>There is a problem with the validity of the remainder
>The theorem states that if f(x) is divided by x-a, the
>remainder is f(a), where f is any rational integral
We use the letter-convention:
Letter 'x' indicate a Variable.
Letter 'a' indicate a Constant.
Letter 'f' indicate a Function.
So 'f(x)' means the function 'f' , not the number,
while 'f(a)' means the function value in the point 'a'.
'x-a' means a function of 'x', not a number.
This convention is often used but rarely spoken of,
because it cannot easily be avoided even if it'stinks.
So, dividing f(x) by (x-a) means writing:
f(x)=(x-a)g(x) + b .
This is an equality of functions of x .
f(x) is the dividend.
x-a is the divisor.
g(x) is the quotient.
b is the remainder.
Now substitute x=a to get
f(a)=(a-a)g(a)+b = 0*g(a)+b = b
So, the remainder, b, of a function, f(x),
divided by the function (x-a),
equals the function value f(a).
This was the theorem.
>But supposeing f(x) = x^2 + x + 5,
>f(4) will be 25.
>Supposing we divide f(4) with (4 - 3) i.e 1, the remainder
>should be f(3).
No sir, you should divide the function f(x)
with the function x-4 :
f(x) = (x-4)(x+5)+25
Check that (x-4)(x+5)+25 = x^2+x+5 .
The dividend is x^2+x+5.
The divisor is x-4.
The quotient is x+5.
The remainder is 25.
The theorem then gives f(4)=25, which is correct.