Answer 1

Since (x-2) is a factor of polynomial  `2x^3+ax^2+bx-14`, we have 

`2(2)^3+a(2)^2+b(2)-14=0` 

⇒ `16+4a+2b-14=0`

⇒`4a+2b+2=0` 

⇒`2a+b+1=0` 

⇒`2a+b=-1`          .............(1) 

On dividing by (x-3), the polynomial `2x^3+ax^2+bx-14` leaves remainder 52, 

⇒`2(3)^3+a(3)^2+b(3)-14=52` 

⇒`54+9a+3b-14=52`

⇒`9a+3b+40=52` 

⇒`9a+3b=12` 

⇒`3a+b=4`       ...............(2) 

Subtracting (1) and (2), we get 

a=5 

substituting a =5 in (1), we get 

`2xx5+b=-1` 

⇒`10+b=-1` 

⇒`b=-11` 

Hence , a=5 and b=-11.