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

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 = x² + 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 a² + 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)
⇒    
⇒ 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

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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)

a³ – 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)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)

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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 + 4y² + 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)]
= x³ + 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 + c² + 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.

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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. (x² + 3x)² – 5(x² + 3x) – y(x² + 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 + b² – ²(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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