - 1 = 26730899$ .

Therefore $(- 299) 3 = - 26730899$ .

Question

3. Factor the following algebraic expressions:

(i)

4 𝑦 2 + 1 + 1 16 𝑦 2

Solution

Compare with $𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ .

Take $𝑎 = 2 𝑦$ and $𝑏 = 1 4 𝑦$ . Then $𝑎 2 = 4 𝑦 2$ and $𝑏 2 = 1 16 𝑦 2$ .

Middle term: $2 𝑎 𝑏 = 2 (2 𝑦) (1 4 𝑦) = 1$ , which matches.

Therefore $4 𝑦 2 + 1 + 1 16 𝑦 2 = (2 𝑦 + 1 4 𝑦) 2$ .

(ii)

9 𝑚 2 - 1 25 𝑛 2

Solution

Write as a difference of squares: $9 𝑚 2 - 1 25 𝑛 2 = (3 𝑚) 2 - (1 5 𝑛) 2$ .

Using $𝑎 2 - 𝑏 2 = (𝑎 - 𝑏) (𝑎 + 𝑏)$ :

$9 𝑚 2 - 1 25 𝑛 2 = (3 𝑚 - 1 5 𝑛) (3 𝑚 + 1 5 𝑛)$ .

(iii)

27 𝑏 3 - 1 64 𝑏 3

Solution

Write as a difference of cubes: $27 𝑏 3 - 1 64 𝑏 3 = (3 𝑏) 3 - (1 4 𝑏) 3$ .

Using $𝑥 3 - 𝑦 3 = (𝑥 - 𝑦) (𝑥 2 + 𝑥 𝑦 + 𝑦 2)$ with $𝑥 = 3 𝑏$ , $𝑦 = 1 4 𝑏$ :

$𝑥 2 + 𝑥 𝑦 + 𝑦 2 = 9 𝑏 2 + (3 𝑏) (1 4 𝑏) + 1 16 𝑏 2 = 9 𝑏 2 + 34 + 1 16 𝑏 2$ .

Therefore $27 𝑏 3 - 1 64 𝑏 3 = (3 𝑏 - 1 4 𝑏) (9 𝑏 2 + 34 + 1 16 𝑏 2)$ .

(iv)

𝑥 2 + 5 𝑥 6 + 16

Solution

Compare $𝑥 2 + 56 𝑥 + 16$ with $𝑥 2 + (𝑎 + 𝑏) 𝑥 + 𝑎 𝑏$ .

We need $𝑎 + 𝑏 = 56$ and $𝑎 𝑏 = 16$ .

The pair $𝑎 = 12$ , $𝑏 = 13$ works: $12 + 13 = 56$ and $12 \cdot 13 = 16$ .

Therefore $𝑥 2 + 56 𝑥 + 16 = (𝑥 + 12) (𝑥 + 13)$ .

(v)

27 𝑢 3 - 1125 - 27 𝑢 2 5 + 9 𝑢 25

Solution

Rearrange in descending powers of $𝑢$ :

$27 𝑢 3 - 1125 - 27 𝑢 2 5 + 9 𝑢 25 = 27 𝑢 3 - 27 𝑢 2 5 + 9 𝑢 25 - 1125$ .

Compare with $(𝑎 - 𝑏) 3 = 𝑎 3 - 3 𝑎 2 𝑏 + 3 𝑎 𝑏 2 - 𝑏 3$ .

Take $𝑎 = 3 𝑢$ , $𝑏 = 15$ . Then $𝑎 3 = 27 𝑢 3$ , $- 3 𝑎 2 𝑏 = - 27 𝑢 2 5$ , $3 𝑎 𝑏 2 = 9 𝑢 25$ , $- 𝑏 3 = - 1125$ .

All match. Therefore the expression is $(3 𝑢 - 15) 3$ .

(vi)

64 𝑦 3 + 1125 𝑧 3

Solution

Write as a sum of cubes: $64 𝑦 3 + 1125 𝑧 3 = (4 𝑦) 3 + (𝑧 5) 3$ .

Using $𝑎 3 + 𝑏 3 = (𝑎 + 𝑏) (𝑎 2 - 𝑎 𝑏 + 𝑏 2)$ with $𝑎 = 4 𝑦$ , $𝑏 = 𝑧 5$ :

$𝑎 2 - 𝑎 𝑏 + 𝑏 2 = 16 𝑦 2 - (4 𝑦) (𝑧 5) + 𝑧 2 25 = 16 𝑦 2 - 4 𝑦 𝑧 5 + 𝑧 2 25$ .

