it was once asked in a zimsec November 2015 question paper

f(x)=a(x)(x-3)-9; for the remainder to be -9, the |-9|<x-3 (remainder is always less than the divisor), so x>12;

f(x)=b(x)(2x-1)-6 where a and b are arbitrary functions which evaluate to integers. |-6|<2x-1, 2x>7. this condition is already satisfied by x>12.

f(x)=a(x)(x-3)=b(x)(2x-1)+3; a(x)=(b(x)(2x-1)+3)/(x-3).

[Let x=15, then a(x)=(29b(x)+3)/12=24b(x)/12+(5b(x)+3)/12=2b(x)+(5b(x)+3)/12.

If b(x)=9, then a(x)=18+4=22. Then f(x)=22*12-9=255=9*29-6. 255/(x-3)(2x-1)=255/(12*29)=255/348. So the remainder is 255.

This example used arithmetic to show how the remainder is found, but we now return to the algebraic method to relate the remainder to x.]

(2x-1)/(x-3)=2+5/(x-3); a(x)=(b(x)(2x-1)+3)/(x-3)=b(x)(2+5/(x-3))+3/(x-3); so a(x)=2b(x)+(5b(x)+3)/(x-3).

For the fraction to evaluate as an integer n=(5b(x)+3)/(x-3), nx-3n=5b(x)+3; nx=5b(x)+3(n+1); b(x) (also an integer)=(nx-3(n+1))/5.

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