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.

  • $\begingroup$ 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 $\frac 5{x-1}$ $\endgroup$
    – lulu
    May 21 at 11:46
  • $\begingroup$ No, if you divide the polynomial $5$ by $x-1$ you get a remainder of $5$. $\endgroup$ May 21 at 12:52

1 Answer 1


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 $3$.


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