Polynomial division.

Question

If an f(x) is divided by x²+4, has 2x+3 of remainder.

If an f(x) is divided by x²+6, has 6x-1 of remainder.

If an f(x) is divided by (x²+4)(x²+6), has S(x) of remainder, then find S(4)!

I have this. I wrote it in polynomial formula.

$f(x)=(x²+4).R(x)+(2x+3)$

And yes, i have x²=-4.

But, i had to express the 2x+3 function into the (x²) one. So, basically if we have p(x)=2x+3, then p(x²)=2x²+3. and i can put -4 to the x, and i got :

f(-4)=2(-4)+3 = -5 ( and the same thing happened with x²+6 ). Is that true so far?

Sorry for bad english.

It would be great if you could add your own thoughts on the question (and your efforts to address/answer it). — amWhy, Mar 25 '17 at 17:10

Bernard · Answer 1 · 2017-03-25T18:18:27.343

2

Use the Chinese remainder theorem for polynomials:

A Bézout's relation between the coprime polynomials $x^2+4$ and $x^2+6$ is $$\tfrac12(x^2+6)-\tfrac12(x^2+4)=1,$$ hence the solution of the simultaneous congruences $\quad\begin{cases}f(x)\equiv 2x+3\mod x^2+4\\f(x)\equiv 6x-1\mod x^2+6\end{cases}\quad$ is $$S(x)\equiv\tfrac12\bigl[(2x+3)(x^2+6)-(6x-1)(x^2+4)\bigr]\mod(x^2+4)(x^2+6),$$so that $\;S(4)=-109$.

edited Mar 25 '17 at 18:18

answered Mar 25 '17 at 17:53

Bernard

175,478

Why should it be the case that $S(x)$ be systematically $0$? – Bernard Mar 25 '17 at 18:05
Nevermind, I am not thinking straight this morning. I will withdraw my comment. – John Wayland Bales Mar 25 '17 at 18:12
You have the general remainder $S(x)$. Maybe you're confusing $f(x)$ and $S(x)$? – Bernard Mar 25 '17 at 18:13
@kingW: Oh! yes. Thanks for pointing it. It's fixed now. – Bernard Mar 25 '17 at 18:19
Yes, my conclusion should have been that in the degree three case, $f(x)=S(x)=-2x^3+2x^2-6x+11$ and that in that case we get $S(4)=-109$ which agrees with your result when the typo is corrected. – John Wayland Bales Mar 25 '17 at 18:19
It's exactly that. – Bernard Mar 25 '17 at 18:25
Guys, should the s(x) is linear like the example (2x+3) and (6x-1)? – Fauzy Nur Noviansyah Mar 25 '17 at 20:34
No: you divide by a degree $4$ polynomial, so the remainder has degree at most $3$ – which is the case here. – Bernard Mar 25 '17 at 20:38
But we don't know the degree of the f(x) – Fauzy Nur Noviansyah Mar 25 '17 at 22:42
So what? If $\deg f(x)<4$, the quotient is $0$ and $f(x)=S(x)$, that's all. – Bernard Mar 25 '17 at 23:08

score 0 · Answer 2 · answered Mar 25 '17 at 18:11

0

Approach using complex numbers. Notice that $x^2+4=0$ when $x=\pm 2i$ and $x^2+6=0$ when $x=\pm i\sqrt{6}$ since we have that $$f(x)=(x^2+4)R(x)+2x+3$$ From this we have that $$f(2i)=4i+3\\f(-2i)=-4i+3$$ Also we have that $$f(x)=(x^2+4)Q(x)+6x-1\\f(i\sqrt{6})=6i\sqrt{6}-1\\f(-i\sqrt{6})=-6i\sqrt{6}-1$$ From this we have that $$f(x)=(x^2+4)(x^2+6)T(x)+ax^3+bx^2+cx+d\\f(2i)=-8ai-4b+2ci+d\\f(-2i)=8ai-4b-2ci+d\\f(i\sqrt{6})=-6a\sqrt{6}i-6b+ci\sqrt{6}+d\\f(-i\sqrt{6})=6a\sqrt{6}i-6b-ci\sqrt{6}+d$$ Now $$f(2i)+f(-2i)=2d-8b=6\\ f(i\sqrt{6})+f(-i\sqrt{6})=2d-12b=-2$$ from this we have that $b=2$ and $d=11$. We also have that $$f(2i)-f(-2i)=4ci-16ai=8i\\c-4a=2\\f(i\sqrt{6})-f(-i\sqrt{6})=2ci\sqrt{6}-12ai\sqrt{6}=12\sqrt{6}i\\c-6a=6$$ Solving this we get $a=-2$ and $c=-6$ This implies that $S(x)=-2x^3+2x^2-6x+11$ and $S(4)=-128+32-24+11=-109$

answered Mar 25 '17 at 18:11

kingW3

13,496

I think the s(x) function will be linear. – Fauzy Nur Noviansyah Mar 25 '17 at 20:31
@FauzyNurNoviansyah Since $(x^2+4)(x^2+6)$ is a quartic,it's natural that the remainder be 1 degree less;cubic.Anyway why do you think that s(x) is linear?You had a linear remainder because you divided with a quadratic – kingW3 Mar 25 '17 at 20:34
from what i know, if you divide a 3-degrees function with quadratic funtion, the remainder will be linear. same thing happen when 4 degrees is divided by 3-degrees function. So, i think it doesn't matter whether how many degrees you have. it only affect the final result, not the remainder function. Correct me if i wrong. thanks :) – Fauzy Nur Noviansyah Mar 25 '17 at 20:42
@FauzyNurNoviansyah Here you have that a $4$ degree polynomial is multiplied by some polynomial which has a arbitrary degree lets say $n$ if $n=7,11,15$ etc. then the degree of the remainder polynomial is $3$ this is the most degree that the remainder polynomial can give,if for example instead we had $a=0$ then the degree would be $2$ etc. – kingW3 Mar 25 '17 at 21:29

Polynomial division.

2 Answers2