3

According to the remainder theorem, if we divide $p(x)$ by $(ax+b)$, the remainder will be $p(-b/a)$

Let $p(x) = 3x^3+ 5x^2+ 8x +5$ and let the divider be $2x + 7.$

So, $a = 2$ and $b =7$. Thus, the remainder should be $p(-7/2)$ $ = 3×(-7/2)^3 + 5×(-7/2)^2 + 8×(-7/2) + 5$ $= -90.375$

If the remainder theorem is true for all values of $x$, then the remainder should always be $-90.375$.

However, for $x=1$ , $p(x)$ = $p(1)$ = 21 and $2x+7 = 9$. So, the remainder is not $-90.375$. Again, for $x =2,3,4$ its not $-90.375$.

Why is it equal to the estimated value from the remainder theorem? Did I make any mistakes?

Any help will be appreciated.

74H54N3
  • 69
  • 8
  • 1
    Saying that $r(x)$ is the remainder for the polynomial division does not mean that it is the remainder (in the sense of integer division) at every particular value for $x$. Is that what you are asking? – lulu Jan 23 '23 at 15:02

1 Answers1

4

It is not true that the remainer is polynomial division must equal the remainder in the associated integer divisions (when we fix a value for the variable).

Here is a simpler example. Let $p(x)=x^2+10$ and consider division by $x$. Clearly we have $$p(x)=x\times x+10$$ so the remainder is $r(x)=10$, a constant as it should be. We note that $10=p(0)$, which is again consistent with the remainder theorem.

And yet, if we pick specific values, $10$ is not the remainder for the associated integer division. For instance, taking $x=2$ we get $p(2)=14$ which is divisible by $2$, so the remainder in that case is $0$.

Worth remarking: For large enough arguments, there certainly is some connection. For instance, sticking with $x^2+10$ as an example, if $x>10$ then $10$ is indeed the numerical remainder as well as the polynomial remainder. A similar claim holds generally though you have to take care of the sign (the numerical remainder must be non-negative, though the polynomial remainder can certainly be negative).

lulu
  • 70,402
  • Thank you so much for answering. I really appreciate the help! However, I'm still curious about two things: 1. In that case, what does r(x) = 10 mean? If the remainder is not always r(x), when it's not 0, does the value of r(x) actually mean anything? As r(x) might not be the actual remainder. So, when it is different, like your example of the remainder being 0, does the r(x) = 10 mean anything? 2. For which values of x will the remainder be the same to the estimated remainder considering the p(x) and divider from my question? Again, thanks! – 74H54N3 Jan 23 '23 at 16:19
  • It's the remainder under polynomial division. That is to say, given any polynomials $p(x), d(x)$ there are unique poilynomials $q(x), r(x)$ with $p(x)=q(x)\times d(x)+r(x)$ and with degree $r(x)<$ degree $d(x)$. That's the Remainder Theorem. It does indeed resemble the usual integer division remainder, but it is for polynomials. – lulu Jan 23 '23 at 16:46
  • In the case where $r(x)$ is a constant (not always true, of course) then I would expect that the polynomial remainder is the numerical remainder for large enough arguments (at least if it is positive). That's certianly true for the example I gave, for instance. A similar claim can be made even for non constant remainders. – lulu Jan 23 '23 at 16:52
  • If you don't like attaching $\infty$ or $-\infty$ to the integers and I don't ,then deg: non-zero polynomials $ \to $non-negative integers and for $g(x), f(x) \ne 0$there exist $q(x),r(x)$ such that $r(x) \ne 0, g(x)=q(x)f(x)+r(x), \deg(r(x))<\deg(f(x))$ or there exists $q(x)$ such that $g(x)=q(x)f(x)$. – P. Lawrence Jan 23 '23 at 18:02