Question

Find the value of a and b so that (x² -4) is a factor of ax^4+2x³ -3x²+bx-4 . Show the working also

Answer 1

factor of ax^4+2x³-3x²+bx-4 is (x²-4) or (x-2)(x+2) so the value of the given equation is zero at x = 2 and x = -2, on putting the value of x = 2
16a+16-12+2b-4 = 0
16a+2b = 0
8a+b = 0  ---------------(1)
on putting the value of x = -2
16a-16-12-2b-4 = 0
16a-2b = 32
8a-b = 16  -----------------(2)
(1) + (2)
16a = 16
a = 1
put a = 1 in equation (1) we get 
b = -8
I hope you can understand .

Answer 2

x²-4 = (x-2)(x+2)      roots are 2 , -2    So divide by 2, and -2

       |  a          2          -3            b            -4 
       |
2     |            2a        4+4a        2+8a        2b+16a+4
      | ____________________________________
       |  a      2+2a      1+4a        b+8a+2      2b+16a

So  reminder 2b+16 a = 0     =>    b = -8 a

Similarly, for root -8 :

       reminder = -2b-32+16a  = 0    =>  8a - b = 16

 solving a = 1 b = -8