Using factor theorem on non-polynomials

[Some1]

I have seen the reason why the factor theorem works:

f(x) = d(x)q(x) + r(x)

If d(x) = (x - a) then
f(a) = 0*q(a) + r(a)

Therefore f(a) = r(a)

However there is no indication that this will not work with equations other than polynomials (f(x) could be sine(x)) so why does this only work on polynomials?

 

[Steven G]

I have seen the reason why the factor theorem works:

f(x) = d(x)q(x) + r(x)

If d(x) = (x - a) then
f(a) = 0*q(a) + r(a)

Therefore f(a) = r(a)

However there is no indication that this will not work with equations other than polynomials (f(x) could be sine(x)) so why does this only work on polynomials?

Not exactly true. If d(v)=0 and f(x) = d(x)q(x) + r(x), then f(v) = r(v) even if no polynomials.

We know that sin(pi)=0. If f(x) = sin(x)*q(x) + r(x), then f(pi) = r(pi)

 

[Ishuda]

I have seen the reason why the factor theorem works:

f(x) = d(x)q(x) + r(x)

If d(x) = (x - a) then
f(a) = 0*q(a) + r(a)

Therefore f(a) = r(a)

However there is no indication that this will not work with equations other than polynomials (f(x) could be sine(x)) so why does this only work on polynomials?

As Jomo indicated, the factor theorem can be applied for any function f. Just choose a function d(x) with a zero at some value of x=a and some function r(x) such that
\(\displaystyle g(x) = \frac{f(x)-r(x)}{d(x)}\)
has a removable singularity at x=a and we see that
f(x) = d(x) g_s(x) + r(x)
where g_s is g with the singularity removed and we have
f(a) = r(a)

What you lose is the general restriction on r(x) if f(x) (and d(x)) is not a polynomial. i.e. something like r(x) is of a degree less that d(x) in much the same way that the remainder after dividing a number, say m, by another number, say n, the remainder is always less than n.

 

[Ishuda]

I have seen the reason why the factor theorem works:

f(x) = d(x)q(x) + r(x)

If d(x) = (x - a) then
f(a) = 0*q(a) + r(a)

Therefore f(a) = r(a)

However there is no indication that this will not work with equations other than polynomials (f(x) could be sine(x)) so why does this only work on polynomials?

sec²(x) = [tan(x) - 1] [tan(x) + 1] + 2