Answer 1

Lef f(x)=`2x^3-7x^2+ax-6`  

x-2=0 ⇒  x=2 

When f (x) is divided by (x-2), remainder = f(2) 

∴ `f(2)=2(2)^3-7(2)^2+a(2)-6` 

          =`16-28-2a-6`    

          = 2a-18 

Let `g(x)=x^3-8x^2+(2a+1)x-16` 

When g (x) is dividend by (x-2), remainder=g(2) 

∴ `g(2)=(2)^3-8(2)^2+(2a+1)(2)-16` 

          = 8-32+4a+2-16 

        = 4a-38  

By the given condition, we have: 

f(2)=g(2) 

2a-18=4a-38 

4a-2a=38-18  

2a=20 

a=10 

Thus, the value of a is 10