# Find the remainder without using long division 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$$. • I'm not sure where you received$x^2 + ax + b$from? – Stuy Oct 23 '18 at 18:02 • It's the quotient. For every division there is a quotient and there is a remainder, right?! – José Carlos Santos Oct 23 '18 at 18:03 • Right, I'm just confused how you got that exact form. – Stuy Oct 23 '18 at 18:03 • The only other thing that I used was that the degree of the remainder had to be smaller than the degree of the divisor ($x-2$). – José Carlos Santos Oct 23 '18 at 18:05 • 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\$? – Stuy Oct 23 '18 at 18:07