Therefore $64 𝑦 3 + 1125 𝑧 3 = (4 𝑦 + 𝑧 5) (16 𝑦 2 - 4 𝑦 𝑧 5 + 𝑧 2 25)$ .

(vii)

𝑝 3 + 27 𝑞 3 + 𝑟 3 - 9 𝑝 𝑞 𝑟

Solution

Use $𝑎 3 + 𝑏 3 + 𝑐 3 - 3 𝑎 𝑏 𝑐 = (𝑎 + 𝑏 + 𝑐) (𝑎 2 + 𝑏 2 + 𝑐 2 - 𝑎 𝑏 - 𝑏 𝑐 - 𝑐 𝑎)$ with $𝑎 = 𝑝$ , $𝑏 = 3 𝑞$ , $𝑐 = 𝑟$ .

Then $- 3 𝑎 𝑏 𝑐 = - 3 (𝑝) (3 𝑞) (𝑟) = - 9 𝑝 𝑞 𝑟$ , which matches.

Also $𝑎 + 𝑏 + 𝑐 = 𝑝 + 3 𝑞 + 𝑟$ ,

and $𝑎 2 + 𝑏 2 + 𝑐 2 - 𝑎 𝑏 - 𝑏 𝑐 - 𝑐 𝑎 = 𝑝 2 + 9 𝑞 2 + 𝑟 2 - 3 𝑝 𝑞 - 3 𝑞 𝑟 - 𝑝 𝑟$ .

Therefore $𝑝 3 + 27 𝑞 3 + 𝑟 3 - 9 𝑝 𝑞 𝑟 = (𝑝 + 3 𝑞 + 𝑟) (𝑝 2 + 9 𝑞 2 + 𝑟 2 - 3 𝑝 𝑞 - 3 𝑞 𝑟 - 𝑝 𝑟)$ .

(viii)

9 𝑚 2 - 12 𝑚 + 4

Solution

Compare $9 𝑚 2 - 12 𝑚 + 4$ with $𝑎 2 - 2 𝑎 𝑏 + 𝑏 2$ .

$9 𝑚 2 = (3 𝑚) 2$ so $𝑎 = 3 𝑚$ ; $4 = 22$ so $𝑏 = 2$ .

Middle term: $- 2 𝑎 𝑏 = - 2 (3 𝑚) (2) = - 12 𝑚$ , which matches.

Therefore $9 𝑚 2 - 12 𝑚 + 4 = (3 𝑚 - 2) 2$ .

(ix)

9 𝑥 3 - 83 𝑦 3 + 𝑧 3 3 + 6 𝑥 𝑦 𝑧

Solution

Factor out $13$ :

$9 𝑥 3 - 83 𝑦 3 + 13 𝑧 3 + 6 𝑥 𝑦 𝑧 = 13 (27 𝑥 3 - 8 𝑦 3 + 𝑧 3 + 18 𝑥 𝑦 𝑧)$ .

Inside, compare with $𝑎 3 + 𝑏 3 + 𝑐 3 - 3 𝑎 𝑏 𝑐$ using $𝑎 = 3 𝑥$ , $𝑏 = - 2 𝑦$ , $𝑐 = 𝑧$ :

$𝑎 3 + 𝑏 3 + 𝑐 3 = 27 𝑥 3 - 8 𝑦 3 + 𝑧 3$ , and $- 3 𝑎 𝑏 𝑐 = - 3 (3 𝑥) (- 2 𝑦) (𝑧) = 18 𝑥 𝑦 𝑧$ .

Now $𝑎 + 𝑏 + 𝑐 = 3 𝑥 - 2 𝑦 + 𝑧$ , and

$𝑎 2 + 𝑏 2 + 𝑐 2 - 𝑎 𝑏 - 𝑏 𝑐 - 𝑐 𝑎 = 9 𝑥 2 + 4 𝑦 2 + 𝑧 2 + 6 𝑥 𝑦 + 2 𝑦 𝑧 - 3 𝑥 𝑧$ .

Therefore the expression equals $13 (3 𝑥 - 2 𝑦 + 𝑧) (9 𝑥 2 + 4 𝑦 2 + 𝑧 2 + 6 𝑥 𝑦 + 2 𝑦 𝑧 - 3 𝑥 𝑧)$ .

