NCERT Solutions Class 8 Mathematics Solutions for Exercise 14.2 in Chapter 14 - Factorisation

Question 16 Exercise 14.2

Factorise:

$16 x^{5}-144 x^{3}$

Answer:

Solution:

In this question, firstly we need to take the common factor out and than we can apply the identity a²-b²= (a+b)(a-b)

$\begin{array}{l} =16 \mathrm{x}^{3}\left(\mathrm{x}^{2}-9\right) \\ =16 \mathrm{x}^{3}\left(\mathrm{x}^{2}-9\right) \\ =16 \mathrm{x}^{3}(\mathrm{x}-3)(\mathrm{x}+3) \\ \text { Using Identity: } x^{2}-y^{2}=(x+y)(x-y) \end{array}$

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Exercises

Exercise 14.1

Exercise 14.2

Exercise 14.3

Exercise 14.4

Chapters

Rational Numbers

Linear Equations in One Variable

Understanding Quadrilaterals

Practical Geometry

Data Handling

Squares and Square Roots

Cubes and Cube Roots

Comparing Quantities

Algebraic Expressions and Identities

Visualising Solid Shapes

Mensuration

Exponents and Powers

Direct and Inverse Proportions

Factorisation

Introduction to Graphs

Playing with Numbers