Ncert solutions

Ncert solutions

Grade 8

Factorisation | Exercise 14.4

Question 10

Q10)Substituting x = – 3 in

(a) x^2+5x+4\ gives\ \left(-3\right)^2+5\left(-3\right)+4\ =\ 9\ +2+4\ =15

(b) x^2-5x+4\ ggives\ \left(-3\right)^2-5\left(-3\right)+4\ =9-15+4=-2

(c) x^2+5x\ gives\ \left(-3\right)^2+5\left(-3\right)\ =\ -9-15=-24

Solution 10:

(a) \left(-3\right)^2+5\left(-3\right)+4

9+\left(-15\right)+4

9-15+4

= -2

(b) \left(-3\right)^2-5\left(-3\right)+4

= 9+15+4

=28

(c) \left(-3\right)^2+5\left(-3\right)

=9-15

= -6

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subject-cta

Question 10

Q10)Substituting x = – 3 in

(a) x^2+5x+4\ gives\ \left(-3\right)^2+5\left(-3\right)+4\ =\ 9\ +2+4\ =15

(b) x^2-5x+4\ ggives\ \left(-3\right)^2-5\left(-3\right)+4\ =9-15+4=-2

(c) x^2+5x\ gives\ \left(-3\right)^2+5\left(-3\right)\ =\ -9-15=-24

Solution 10:

(a) \left(-3\right)^2+5\left(-3\right)+4

9+\left(-15\right)+4

9-15+4

= -2

(b) \left(-3\right)^2-5\left(-3\right)+4

= 9+15+4

=28

(c) \left(-3\right)^2+5\left(-3\right)

=9-15

= -6

Still have questions? Our expert teachers can help you out

Book a free class now

Want to top your mathematics exam ?

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Book a free class now
subject-cta
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