(x)

4 𝑥 2 + 9 𝑦 2 + 36 𝑧 2 + 12 𝑥 𝑧 + 36 𝑦 𝑧 + 24 𝑥 𝑦

Solution

The square terms are $4 𝑥 2 = (2 𝑥) 2$ , $9 𝑦 2 = (3 𝑦) 2$ , $36 𝑧 2 = (6 𝑧) 2$ . The intended perfect square is $(2 𝑥 + 3 𝑦 + 6 𝑧) 2$ , whose expansion is $4 𝑥 2 + 9 𝑦 2 + 36 𝑧 2 + 12 𝑥 𝑦 + 36 𝑦 𝑧 + 24 𝑥 𝑧$ .

Note: As printed in the textbook, the cross terms are

12 𝑥 𝑧

and

24 𝑥 𝑦

(i.e. the coefficients of

𝑥 𝑦

and

𝑥 𝑧

are swapped relative to

(2 𝑥 + 3 𝑦 + 6 𝑧) 2

). Treating that as a typographical swap of those two coefficients, the factorisation is

(2 𝑥 + 3 𝑦 + 6 𝑧) 2

With the corrected/intended coefficients: $(2 𝑥 + 3 𝑦 + 6 𝑧) 2$ .

(xi)

27 𝑢 3 - 1216 - 9 𝑢 2 2 + 𝑢 4

Solution

Compare with $(𝑎 - 𝑏) 3 = 𝑎 3 - 3 𝑎 2 𝑏 + 3 𝑎 𝑏 2 - 𝑏 3$ .

Take $𝑎 = 3 𝑢$ , $𝑏 = 16$ . Expand:

$(3 𝑢 - 16) 3 = 27 𝑢 3 - 3 (9 𝑢 2) (16) + 3 (3 𝑢) (136) - 1216$

$= 27 𝑢 3 - 92 𝑢 2 + 14 𝑢 - 1216$ .

This matches the given expression (after rearranging). Factored form: $(3 𝑢 - 16) 3$ .

Question

4. Simplify the following:

(i)

4 𝑥 2 + 4 𝑥 + 1 4 𝑥 2 - 1

Solution

Numerator: $4 𝑥 2 + 4 𝑥 + 1 = (2 𝑥 + 1) 2$ .

Denominator: $4 𝑥 2 - 1 = (2 𝑥) 2 - 12 = (2 𝑥 - 1) (2 𝑥 + 1)$ .

So $4 𝑥 2 + 4 𝑥 + 1 4 𝑥 2 - 1 = (2 𝑥 + 1) 2 (2 𝑥 - 1) (2 𝑥 + 1)$ .

Cancel the common factor $(2 𝑥 + 1)$ (assuming $2 𝑥 + 1 \neq 0$ ): $2 𝑥 + 1 2 𝑥 - 1$ .

(ii)

9 (3 𝑎 3 - 24 𝑏 3) 9 𝑎 2 - 36 𝑏 2

Solution

Numerator: $9 (3 𝑎 3 - 24 𝑏 3) = 27 (𝑎 3 - 8 𝑏 3) = 27 (𝑎 3 - (2 𝑏) 3)$ .

Using $𝑎 3 - 𝑏 3 = (𝑎 - 𝑏) (𝑎 2 + 𝑎 𝑏 + 𝑏 2)$ : $𝑎 3 - 8 𝑏 3 = (𝑎 - 2 𝑏) (𝑎 2 + 2 𝑎 𝑏 + 4 𝑏 2)$ .

So numerator $= 27 (𝑎 - 2 𝑏) (𝑎 2 + 2 𝑎 𝑏 + 4 𝑏 2)$ .

Denominator: $9 𝑎 2 - 36 𝑏 2 = 9 (𝑎 2 - 4 𝑏 2) = 9 (𝑎 - 2 𝑏) (𝑎 + 2 𝑏)$ .

Therefore $27 (𝑎 - 2 𝑏) (𝑎 2 + 2 𝑎 𝑏 + 4 𝑏 2) 9 (𝑎 - 2 𝑏) (𝑎 + 2 𝑏) = 3 (𝑎 2 + 2 𝑎 𝑏 + 4 𝑏 2) 𝑎 + 2 𝑏$ .

