ujjwal
  • ujjwal
find the remainder of each of the following polynomials divided by \(x^2+1\) 1)\((1+x)^4\) 2)\((1-x)^{32}\)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1) (-4) 2) (-4)(x-1)
anonymous
  • anonymous
^ dont give the final "ansur
ujjwal
  • ujjwal
@trader , #2 was a typo, i just corrected it..

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ujjwal
  • ujjwal
And yeah. just the final answer doesn't help much..
experimentX
  • experimentX
(ab) mod c = ((a mod c)(b mod c)) mod c
Loser66
  • Loser66
@experimentX but the second one is exponent to 32.
ujjwal
  • ujjwal
i don't the second one much..
ujjwal
  • ujjwal
is there a property like \(a\equiv b \mod c\) then \(a^n \equiv b^n \mod c\) ?
experimentX
  • 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
  • ujjwal
\((1-x)^4 \equiv-4\mod x^2+1 \) \({[(1-x)^4]}^8\equiv (-4)^8 \mod x^2+1\)
experimentX
  • experimentX
yes you can do that ... except the result of mod operation is not negative.
experimentX
  • 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
  • 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
  • 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
  • experimentX
* for x>sqrt(3)
blockcolder
  • 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
  • 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
  • ujjwal
@blockcolder I don't get almost anything in your solution.. :\
blockcolder
  • 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
  • ujjwal
How do we know that \(R(x)\) has a maximum degree of 1 ?
blockcolder
  • 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
  • ujjwal
ok! I get it so far!
blockcolder
  • 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
  • ujjwal
yep!
blockcolder
  • blockcolder
After we plug in \(x=\pm i\), we are left with evaluating \((1\pm i)^4\). Are you familiar with polar coordinates?
ujjwal
  • ujjwal
kind of familiar..
blockcolder
  • 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
  • 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
  • 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
  • ujjwal
is it like \((a+bi)=\sqrt{a^2+b^2}\cos[\tan^{-1}(\frac{b}{a})]\) ?
blockcolder
  • blockcolder
Not \(\cos\), \(\operatorname{cis}\).
ujjwal
  • ujjwal
Is \(cis \theta=\cos\theta+i\sin\theta\) ?
blockcolder
  • blockcolder
Yep
ujjwal
  • ujjwal
I got it now!! Learned it by a totally new approach and it seems easier.. Thanks !! :)
blockcolder
  • 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
  • ujjwal
yeah! add them up!
blockcolder
  • blockcolder
That's right, and you can adapt this solution depending on the dividend and divisor.
ujjwal
  • ujjwal
I will just try and solve a few other problems by this method.. Thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.