NCERT Solutions Class 8 Mathematics Solutions for Exercise 14.4 in Chapter 14 - Factorisation

Question 19 Exercise 14.4

Substituting x = – 3 in

$\text { (a) } x^{2}+5 x+4 \text { gives }(-3)^{2}+5(-3)+4=9+2+4=15$

$\text { (b) } x^{2}-5 x+4 \text { gives }(-3)^{2}-5(-3)+4=9-15+4=-2$

$\text { (c) } x^{2}+5 x \text { gives }(-3)^{2}+5(-3)=-9-15=-24$

Answer:

Solution: $\text { (a) Substituting } x=-3 \text { in } x^{2}+5 x+4, \text { we have }$ $x^{2}+5 x+4=(-3)^{2}+5(-3)+4=9-15+4=-2 . \text { This is the correct answer. }$

$\text { (b) Substituting } x=-3 \text { in } x^{2}-5 x+4$ $x^{2}-5 x+4=(-3)^{2}-5(-3)+4=9+15+4=28 . \text { This is the correct answer }$

$\text { (c) Substituting } x=-3 \text { in } x^{2}+5 x$ $x^{2}+5 x=(-3)^{2}+5(-3)=9-15=-6 . \text { This is the correct answer }$

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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