1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

7
$\begingroup$

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 $3$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.