>There is a problem with the validity of the remainder

>theorem.

>The theorem states that if f(x) is divided by x-a, the

>remainder is f(a), where f is any rational integral

>function. 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.