Using Remainder Theorem to find remainder

[Schaus]

Homework Statement

(y⁴ - 5y² + 2y - 15) / (3y - √(2))
The answer says (2√(2)/3)-(1301/81)
https://lh3.googleusercontent.com/tdxTK9jIztkS5_7q9x0jP_D3HdkcDcKO9K_T2glYcdlF5MW71SiVq0xNjj09bkX3yUTi0lVS5aehQcTCpjVMgEyIEySNwanKwnxkfuh-l9wGnecvNy11IkibZd5nNgn4eiv4l3I-SGY89O9V_M2uzlcRBLfMVMkS_gybhaz3be3-eDmspF47-z1thzMs6iseX2eU2IjJMpAmx0YjN4mrOY4Bx7hQtrUnyMA4GSXyavCjbLx89CaKmAU6bbqm36Qqm3ieAW1299QOvcXrBqH2XaT_u9T_THz1ktwI9a_Mw5ozQurH0RnZvOIWbArej-6PU9mOWIpvasISRaLkwCWpG_w76wjFjZGISRXhq33LvWxM4bGFbvirN6c--KFp5yLWC7djePdwbpb69UF5FbPYJxOuBIWoaVg1QPwwg0V7im1Zp0fThljuNWHmwyM1vA6tXhnc5zkx7WnZ8BCiP2feBXWOdal8yvnTA0BXY8O0_Ly1YSO8fqX1mATZ8RY8YlawtMB_FMwuNsAYVJ6Vde0tdPFQEYlQ3yYERXpeo-31kNJRs_Y9wP5pWUp58FTVgyBsOmwiztqNVUVJyQfnMiKjDKJeR9_wevbfnEStla1ntJRTakXqP8dG=w449-h311-no

Homework Equations

The Attempt at a Solution

Using synthetic division
3y - √(2) = 0
3y = √(2)
y = √(2)/3
View attachment 111547

My final answer that I keep getting is (2√(2)/3)-21. I can't seem to get (2√(2)/3)-(1301/81). I was just wondering if someone could show me where I'm going wrong? Sorry writing this out was very difficult so I hope you can still understand it.

 

[fresh_42]

I don't actually can follow your calculations. If I compare my calculation with your result, then I think you simply lost some denominators.
I get $\frac{4}{81}-\frac{10}{9}-15+\frac{2\sqrt{2}}{3}$ whereas you seem to have added only the nominators $4-10-15+\frac{2\sqrt{2}}{3}$.

I once gave an example on PF here:
https://www.physicsforums.com/threa...r-a-polynomial-over-z-z3.889140/#post-5595083

Maybe it helps.

 

[Schaus]

Sorry, I should have just posted this to begin with.

 

[Ray Vickson]

View attachment 111547 Sorry, I should have just posted this to begin with.

Your long-division process is all wrong. Let $p(y) = y^4 - 5y^2 + 2y - 15$ and $d(y) = 3y - \sqrt{2}$. You want to find a "quotient" $q(y)$ and a number $r$ (the "remainder") that give you $p(y) = q(y) d(y) + r.$ The question is asking you to find $r$.

Step 1: see how many times the leading term of $d(y)$ will go into the leading term of $p(y)$; that is the number of times $3y$ goes into $y^4$. The answer in this case is $y^4/(3y) = (1/3)y^3$, so that is the first term in your quotient $q(y)$. Now we have
$$ p(y) - \frac{1}{3} y^3 d(y) = \frac{\sqrt{2}}{3} y^3 -5 y^2 + 2y - 15 \equiv p_1(y).$$

However, all that would be doing it the hard way. The easy way would be to go ahead and use the "remainder theorem".

Step 2: see how many times the leading term of $d(y)$ goes into the leading term of $p_1(y)$. The answer is $(\sqrt{2}/3)/(3 y) = (\sqrt{2}/9) y^2$, so that is the next term in your quotient $q(y)$. We have
$$p_2(y) \equiv p_1(y) - \frac{\sqrt{2}}{9} y^2 d(y) = -\frac{43}{9} y^2 + 2y - 15.$$

So, up to now we have
$$p(y) = \left( \frac{1}{3} y^3 + \frac{\sqrt{2}}{9} y^2 \right) d(y) + p_2(y).$$

Keep going like that; the next term in the quotient $q(y)$ will be the number of times the leading term of $d(y)$ goes into the leading term of $p_2(y)$, so is the number of times $3y$ goes into $-(43/9) y^2$, etc., etc.

The remainder will be what you have left when the process comes to an end.

However, all that would be doing it the hard way; I included the material just because you attempted to do it, but did it all wrong. The easy way would be to just use the so-called "remainder theorem".

 

[Schaus]

I was using synthetic division which worked on every other question but I'll try it the way you're suggesting it.

 

[Schaus]

I got the same result as the answer. Thanks for your help!

 

[PetSounds]

I was using synthetic division which worked on every other question but I'll try it the way you're suggesting it.

It looks to me like you simply made a couple multiplication errors when working with fractions. I worked the problem with synthetic division and got the right answer.