NCERT Solutions
Class 8 Maths
Algebraic Expressions and Identities

Ex.9.5 Q.1
Use a suitable identity to get each of the following products:
1. (x + 3) × (x + 3)
2. (2y + 5) × (2y + 5)
3. (2a - 7) × (2a - 7)
4. (3a – ) × (3a –
)
5. (1.1m – 0.4) × (1.1m + 0.4)
6. (a2 + b2) × (-a2 + b2)
7. (6x - 7) × (6x + 7)
8. (-a + c) × (-a + c)
9. ( +
) × (
+
)
10. (7a – 9b) × (7a – 9b)
1. (x + 3) × (x + 3) = (x + 3)2
= x2 + 2 × x × 3 + 32
[Using (a + b)2 = a2 + 2ab + b2]
= x2 + 6x + 9
2. (2y + 5) × (2y + 5) = (2y + 5)
= (2y)2 + 2 × 2y × 5 + 52
[Using (a + b)2 = a2 + 2ab + b2]
= 4y2 + 20y + 25
3. (2a - 7) × (2a - 7) = (2a - 7)2
= (2y)2 - 2 × 2y × 7 + 72
[Using (a - b)2 = a2 - 2ab + b2]
= 4y2 - 28y + 49
4. (3a – 12 ) × (3a – 12
) = (3a – 12
)2
= (3a)2 - 2 × 3a × 12 + (12)
2
[Using (a - b)2 = a2 - 2ab + b2]
= 9a2 – 3a + 1/4
5. (1.1m – 0.4) × (1.1m + 0.4) = (1.1m)2 – (0.4)2
[Using (a - b) × (a + b) = a2 – b2]
= 1.21 m2 – 0.16
6. (a2 + b2) × (-a2 + b2) = (b2 + a2) × (b2 – a2)
= (b2)2 – (a2)2
[Using (a - b) × (a + b) = a2 – b2]
= b4 – a4
7. (6x - 7) × (6x + 7) = (6x)2 – (7)2
[Using (a - b) × (a + b) = a2 – b2]
= 36 x2 – 49
8. (-a + c) × (-a + c) = (c - a) × (c - a)
= (c – a)2
= (c)2 - 2 × c × a + a2
[Using (a - b)2 = a2 - 2ab + b2]
= c2 – 2ac + a2
9. (x2 + 3y4
) × (x2
+ 3y2
) = (x2
+ 3y4
)2
= (x2 )2 + 2 × x2
× 3y4
+ (3y4
)2
[Using (a + b)2 = a2 + 2ab + b2]
= x24 + 3xy4
+ 9y216
10. (7a – 9b) × (7a – 9b) = (7a – 9b)2
= (7a)2 - 2 × 7a × 9b + (9b)2
[Using (a - b)2 = a2 - 2ab + b2]
= 49a2 – 126ab + 91b2