Question

find the remainder of each of the following polynomials divided by \(x^2+1\)
1)\((1+x)^4\)
2)\((1-x)^{32}\)

Answer 1

1) (-4)
2) (-4)(x-1)

Answer 2

^
dont give the final "ansur

Answer 3

@trader , #2 was a typo, i just corrected it..

Answer 4

And yeah. just the final answer doesn't help much..

Answer 5

(ab) mod c = ((a mod c)(b mod c)) mod c

Answer 6

@experimentX but the second one is exponent to 32.

Answer 7

i don't the second one much..

Answer 8

is there a property like \(a\equiv b \mod c\) then \(a^n \equiv b^n \mod c\) ?

Answer 9

it's natural that
a^n = b^n mod c follows from a=b mod c, it's just consequence of geometric series sum

the result of mod operator is +ve
should be 4x^2 mod (1+x^2) for first, I am not sure how to reduce this.

Answer 10

\((1-x)^4 \equiv-4\mod x^2+1 \)
\({[(1-x)^4]}^8\equiv (-4)^8 \mod x^2+1\)

Answer 11

yes you can do that ... except the result of mod operation is not negative.

Answer 12

that gives -4 is not the answer of first. If you can do that the second one must be correct. also it is always negative.

Answer 13

Here is a list calculated by Wolf.
http://www.wolframalpha.com/input/?i=Table%5BMod%5B%281+%2B+x%29%5E4%2C+1+%2B+x%5E2%5D%2C+%7Bx%2C+0%2C+10%7D%5D

Answer 14

most likely you should reduce it this way
4x^2 = -4 mod x^2 + 1
4x^2 + 4 = 0 mod x^2 + 1 <-- the result of division is 4
so remainder is 4x^2 - 3(x^2 + 1) = x^2 - 3
http://www.wolframalpha.com/input/?i=Table%5Bx%5E2+-1%2C+%7Bx%2C+0%2C+10%7D%5D
http://www.wolframalpha.com/input/?i=Table%5BMod%5B%281+%2B+x%29%5E4%2C+1+%2B+x%5E2%5D%2C+%7Bx%2C+0%2C+10%7D%5D

Answer 15

* for x>sqrt(3)

Answer 16

Here's an alternate solution if you're still interested:
\[(1+x)^4=(ax^2+bx+c)(1+x^2)+(dx+e)\]
We only need to find \(d\) and \(e\). Plugging in \(x=i\) and \(x=-i\),:
\[(1+i)^4=\left(\sqrt{2} \operatorname{cis}\frac{\pi}{4}\right )^4=4 \operatorname{cis} \pi=-4\]
\[(1-i)^4=\left(\sqrt{2} \operatorname{cis}\frac{-\pi}{4}\right )^4=4 \operatorname{cis} -\pi=-4\]

Answer 17

You get two equations\(-4=di+e;-4=-di+e\), from which you can see that \(d=0\) and \(e=-4\).
You can do the same for \((1-x)^{32}\).
Just tell me if you didn't understand anything in the solution.

Answer 18

@blockcolder I don't get almost anything in your solution.. :\

Answer 19

Sorry bout that. I was kinda in a hurry to get that all out a while ago.
First, I expressed \((1+x)^4\) as \(Q(x)(1+x^2)+R(x)\), where \(Q(x)\) and \(R(x)\) are the quotient and remainder, respectively.
The first observation to make is that \(R(x)\) can have a maximum degree of 1.
Follow so far?

Answer 20

How do we know that \(R(x)\) has a maximum degree of 1 ?

Answer 21

If we had an \(x^2\) term in the remainder, we could divide further, since we are dividing by \(1+x^2\). Once we do that, we will be left with at most an \(x\) term and a constant.

Answer 22

ok! I get it so far!

Answer 23

Now that we know \(R(x)\) can only be a linear or a constant, we know that \(R(x)=dx+e\), for some d and e.
Thus, we have \((1+x)^4=Q(x)(1+x^2)+(dx+e)\), and we need to know what d and e are.
Now, we need to get rid of the annoying \(Q(x)\) term. We can do this by plugging in \(x=\pm i\), so that that term will become \(Q(x)\cdot 0\) which simply becomes 0.
In short, we plug in the roots of the divisor.
Follow so far?

Answer 24

yep!

Answer 25

After we plug in \(x=\pm i\), we are left with evaluating \((1\pm i)^4\).
Are you familiar with polar coordinates?

Answer 26

kind of familiar..

Answer 27

Here's a review. To convert from polar (\(r \operatorname{cis} \theta \)) to coordinate (\(a+bi\)) and vice-versa, we can use the formulas:
\[r=\sqrt{a^2+b^2}\ ;\theta=\arctan\left(\frac{b}{a}\right)\\
a=r \cos\theta\ ; b=r\sin\theta\]
For now, I'll call the first row of formulas Set 1, and the second row Set 2.

Answer 28

We'll also need de Moivre's formula:
\[(r \operatorname{cis} \theta)^n=r^n \operatorname{cis} n\theta\]
for integer n and complex number \(r \operatorname{cis} \theta\).

Answer 29

Now, we use these to simplify \((1+i)^4\).
I'll leave it up to you to verify that \(1+i=\sqrt{2} \operatorname{cis} \frac{\pi}{4}\) by using Set 1.
Using de Moivre's formula, \((1+i)^4=\left(\sqrt{2} \operatorname{cis} \frac{\pi}{4}\right)^4=4 \operatorname{cis} \pi\).
This last one is equal to \(-4\), by Set 2.
Clear so far?

Answer 30

is it like \((a+bi)=\sqrt{a^2+b^2}\cos[\tan^{-1}(\frac{b}{a})]\) ?

Answer 31

Not \(\cos\), \(\operatorname{cis}\).

Answer 32

Is \(cis \theta=\cos\theta+i\sin\theta\) ?

Answer 33

Yep

Answer 34

I got it now!! Learned it by a totally new approach and it seems easier.. Thanks !! :)

Answer 35

The thing about this solution is that as long as you know the roots of the divisor, you can find out the remainder by solving a system, so you can use this even when the divisor is different.
BTW, that solution was not totally finished yet. You still need to solve the system:
\[\begin{cases}
-4=di+e\\
-4=-di+e
\end{cases}\]
for d and e but I guess you know how to do this, ryt? :D

Answer 36

yeah! add them up!

Answer 37

That's right, and you can adapt this solution depending on the dividend and divisor.

Answer 38

I will just try and solve a few other problems by this method.. Thanks!