KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.3
KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.3
Karnataka Board Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.3
Question 1.
Complete the following table of products of two monomials
| First → Second ↓ |
3x | -6y | 4x2 | – 8xy | 9x2y | -11 x3y2 |
| 3x | 9x2 | -18xy | 12x3 | -24x:y | 27x3y | -33x4y2 |
| -6y | -18xy | 36y2 | -24x2y | 48xy- | -54x2y2 | 66x3y3 |
| 4x- | I2x3 | -24x2y | I6x4 | -32x3y2 | 36x4y | -44x3y2 |
| -8xy | -24x2y | 48xy2 | -32x3y | 64x2y2 | -72x3y2 | 88x4y3 |
| 9x2y | 27x3y | -54x2y2 | 36x4y | -72xy | 81x4y2 | -99 x3y3 |
| -11 x3y2 | -33x4y3 | 66 x2y3 | -44 x5y2 | 88 x4y3 | -99 x5y3 | 121 x6y4 |
Question 2.
Find the products
Question 3.
Simplify the following :
i. (2xy -xy) (3xy – 5)
ii. (3xy² +1) (4xy – 6xy²)
iii. (3x² + 2x) (2x² + 3)
iv. (2m3 + 3m) (5m – 1)
Answer:
i. 2x y (3xy – 5) – 2xy(3xy – 5) = 6x²y² – 10xy – 6x²y² + 10xy
ii. 3xy² (4xy – 6xy²) +1 (4xy – 6xy²) = 12 x²y3 – 18x²y4 + 4xy – 6xy²
iii. 3x²(2x² + 3) + 2x(2x² + 3] = 6x4 + 9x² + 4x3 + 6x
iv. 2m3 (5m – 1) + 3m(5m – 1) = 10m4 – 2m3 + 15m² – 3m
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