(iii)

𝑠 3 + 125 𝑡 3 𝑠 2 - 2 𝑠 𝑡 - 35 𝑡 2

Solution

Numerator is a sum of cubes: $𝑠 3 + 125 𝑡 3 = 𝑠 3 + (5 𝑡) 3 = (𝑠 + 5 𝑡) (𝑠 2 - 5 𝑠 𝑡 + 25 𝑡 2)$ .

Denominator: $𝑠 2 - 2 𝑠 𝑡 - 35 𝑡 2$ . Numbers with product $- 35$ and sum $- 2$ are $- 7$ and $5$ .

So $𝑠 2 - 2 𝑠 𝑡 - 35 𝑡 2 = 𝑠 2 - 7 𝑠 𝑡 + 5 𝑠 𝑡 - 35 𝑡 2 = 𝑠 (𝑠 - 7 𝑡) + 5 𝑡 (𝑠 - 7 𝑡) = (𝑠 - 7 𝑡) (𝑠 + 5 𝑡)$ .

Therefore $(𝑠 + 5 𝑡) (𝑠 2 - 5 𝑠 𝑡 + 25 𝑡 2) (𝑠 - 7 𝑡) (𝑠 + 5 𝑡) = 𝑠 2 - 5 𝑠 𝑡 + 25 𝑡 2 𝑠 - 7 𝑡$ .

Question

5. Find possible expressions for the length and breadth of each of the following rectangles whose areas are given:

(i)

25 𝑎 2 - 30 𝑎 𝑏 + 9 𝑏 2

Solution

Compare $25 𝑎 2 - 30 𝑎 𝑏 + 9 𝑏 2$ with $𝑎 2 - 2 𝑎 𝑏 + 𝑏 2$ :

$25 𝑎 2 = (5 𝑎) 2$ , $9 𝑏 2 = (3 𝑏) 2$ , and $- 2 (5 𝑎) (3 𝑏) = - 30 𝑎 𝑏$ , which matches.

So $25 𝑎 2 - 30 𝑎 𝑏 + 9 𝑏 2 = (5 𝑎 - 3 𝑏) 2 = (5 𝑎 - 3 𝑏) (5 𝑎 - 3 𝑏)$ .

Possible length and breadth: both equal to $(5 𝑎 - 3 𝑏)$ .

(ii)

36 𝑠 2 - 49 𝑡 2

Solution

Write as a difference of squares: $36 𝑠 2 - 49 𝑡 2 = (6 𝑠) 2 - (7 𝑡) 2$ .

Using $𝑎 2 - 𝑏 2 = (𝑎 - 𝑏) (𝑎 + 𝑏)$ : $36 𝑠 2 - 49 𝑡 2 = (6 𝑠 - 7 𝑡) (6 𝑠 + 7 𝑡)$ .

Possible length $= 6 𝑠 + 7 𝑡$ , breadth $= 6 𝑠 - 7 𝑡$ (or vice versa).

Question

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

(i)

6 𝑎 2 - 24 𝑏 2

Solution

First factor out $6$ : $6 𝑎 2 - 24 𝑏 2 = 6 (𝑎 2 - 4 𝑏 2)$ .

Then $𝑎 2 - 4 𝑏 2 = 𝑎 2 - (2 𝑏) 2 = (𝑎 - 2 𝑏) (𝑎 + 2 𝑏)$ .

So $6 𝑎 2 - 24 𝑏 2 = 6 (𝑎 - 2 𝑏) (𝑎 + 2 𝑏)$ .

Possible dimensions of the cuboid: $6$ , $(𝑎 - 2 𝑏)$ , and $(𝑎 + 2 𝑏)$ .

(ii)

3 𝑝 𝑠 2 - 15 𝑝 𝑠 + 12 𝑝

Solution

Factor out the common factor $3 𝑝$ : $3 𝑝 𝑠 2 - 15 𝑝 𝑠 + 12 𝑝 = 3 𝑝 (𝑠 2 - 5 𝑠 + 4)$ .

Factor the quadratic: numbers with product $4$ and sum $- 5$ are $- 1$ and $- 4$ .

$𝑠 2 - 5 𝑠 + 4 = 𝑠 2 - 4 𝑠 - 𝑠 + 4 = 𝑠 (𝑠 - 4) - 1 (𝑠 - 4) = (𝑠 - 1) (𝑠 - 4)$ .

So $3 𝑝 𝑠 2 - 15 𝑝 𝑠 + 12 𝑝 = 3 𝑝 (𝑠 - 1) (𝑠 - 4)$ .

