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

Question 26 Exercise 14.2

Factorise:

$q^{2}-10 q+121$

Answer:

Solution:

The middle term is split into two elements using the "Splitting the Middle Term by Factorization" approach and further by grouping the pairs, given quadratic polynomial can be factorized.

Observed that, 21 = -7 x -3 and -7 + (-3) = -10

$\begin{array}{l} q^{2}-10 q+21=q^{2}-3 q-7 q+21 \\ =q(q-3)-7(q-3) \end{array}$

$\begin{array}{l} =(q-7)(q-3) \\ \text { This implies } q^{2}-10 q+21=(q-7)(q-3) \end{array}$

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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