Question

HELPP!!

LET:
\[P(X)=X ^{2011}+X ^{1783}-3X ^{1707}+2X ^{341}-3X ^{2}-3\]. Find the remainder, without using a calculator, when you divide P(x) by
\[X ^{3}-X\].

Answer 1

Okay have you divided polynomials before?

Answer 2

If you like we can go through this step by step.

Answer 3

Hmmm... don't think so :)

Answer 4

yes, that would be helpful!!!

Answer 5

Okay the first step is to take the X^3 and find out how many times it goes into x^{2011}.

_____________________________________________________
X^3−X |X^{2011}+X^{1783}−3X^{1707}+2X^{341}−3X^2−3

Do you know what the answer is?

Answer 6

\[x^mx^n=x^{m+n}\]

Answer 7

not sure what the answer is; honestly...

Answer 8

that's okay,
\[x^3x^n=x^{2011}\]\[x^n=x^{2011-3}=x^{2008}\]

So it is x^{2008}

X^{2008}
_____________________________________________________
X^3−X |X^{2011}+X^{1783}−3X^{1707}+2X^{341}−3X^2−3

Now that we know this we need to times x^{2008} by x^3-x, can you do this step?

Answer 9

if \[x ^{3}-x ^{n}=2008\]
then \[x ^{n}=2005\]?

Answer 10

not x3-xn, but x3xn... sorry

Answer 11

\[x ^{3}x ^{n}=2008\]

Answer 12

yes that working is correct but we need to multiply

\[x^{2008}(x^3-x)\]

Answer 13

I'm not sure if you've done long division before but we are using this method. it's the easiest way to do these polynomial questions :)

Answer 14

oh... ok

Answer 15

would it be better if I did another example first and then continue with this one so you know the method?

Answer 16

yeah, that would would be cool... thanks

Answer 17

okay let's do this one

__________________
X^2+3 |X^3-5X^2+3X-15

Answer 18

Step 1 : find out how many x^2 go into x^3

X^{3-2}=X^1 or X

X
__________________
X^2+3 |X^3-5X^2+3X-15

Step 2: multiply x by X^2+3, put this underneath the division

X(X^2+3)=X^3+3X


X
__________________
X^2+3 |X^3-5X^2+3X-15
X^3+3X

Step 3: minus the two under the division sign

X
__________________
X^2+3 |X^3-5X^2+3X-15
-( X^3 +3X)
----------------------
0 - 5X^2 +0 -15
-5X^2-15

Answer 19

understand so far... thanks :)

Answer 20

Step 4: Redo step 1-3 but for -5X^2

-5X^{2-2}=-5X^0 or -5

X -5
__________________
X^2+3 |X^3-5X^2+3X-15


-5(X^2+3)=-5X^2-15

X - 5
__________________
X^2+3 |X^3-5X^2+3X-15
-( X^3 + 3X)
----------------------
0 - 5X^2 +0 -15
-5X^2-15
-5X^2-15
------------------------
0+0

So therefore X^2+3 goes into X^3-5X^2+3X-15, X - 5 times with no remainder.

Answer 21

Would you like to try your question again first and I'll see if it's right or if you're on the right track.

Answer 22

yeah, i'll do that :)

Answer 23

\[x ^{2011}\]\[x ^{2011}+x ^{1783}-3x ^{1707}+3x ^{2}-3\]

Answer 24

\[x ^{2011-3}=x ^{2008}\]

Answer 25

Yes that's good but I think we will have to try another method to solve this because you will have to do long division by hand way too many times.

Just give me a minute to think about it

Answer 26

What we are going to do is separate this big question into smaller ones and do the long division on those.

Answer 27

okay

Answer 28

\[\frac{X^{2011}}{X^3-X}+\frac{X^{1783}}{X^3-X}-
\frac{3X^{1707}}{X^3-X}+\frac{2X^{341}}{X^3-X}-\frac{3X^2}{X^3-X}-\frac{3}{X^3-X}\]

So are problem is now like this, much easier to handle. We are doing the exact same method of solving just smaller scale. try the first fraction

Answer 29

okay

Answer 30

hold on I'm going to see if anyone has an idea on how to do this one. I thought separating it would help but we have the same problem. Sorry