Possible dimensions: $3 𝑝$ , $(𝑠 - 1)$ , and $(𝑠 - 4)$ .

Question

7. The village playground is shaped as a square of side 40 metres. A path of width

𝑠

metres is created around the playground for people to walk. Find an expression for the area of the path in terms of

𝑠

Solution

The path of width $𝑠$ surrounds the playground on all sides, so the outer square has side $40 + 2 𝑠$ .

Area of path $=$ outer area $-$ playground area $= (40 + 2 𝑠) 2 - 402$ .

Using $𝑎 2 - 𝑏 2 = (𝑎 - 𝑏) (𝑎 + 𝑏)$ with $𝑎 = 40 + 2 𝑠$ , $𝑏 = 40$ :

$(40 + 2 𝑠 - 40) (40 + 2 𝑠 + 40) = (2 𝑠) (80 + 2 𝑠) = 2 𝑠 \cdot 2 (40 + 𝑠) = 4 𝑠 (40 + 𝑠)$

$= 160 𝑠 + 4 𝑠 2$ .

So the area of the path is $4 𝑠 2 + 160 𝑠$ square metres.

Question

8. If a number plus its reciprocal equals

103

, find the number.

Solution

Let the number be $𝑥$ . $𝑥 + 1 𝑥 = 103$ .

Multiply by $3 𝑥$ : $3 𝑥 2 + 3 = 10 𝑥 ⟹ 3 𝑥 2 - 10 𝑥 + 3 = 0$ .

Factor: $3 𝑥 2 - 9 𝑥 - 𝑥 + 3 = 3 𝑥 (𝑥 - 3) - 1 (𝑥 - 3) = (3 𝑥 - 1) (𝑥 - 3) = 0$ .

The number is $3$ or $13$ .

Question

9. A rectangular pool has area

2 𝑥 2 + 7 𝑥 + 3

square hastas. If its width is

2 𝑥 + 1

hastas, find its length.

Solution

Length $=$ Area $\div$ Width $= 2 𝑥 2 + 7 𝑥 + 3 2 𝑥 + 1$ .

Factor the numerator. For $2 𝑥 2 + 7 𝑥 + 3$ : $𝑎 𝑐 = 6$ , numbers with product $6$ and sum $7$ are $6$ and $1$ .

Split: $2 𝑥 2 + 7 𝑥 + 3 = 2 𝑥 2 + 6 𝑥 + 𝑥 + 3 = 2 𝑥 (𝑥 + 3) + 1 (𝑥 + 3) = (2 𝑥 + 1) (𝑥 + 3)$ .

So $(2 𝑥 + 1) (𝑥 + 3) 2 𝑥 + 1 = 𝑥 + 3$ (assuming $2 𝑥 + 1 \neq 0$ ).

The length is $(𝑥 + 3)$ hastas.

Question

*10. If both

𝑥 - 2

and

𝑥 - 12

are factors of

𝑝 𝑥 2 + 5 𝑥 + 𝑟

, show that

𝑝 = 𝑟

Solution

By Factor Theorem, if $𝑥 - 2$ is a factor, substituting $𝑥 = 2$ gives 0: $𝑝 (4) + 5 (2) + 𝑟 = 0 ⟹ 4 𝑝 + 10 + 𝑟 = 0$ .

If $𝑥 - 12$ is a factor, substituting $𝑥 = 12$ gives 0: $𝑝 (14) + 5 (12) + 𝑟 = 0 ⟹ 𝑝 4 + 52 + 𝑟 = 0$ . Multiply by 4: $𝑝 + 10 + 4 𝑟 = 0$ .

From the two equations: $4 𝑝 + 𝑟 = - 10$ (1) and $𝑝 + 4 𝑟 = - 10$ (2).

Since both right-hand sides equal $- 10$ , we have $4 𝑝 + 𝑟 = 𝑝 + 4 𝑟$ .

Bring like terms together: $4 𝑝 - 𝑝 = 4 𝑟 - 𝑟$

$3 𝑝 = 3 𝑟$

Divide both sides by $3$ : $𝑝 = 𝑟$ .

Question

*11. If

𝑎 + 𝑏 + 𝑐 = 5

and

𝑎 𝑏 + 𝑏 𝑐 + 𝑐 𝑎 = 10

, then prove that

𝑎 3 + 𝑏 3 + 𝑐 3 - 3 𝑎 𝑏 𝑐 = - 25

Solution

We need $𝑎 2 + 𝑏 2 + 𝑐 2$ .

