# 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

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+4x^{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 xx^{2}+5 x=(-3)^{2}+5(-3)=9-15=-6 . \text { This is the correct answer }

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