KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.3

In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.3 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.3 pdf, free KSEEB SSLC Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.3 pdf download. Now you will get step by step solution to each question.

Karnataka State Syllabus 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-6y4x2– 8xy9x2y-11 x3y2
3x9x2-18xy12x3-24x:y27x3y-33x4y2
-6y-18xy36y2-24x2y48xy--54x2y266x3y3
4x-I2x3-24x2yI6x4-32x3y236x4y-44x3y2
-8xy-24x2y48xy2-32x3y64x2y2-72x3y288x4y3
9x2y27x3y-54x2y236x4y-72xy81x4y2-99 x3y3
-11 x3y2-33x4y366 x2y3-44 x5y288 x4y3-99 x5y3121 x6y4

Question 2.
Find the products
i. (5x + 8) 3x
ii. (-3 pq)(-15 p3q² – q3)
iii. 2×5(3a3−3b3)
iv. -x²(x – 15)
Answer:
i. 5x . 3x + 8.3x = 15x² + 24x

ii. (- 3pq) (- 15p3q² – q3)
= (- 3pq) (- 15p3q²) – (- 3pq)(q3)
= 45 p4q3 + 3 pq4

iii. 2×5×3a3−2×5×3b3=6xa35−6xb33

iv. (-x²)x – (-x²) 15
= -x3 + 15x²

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. 2m(5m – 1) + 3m(5m – 1) = 10m4 – 2m+ 15m² – 3m

All Chapter KSEEB Solutions For Class 8 maths

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