# NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.4

NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.4 Factorization in English Medium and Hindi Medium free to use online. These Solutions are based on latest and updated NCERT Books following the CBSE Syllabus.

## NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.4

If you need solutions in Hindi, Click for Hindi Medium solutions of 8 Maths Exercise 14.4       Exercise 14.1

Exercise 14.2

Exercise 14.3

8 Maths Solutions: Main Page

### 8 Maths Chapter 14 Exercise 14.4 in Hindi      Exercise 14.1

Exercise 14.2

Exercise 14.3

8 Maths Solutions: Main Page

To get the solutions in English, Click for English Medium solutions.

##### Important Questions for Practice

1. The height of a triangle is x4 + y4 and its base is 14xy. Find the area of the triangle.
2. The radius of a circle is 7ab – 7bc – 14ac. Find the circumference of the circle.
3. The sum of first n natural numbers is given by the expression x²/2+x/2. Factorise this expression.
4. If a + b = 25 and a² + b² = 225, then find ab.
5. The product of two expressions is x5 + x³ + x. If one of them is x² + x + 1, find the other.
6. Subtract b (b² + b – 7) + 5 from 3b² – 8 and find the value of expression obtained for b = – 3.
7. The curved surface area of a cylinder is 2π (y² – 7y + 12) and its radius is (y – 3). Find the height of the cylinder (C.S.A. of cylinder = 2πrh).
8. What should be added to 4c (– a + b + c) to obtain 3a (a + b + c) – 2b (a – b + c)?
9. Build a square garden. Divide the square garden into four rectangular flower beds in such a way that each flower bed is as long as one side of the square. The perimeter of each flower bed is 40 m. (a) Draw a diagram to represent the above information. (b) Mention the expression for perimeter of the entire garden.
10. The sum of (x + 5) observations is x4 – 625. Find the mean of the observations.

###### Try These

1. Divide 15 (y + 3) (y² – 16) by 5 (y² – y – 12) and find the quotient and remainder.
2. Find the value of x, if 10000x = (9982)² – (18)²