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

- ujjwal

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- anonymous

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

- anonymous

^
dont give the final "ansur

- ujjwal

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

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## More answers

- ujjwal

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

- experimentX

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

- Loser66

@experimentX but the second one is exponent to 32.

- ujjwal

i don't the second one much..

- ujjwal

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

- experimentX

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.

- ujjwal

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

- experimentX

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

- experimentX

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.

- experimentX

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

- experimentX

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

- experimentX

* for x>sqrt(3)

- blockcolder

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 \(x=-i\),:
\[(1+i)^4=\left(\sqrt{2} \operatorname{cis}\frac{\pi}{4}\right )^4=4 \operatorname{cis} \pi=-4\]
\[(1-i)^4=\left(\sqrt{2} \operatorname{cis}\frac{-\pi}{4}\right )^4=4 \operatorname{cis} -\pi=-4\]

- blockcolder

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.

- ujjwal

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

- blockcolder

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?

- ujjwal

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

- blockcolder

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.

- ujjwal

ok! I get it so far!

- blockcolder

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?

- ujjwal

yep!

- blockcolder

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

- ujjwal

kind of familiar..

- blockcolder

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.

- blockcolder

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 \(r \operatorname{cis} \theta\).

- blockcolder

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?

- ujjwal

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

- blockcolder

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

- ujjwal

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

- blockcolder

Yep

- ujjwal

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

- blockcolder

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

- ujjwal

yeah! add them up!

- blockcolder

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

- ujjwal

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

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