Class 9 Algebraic Identities
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Class 9 chapter - 2 POLYNOMIALS

Algebraic Identities

Topic Covered

♦ Algebraic Identities

Algebraic Identities

From your earlier classes, you may recall that an algebraic identity is an algebraic equation that is true for all values of the variables occurring in it. You have studied the following algebraic identities in earlier classes:

Identity I : ` color {red} { (x + y)^2 = x^2 + 2xy + y^2 } `

Identity II : ` color {blue} { (x – y)^2 = x^2 – 2xy + y^2 } `

Identity III : ` color {orange} { x^2 – y^2 = (x + y) (x – y) } `

Identity IV : ` color { green } { (x + a) (x + b) = x^2 + (a + b)x + ab } `

You must have also used some of these algebraic identities to factorise the algebraic expressions. You can also see their utility in computations.
Q 3240356213

Find the following products using appropriate identities:

(i) `(x + 3) (x + 3)`

(ii)` (x – 3) (x + 5)`

Class 9 Chapter 2 Example 16
Solution:

(i) Here we can use Identity `I : (x + y)^2 = x^2 + 2xy + y^2`. Putting `y = 3` in it,
we get

`(x + 3) (x + 3) = (x + 3)^2 = x^2 + 2(x)(3) + (3)^2`

`= x^2 + 6x + 9`

(ii) Using Identity IV above, i.e., `(x + a) (x + b) = x^2 + (a + b)x + ab`, we have

`(x – 3) (x + 5) = x^2 + (–3 + 5)x + (–3)(5)`

`= x^2 + 2x – 15`
Q 3250356214

Evaluate `105 × 106` without multiplying directly.
Class 9 Chapter 2 Example 17
Solution:

`105 × 106 = (100 + 5) × (100 + 6)`
`= (100)^2 + (5 + 6) (100) + (5 × 6)`, using Identity IV

`= 10000 + 1100 + 30`

`= 11130`

You have seen some uses of the identities listed above in finding the product of some
given expressions. These identities are useful in factorisation of algebraic expressions
also, as you can see in the following examples.
Q 3270356216

Factorise:

(i) `49a^2 + 70ab + 25b^2`

(ii) `25/4 x^2 - y^2/9`


Class 9 Chapter 2 Example 18
Solution:

(i) Here you can see that

`49a^2 = (7a)^2, 25b^2 = (5b)^2, 70ab = 2(7a) (5b)`

Comparing the given expression with `x^2 + 2xy + y^2`, we observe that `x = 7a` and `y = 5b`.

Using Identity I, we get

`49a^2 + 70ab + 25b^2 = (7a + 5b)^2 = (7a + 5b) (7a + 5b)`

(ii) We have `25/4 x^2 - y^2/9 = (5/2 x)^2 - (y/3)^2`

Now comparing it with Identity III, we get

`25/4 x^2 - y^2/9 = (5/2 x)^2 - ( y/3)^2`

`= (5/2 x + y/3) ( 5/2 x - y/3)`

So far, all our identities involved products of binomials. Let us now extend the Identity
I to a trinomial `x + y + z`. We shall compute `(x + y + z)^2` by using Identity `I`.

Let `x + y = t`. Then,

`(x + y + z)^2 = (t + z)^2`

`= t^2 + 2tz + t^2` (Using Identity I)

`= (x + y)^2 + 2(x + y)z + z^2` (Substituting the value of t)

`= x^2 + 2xy + y^2 + 2xz + 2yz + z^2` (Using Identity I)

`= x^2 + y^2 + z^2 + 2xy + 2yz + 2zx` (Rearranging the terms)

So, we get the following identity:

Identity V :

Identity V : `color { red} { (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx } `

`text ( Remark : )` We call the right hand side expression the expanded form of the left hand side expression. Note that the expansion of `(x + y + z)^2` consists of three square terms and three product terms.
Q 3200356218

Write `(3a + 4b + 5c)^2` in expanded form.
Class 9 Chapter 2 Example 19
Solution:

Comparing the given expression with `(x + y + z)^2`, we find that

`x = 3a, y = 4b` and `z = 5c`.

Therefore, using Identity `V`, we have

`(3a + 4b + 5c)^2 = (3a)^2 + (4b)^2 + (5c)^2 + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)`

`= 9a^2 + 16b^2 + 25c^2 + 24ab + 40bc + 30ac`
Q 3210356219

Expand `(4a – 2b – 3c)^2`.
Class 9 Chapter 2 Example 20
Solution:

Using Identity V, we have

`(4a – 2b – 3c)^2 = [4a + (–2b) + (–3c)]^2`

`= (4a)^2 + (–2b)^2 + (–3c)^2 + 2(4a)(–2b) + 2(–2b)(–3c) + 2(–3c)(4a)`

`= 16a^2 + 4b^2 + 9c^2 – 16ab + 12bc – 24ac`
Q 3220478311

Factorise `4x^2 + y^2 + z^2 – 4xy – 2yz + 4xz`.
Class 9 Chapter 2 Example 21
Solution:

We have `4x^2 + y^2 + z^2 – 4xy – 2yz + 4xz = (2x)^2 + (–y)^2 + (z)^2 + 2(2x)(–y)`
`+ 2(–y)(z) + 2(2x)(z)`

`= [2x + (–y) + z]^2` (Using Identity V)

`= (2x – y + z)^2 = (2x – y + z)(2x – y + z)`

So far, we have dealt with identities involving second degree terms. Now let us
extend Identity I to compute `(x + y)^3`. We have:

`(x + y)^3 = (x + y) (x + y)^2`

`= (x + y)(x^2 + 2xy + y^2)`

`= x(x^2 + 2xy + y^2) + y(x^2 + 2xy + y^2)`

`= x^3 + 2x^2 y + xy^2 + x^2 y + 2xy^2 + y^3`

`= x^3 + 3x^2 y + 3xy^2 + y^3`

`= x^3 + y^3 + 3xy(x + y)`


So, we get the following identity:

Identity VI :` (x + y)^3 = x^3 + y^3 + 3xy (x + y)`

Also, by replacing `y` by `–y` in the Identity VI, we get
Identity VII :` (x – y)^3 = x^3 – y^3 – 3xy(x – y)`

`= x^3 – 3x^2y + 3xy^2 – y^3`

Identity

Identity VI `: (x + y)^3 = x^3 + y^3 + 3xy *(x + y)`

Also, by replacing `y` by `–y` in the Identity VI, we get

Identity VII `: (x - y)^3 = x^3 - y^3 - 3xy *(x - y)`
Q 3230478312

Write the following cubes in the expanded form:

(i)` (3a + 4b)^3`

(ii) `(5p – 3q)^3`
Class 9 Chapter 2 Example 22
Solution:

(i) Comparing the given expression with `(x + y)^3`, we find that

`x = 3a` and `y = 4b`.

So, using Identity VI, we have:

`(3a + 4b)^3 = (3a)^3 + (4b)^3 + 3(3a)(4b)(3a + 4b)`

`= 27a^3 + 64b^3 + 108a^2 b + 144ab^2`

(ii) Comparing the given expression with `(x – y)^3`, we find that

`x = 5p, y = 3q`.

So, using Identity VII, we have:

`(5p – 3q)^3 = (5p)^3 – (3q)^3 – 3(5p)(3q)(5p – 3q)`

`= 125p^3 – 27q^3 – 225p^2 q + 135pq^2`
Q 3240478313

Evaluate each of the following using suitable identities:

(i) `(104)^3`

(ii) `(999)^3`
Class 9 Chapter 2 Example 23
Solution:

(i) We have

`(104)^3 = (100 + 4)^3`

`= (100)^3 + (4)^3 + 3(100)(4)(100 + 4)`

(Using Identity VI)

`= 1000000 + 64 + 124800`

`= 1124864`

(ii) We have

`(999)^3 = (1000 – 1)^3`

`= (1000)^3 – (1)^3 – 3(1000)(1)(1000 – 1)`

(Using Identity VII)

`= 1000000000 – 1 – 2997000`

`= 997002999`
Q 3250478314

Factorise `8x^3 + 27y^3 + 36x^2 y + 54xy^2`
Class 9 Chapter 2 Example 24
Solution:

The given expression can be written as

`(2x)^3 + (3y)^3 + 3(4x^2) (3y) + 3(2x)(9y^2)`

`= (2x)^3 + (3y)^3 + 3(2x)^2 (3y) + 3(2x)(3y)^2`

`= (2x + 3y)^3` (Using Identity VI)

`= (2x + 3y)(2x + 3y)(2x + 3y)`

Now consider `(x + y + z)(x^2 + y^2 + z^2 – xy – yz – zx)`

On expanding, we get the product as

`x(x^2 + y^2 + z^2 – xy – yz – zx) + y(x^2 + y^2 + z^2 – xy – yz – zx)`

`+ z(x^2 + y^2 + z^2 – xy – yz – zx) = x^3 + xy^2 + xz^2 – x^2 y – xyz – zx^2 + x^2 y`

`+ y^3 + yz^2 – xy^2 – y^2 z – xyz + x^2 z + y^2 z + z^3 – xyz – yz^2 – xz^2`

`= x^3 + y^3 + z^3 – 3xyz` (On simplification)

Identity

Identity VIII `: x^3 + y^3 + z^3 – 3xyz = (x + y + z)(x^2 + y^2 + z^2 – xy – yz – zx)`
Q 3260478315

Factorise :` 8x^3 + y^3 + 27z^3 – 18xyz`
Class 9 Chapter 2 Example 25
Solution:

Here, we have

`8x^3 + y^3 + 27z^3 – 18xyz`

`= (2x)^3 + y^3 + (3z)^3 – 3(2x)(y)(3z)`

`= (2x + y + 3z)[(2x)^2 + y^2 + (3z)^2 – (2x)(y) – (y)(3z) – (2x)(3z)]`

`= (2x + y + 3z) (4x^2 + y^2 + 9z^2 – 2xy – 3yz – 6xz)`
 
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