$(𝑎 + 𝑏 + 𝑐) 2 = 𝑎 2 + 𝑏 2 + 𝑐 2 + 2 (𝑎 𝑏 + 𝑏 𝑐 + 𝑐 𝑎)$ .

$52 = (𝑎 2 + 𝑏 2 + 𝑐 2) + 2 (10) ⟹ 25 = 𝑎 2 + 𝑏 2 + 𝑐 2 + 20 ⟹ 𝑎 2 + 𝑏 2 + 𝑐 2 = 5$ .

Now use the identity: $𝑎 3 + 𝑏 3 + 𝑐 3 - 3 𝑎 𝑏 𝑐 = (𝑎 + 𝑏 + 𝑐) (𝑎 2 + 𝑏 2 + 𝑐 2 - (𝑎 𝑏 + 𝑏 𝑐 + 𝑐 𝑎))$ .

Substitute values: $(5) (5 - 10) = 5 (- 5) = - 25$ . The statement is proved.

Question

*12. By factoring the expression, check that

𝑛 3 - 𝑛

is always divisible by 6 for all natural numbers

𝑛

. Give reasons.

Solution

Factor $𝑛 3 - 𝑛 = 𝑛 (𝑛 2 - 1) = 𝑛 (𝑛 - 1) (𝑛 + 1) = (𝑛 - 1) 𝑛 (𝑛 + 1)$ .

This represents the product of three consecutive integers. Among any three consecutive integers, at least one must be even (divisible by 2), and exactly one must be a multiple of 3. Therefore, their product is always divisible by $2 \times 3 = 6$ . The statement is correct.

Question

*13. Find the value of
(i)

𝑥 3 + 𝑦 3 - 12 𝑥 𝑦 + 64

, when

𝑥 + 𝑦 = - 4

(ii)

𝑥 3 - 8 𝑦 3 - 36 𝑥 𝑦 - 216

, when

𝑥 = 2 𝑦 + 6

Solution

(i) Given $𝑥 + 𝑦 = - 4$ . Consider the expression $𝑥 3 + 𝑦 3 - 12 𝑥 𝑦 + 64$ .

Note that $64 = 43$ and $12 𝑥 𝑦 = 3 \cdot 𝑥 \cdot 𝑦 \cdot 4$ .

So the expression is $𝑥 3 + 𝑦 3 + 43 - 3 𝑥 𝑦 𝑧$ with $𝑧 = 4$ , i.e. $𝑎 3 + 𝑏 3 + 𝑐 3 - 3 𝑎 𝑏 𝑐$ for $𝑎 = 𝑥$ , $𝑏 = 𝑦$ , $𝑐 = 4$ .

Since $𝑥 + 𝑦 = - 4$ , we have $𝑥 + 𝑦 + 4 = 0$ , i.e. $𝑎 + 𝑏 + 𝑐 = 0$ .

When $𝑎 + 𝑏 + 𝑐 = 0$ , the identity gives $𝑎 3 + 𝑏 3 + 𝑐 3 - 3 𝑎 𝑏 𝑐 = 0$ .

Hence the value of the expression is $0$ .

(ii) Given $𝑥 = 2 𝑦 + 6$ , so $𝑥 - 2 𝑦 - 6 = 0$ , i.e. $𝑥 + (- 2 𝑦) + (- 6) = 0$ .

Consider $𝑥 3 - 8 𝑦 3 - 36 𝑥 𝑦 - 216$ .

Note $- 8 𝑦 3 = (- 2 𝑦) 3$ and $- 216 = (- 6) 3$ .

Also $- 3 (𝑥) (- 2 𝑦) (- 6) = - 3 \cdot 𝑥 \cdot 12 𝑦 = - 36 𝑥 𝑦$ .

So the expression is $𝑎 3 + 𝑏 3 + 𝑐 3 - 3 𝑎 𝑏 𝑐$ with $𝑎 = 𝑥$ , $𝑏 = - 2 𝑦$ , $𝑐 = - 6$ .

Since $𝑎 + 𝑏 + 𝑐 = 0$ , we get $𝑎 3 + 𝑏 3 + 𝑐 3 - 3 𝑎 𝑏 𝑐 = 0$ .

Hence the value is $0$ .

End of Chapter 4 — Exploring Algebraic Identities