# Does the remainder theorem work for constant function?

The remainder theorem stated that " If a polynominal f(x) is divided by (x-k), the remainder is f(k)."

So I think of a constant function ,say, f(x)= 5 and I divide this by (x-1). According to this theorem, the remainder must be f(1). So, f(1)= 5 which just returns the original input value as it is a constant function.

But in reality, let's say for the situation f(5)= 5, if I divide 5 by 4, which also is f(x) divided by (x-1), I get the remainder 1, that contradicts the theorem. So I just want to know whether this theorem really doesn't work for constant function and if there are also other situations where this theorem won't work. Thanks

Edit: I have read that the remainder is always constant and does not depend on the value of "x" in f(x) but I'm curious why this happens.

• this is hard to follow. "If I divide $5$ by $4$, which is also $f(x)$ divided by $(x-1)$, ..." is just wrong. $\frac 54$ is in no way the same as $\frac 5{x-1}$
– lulu
May 21 at 11:46
• No, if you divide the polynomial $5$ by $x-1$ you get a remainder of $5$. May 21 at 12:52

If you divide $$f(x)=5$$ by $$g(x)=4$$ you would write $$f(x)=1.25 \cdot 4 + 0$$, so the remainder would be $$0$$.
If you divide $$f(x)=5$$ by $$g(x)=x-1$$, you would write $$f(x)=0 \cdot (x-1) + 5$$, so the remainder would be $$5$$.
Does that explain it? If not, here's another attempt. Division conceptually is all about writing the dividend, call it $$n$$, as $$qd+r$$, where $$q$$ is the quotient, $$d$$ is the divisor, and $$r$$ is the remainder. We narrow down choices of $$q$$ and $$r$$ by requiring that $$r$$ is "strictly less" than $$d$$ in some sense; in the case of polynomial division, we require the degree of $$r$$ to be strictly less than the degree of $$d$$. This means that if the degree of $$d$$ already exceeds that of $$n$$ then $$q=0,r=n$$. This is just like how if you divide $$3$$ by $$7$$ then you get a remainder of $$3$$.