enter image description here

So I was looking at other posts related to this, and most of them contained polynomials divided by a function with multiple roots. In my question, I only have $g(x) = (x-2)$, assuming $f(x) = x^{3} - 2x + 4 $. So I set it up as such:

$$f(x) = q(x)(x-2) + l(x)$$

where $l(x) = (ax+b)$

And now if I plug in $x=2$ I get $f(x) = 2a+b$ but I'm not sure where to go from here because I only have 1 unknown, and can't solve for $a$ or $b$...


Division $\,\Rightarrow\ f(x) = q(x)(x\!-\!2) + r(x),\ \bbox[5px,border:1px solid red] {\deg r < \deg(x\!-\!2) = 1}\,$ $\,\Rightarrow\,r(x) = r\,$ is constant

So we obtain $\ f(2) = r\ $ by evaluating above at $\, x=2.4

Your confusion stems from assuming that $r(x)$ has higher degree.

Remark $ $ Similarly $\ f(a) = f(x)\bmod x\!-\!a,\ $ the ubiquitous Polynomial Remainder Theorem

Alternatively we can use modular arithmetic for the proof:

$\!\bmod x\!-\!a\!:\,\ x\equiv a\,\Rightarrow\,f(x)\equiv f(a)\ $ by the Polynomial Congruence Rule


The remainder is a constant polynomial $k$. And if$$x^3-2x+4=(x^2+ax+b)(x-2)+k,$$then, putting $x=2$, this becomes $8=0+k$. Therefore, the remainder is the constant polynomial $8$.

  • $\begingroup$ I'm not sure where you received $x^2 + ax + b$ from? $\endgroup$ – Stuy Oct 23 '18 at 18:02
  • $\begingroup$ It's the quotient. For every division there is a quotient and there is a remainder, right?! $\endgroup$ – José Carlos Santos Oct 23 '18 at 18:03
  • $\begingroup$ Right, I'm just confused how you got that exact form. $\endgroup$ – Stuy Oct 23 '18 at 18:03
  • $\begingroup$ The only other thing that I used was that the degree of the remainder had to be smaller than the degree of the divisor ($x-2$). $\endgroup$ – José Carlos Santos Oct 23 '18 at 18:05
  • $\begingroup$ So let's say if I was doing $x^4 -7x^2 + 3 $ is divided by $x+1$, I would write it in the form: $(x^3 + ax^2 + bx + c)(x+1) + k$? $\endgroup$ – Stuy Oct 23 '18 at 18:07

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.