Question

Please answer 8 th and 9th question

Answer 1

8) 1^3 - 6× 1^2 + 11×1 -6

= 1-6+11-6

=12-12 = 0

(3^3)-(6×3^2)+(11×3)-6

=27-54+33-6

60-60 = 0

Yes both 1 and 3 are zeroes

9) AC=BD

AB+BC=BC+CD

Cancel BC on both sides

AB = CD

HENCE PROVED


Hope this helps

Answer 2

Answer:

8) f (x)=x^3-6x^2+11-6

If x=1

F(1)=1-6+11-6

=12-12

=0

Therefore yes

If x=3

f(3)=27-54+33-6

=60-60

=0

Therefore yes

9) AC=BE

AB+BC=BC+CD

AB=CD

Hence proved

Hope you will satisfied with this