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Algebraic Identities - Polynomials, Class 9, Mathematics PDF Download

"Polynomials" is a chapter in Class 9 Mathematics that focuses on the study of algebraic expressions and polynomials, with a particular emphasis on "Algebraic Identities." Through this chapter, students gain an understanding of a range of algebraic identities and the properties of polynomials. 

Algebraic Identities - Polynomials, Class 9, Mathematics

Algebraic Identities

Let us recall the following algebraic identities is earlier.

  1. (a + b)2 = a2 + 2ab + b2 
  2. (a – b)2 = a2 – 2ab + b2
  3. (a + b) (a – b) = a2 – b2
  4. (x + a) (x + b) = x2 + (a + b) x + ab

Ex. Expand each of the following :
(i) (2x + 3y)2
(ii) (4x – 5y)2
(iii) (x + 5) (x + 6)
(iv) (x – 3) (x – 5)
Sol. (i) (2x + 3y)2 = (2x)2 + 2.2x.3y + (3y)2
[∵ (a + b)2 = a2 + 2ab + b2]
= 4x2 + 12xy + 9y2

(ii) (4x – 5y)2 = (4x)2 + 2.4x.5y + (5y)2
[∵ (a – b)2 = a2 – 2ab + b2]
= 16x2 – 40xy + 25y2

(iii) (x + 5) (x + 6) = x2 + (5 + 6)x + 5.6
[∵ (x + a) (x + b) = x2 + (a + b) x + ab]
= x2 + 11x + 30

(iv) (x – 3) (x – 5) = {x + (–3)} {x + (–5)}
[∵ (x + a) (x + b) = x2 + (a + b) x + ab]
= x2 + {(–3) + (–5)}x + (–3).(–5) = x2 + (–3 – 5)x + 15 = x2 – 8x + 15

Ex. Find the product using appropriate identities :
(i) (x + 8) (x + 8)
(ii) (3x – 2y) (3x – 2y)
(iii) (x + 0.1) (x – 0.1)
Sol. (i) (x + 8) (x + 8) = (x + 8)2 = x+ 2(x) × 8 + (8)2 = x2 + 16x + 64.
(ii) (3x – 2y) (3x – 2y) = (3x – 2y)2
[∵ (a – b)2 = a2 – 2ab + b2]
= (3x)2 – 2(3x) × 2y + (2y)2 = 9x2 – 12xy + 4y2.

(iii) (x + 0.1) (x – 0.1) = (x)2 – (0.1)2
[∵ (a + b) (a – b) = a2 – b2]
= (x)2 – 0.01

Ex. Evaluate each of the following by using identities :

(i) 103 × 97
(ii) 103 × 103
(iii) (97)2
(iv) 185 × 185 – 115 × 115

Sol. (i) 103 × 97 = (100 + 3) (100 – 3) = (100)2 – (3)2 = 10000 – 9 = 9991
(ii) 103 × 103 = (103)2 = (100 + 3)2 = (100)2 + 2 × 100 × 3 + (3)2 = 10000 + 600 + 9 = 10609
(iii) (97)2 = (100 – 3)2 = (100)2 – 2 × 100 × 3 + (3)2 = 10000 – 600 + 9 = 9409
(iv)185 × 185 – 115 × 115 = (185)2 – (115)2 = (185 + 115) (185 – 115) = 300 × 70 = 21000


Identity For The Square Of Trinomial

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Ex. Expand each of the following using suitable identities :

(i) (x + 2y + 4z)2
(ii) (– 2x + 5y – 3z)2

Sol. (i) (x + 2y + 4z)2 = (x)2 + (2y)2 + (4z)2 + 2.x.2y + 2.2y.4z + 2.4z.x
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
(ii) (– 2x + 5y – 3z)2 = (–2x)2 + (5y)2 + (–3z)2 + 2.(–2x).5y + 2.5y.(–3z) + 2.(–3z).(–2x)
= 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx  
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]

Ex. If a+ b2 + c2 = 20 and a + b + c = 0, find ab + bc + ca.
Sol. (a2 + b2 + c2) = a2 + b2 + c2 + 2ab + 2bc + 2ca
⇒ (a2 + b2 + c2) = a2 + b2 + c2 + 2(ab + bc + ca)
⇒    02 = 20 + 2(ab + bc + ca)
⇒ –20 = 2(ab + bc + ca)
⇒    Algebraic Identities - Polynomials, Class 9, Mathematics
⇒ ab + bc + ca = – 10

Ex. If a + b + c = 9 and ab + bc + ca = 40, find a2 + b2 + c2.
Sol. We know that, (a2 + b2 + c2) = a2 + b2 + c2 + 2ab + 2bc + 2ca

⇒ 92 = a2 + b2 + c2 + 2 × 40
⇒ 81 = a2 + b2 + c2 + 80
⇒ a2 + b2 + c2 = 81 – 80
⇒ a2 + b2 + c2 = 1


Identity For The Cube Of A Binomial

(a + b)3 = a3 + b3 + 3ab (a + b)
(a – b)3 = a3 – b3 – 3ab (a – b)

Ex. Write the following in the expanded form :
(i) (3x + 4y)3      (ii) (2a – 3b)3

Sol. (i) (3x + 4y)3 = (3x)3 + (4y)3 + 3(3x) × 4y(3x + 4y)
(3x + 4y)3 = (3x)3 + (4y)3 + 3.(3x)2.4y + 3.(3x).(4y)2 
[∵ (a + b)3 = a3 + b3 + 3ab (a + b)]
= 27x3 + 64y3 + 3.9x2.4y + 9x.16y2 = 27x3 + 64y3 + 108x2 y + 144xy2

(ii) (2a – 3b)3 = (2a)3 – (3b)3 – 3 × 2a × 3b(2a – 3b)          
[∵ (a – b)3 = a3 – b3 – 3ab (a – b)]
= 8a3 – 27b3 – 18ab(2a – 3b)
= 8a3 – 27b3 – 18ab × 2a + 18ab × 3b
= 8a3 – 27b3 – 36a2b + 54ab2


Sum And Difference Of Cube

a3 + b3 = (a + b) (a2 – ab + b2)

a– b3 = (a – b) (a2 + ab + b2)

Ex. Factorize each of the following expressions :
1. x3 + 125
2. x3 – 0.027

Sol. 

