Question 440046
Dividing {{{(ax^3+3x^2-3)/(x-4)}}}. We find the remainder:{{{16(3+4a)-3}}}, and 

dividing {{{(2x^3-5x+a)/(x-4)}}}, find the remainder:{{{108+a}}}, since the 

remainders are equal we write:{{{16(3+4a)-3=108+a}}}. Solving this linear 

equation find a=1.

Done.