1. x3 + 125 = (x)3 + (5)3 = (x + 5) (x2 – x.5 + 52)
[∵ a3 + b3 = (a + b) (a2 – ab + b2)]  

(x + 5) (x2 – 5x + 25)

2. x3 – 0.027 = (x)3 – (0.3)= (x – 0.3) [x2 + x × 0.3 + (0.3)2]            
∵ [a3 – b3 = (a – b) (a2 + ab + b2)]

(x – 0.3x) (x2 + 0.3x + 0.09)


Conditional Identity

a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
If a + b + c = 0, then a3 + b3 + c3 = 3abc.

Ex. Find the product : (x + 2y + 3z) (x2 + 4y2 + 9z2 – 2xy – 6yz – 3zx)
Sol. (x + 2y + 3z) (x2 + 4y+ 9z2 – 2xy – 6yz – 3zx)
= (x + 2y + 3z) (x2 + (2y)2 + (3z)2 – x × 2y – 2y × 3z – 3z × x)
= x3 + (2y)3 + (3z)2 – 3 × x × 2y × 3z          
[∵ a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)]
= x+ 8y3 + 27z3 – 18xyz.

Ex. If a + b + c = 8 and ab + bc + ca = 20, find the value of a3 + b3 + c3 – 3abc.
Sol. Since (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
a + b + c = 8 and ab + bc + ca = 20,            
(8)2 = a2 + b2 + c+ 2 × 20  
⇒   64 = a2 + b2 + c2 + 40
∴   a2 + b2 + c2 = 64 – 40 = 24

We know that a3 + b3 + c3 – 3abc = (a + b + c) {a2 + b2 + c2 –(ab + bc + ca)}
∴ a3 + b3 + c3 – 3abc = 8 × (24 – 20) = 4 × 8 = 32
[∴ a + b + c = 8, ab + bc + ca = 20 and a2 + b2 + c2 = 24]
Thus, a3 + b3 + c3 – 3abc = 32.


Type Of Factorization

1.) Factorization by taking out the common factors
Ex. ab(a2 + b2 – c2) + bc (a2 + b2 – c2) – ca(a2 + b2 – c2)
Sol. We have ab(a2 + b2 – c2) + bc (a2 + b2 – c2) – ca(a2 + b2 – c2)
= (a2 + b2 – c2) (ab + bc – ca)

2.) Factorization by grouping the terms
Ex. (x+ 3x)2 – 5(x2 + 3x) – y(x+ 3x) +5y

Sol. We have
(x2 + 3x)2 – 5(x2 + 3x) – y(x2 + 3x) +5y
= (x2 + 3x) {(x2 + 3x) –5} – y{(x2 + 3x) –5} = (x2 + 3x – 5) (x2 + 3x – y)

3.) Factorization by making a perfect square
Ex. a2 + b2 – 2(ab – ac + bc)
Sol. We have
a2 + b22(ab – ac + bc)
= a2 + b2 – 2ab + 2ac – 2bc
= (a – b)2 + 2c (a – b)
= (a – b){(a – b) + 2c} = (a – b) (a – b + 2c)

4.) Factorization the difference of two squares
Ex. x8 – y8
Sol. We have
x8 – y8 = {(x4)2 – (y4)2} = (x4 – y4) (x4 + y4)
= {(x2)2 – (y2)2} (x4 + y4) = (x2 – y2) (x2 + y2) (x4 + y4)
= (x – y)(x + y)(x2 + y2) (x4 + y4)
= (x – y)(x + y)(x2 + y2){(x2)2 + (y2)2 + 2x2y2 – 2x2y2)
= (x – y)(x + y)(x2 + y2){(x2+ y2)2 – (√2xy)2}
= (x – y)(x + y)(x2 + y2)(x2+ y2 – √2xy)(x2 + y2 + √2xy)


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FAQs on Algebraic Identities - Polynomials, Class 9, Mathematics

1. What are algebraic identities?
Ans. Algebraic identities are mathematical equations that are true for all values of the variables in the equation. These identities are used to simplify and solve complex mathematical expressions and equations. Some common algebraic identities include the identity for the square of a trinomial, the identity for the cube of a binomial, and the sum and difference of cubes.
2. What is the identity for the square of a trinomial?
Ans. The identity for the square of a trinomial is (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc, where a, b, and c are any real numbers. This identity can be used to easily expand and simplify expressions involving square trinomials.
3. What is the sum and difference of cubes identity?
Ans. The sum and difference of cubes identity state that a^3 + b^3 = (a+b)(a^2 - ab + b^2) and a^3 - b^3 = (a-b)(a^2 + ab + b^2), where a and b are any real numbers. These identities can be used to factorize and simplify expressions involving cube polynomials.
4. What is the identity for the cube of a binomial?
Ans. The identity for the cube of a binomial is (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3, where a and b are any real numbers. This identity can be used to expand and simplify expressions involving cube binomials.
5. What is a conditional identity?
Ans. A conditional identity is a mathematical equation that is only true under certain conditions or for a specific range of values of the variables in the equation. For example, the identity (a+b)^2 = a^2 + 2ab + b^2 is only true for all values of a and b if a and b are real numbers. If a and b are complex numbers, this identity may not hold true.
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