How Do You Solve Complex Polynomial Long Division Problems?

AI Thread Summary
To solve the polynomial long division problem of dividing (-9x^6 + 7x^4 - 2x^3 + 5) by (3x^4 - 2x + 1), the first step involves finding the first partial quotient, which is -3x^2. After multiplying this by the divisor and subtracting from the dividend, the revised dividend becomes 7x^4 - 8x^3 + 3x^2 + 5. The discussion highlights the challenge of managing higher degree polynomials and the need to include all terms in the divisor for accurate subtraction. The final result after simplification includes a remainder, prompting questions about correctness and the process involved. Verifying the result by multiplying it back with the divisor is suggested to ensure it equals the original dividend.
acuraintegra9
Messages
7
Reaction score
0
Divide the polynomials by using long division.
(-9x^6+7x^4-2x^3+5)/(3x^4-2x+1)



When I attempted it I started by pulling using 3x^2 . multiplied that by the (3x^4-2x+1) and from there I had to use a fraction of 7/3 or something and then couldn't divide into x cubed.


If anyone can help me that would help out so much.. All the examples in my book, that we talked about in class, and the ones online are mostly simple ones that work out evenly, are not that high of exponents etc.
 
Physics news on Phys.org
Use the polynomials in their expanded forms. Your first partial quotient should be -3x^2. Multiply this by the divisor and subtract the result from the dividend to obtain your revised dividend in order to continue the process.

(note carefully, "-3x^2" means "negative three times x squared")
 
Last edited:
I did that much, but then the tricky part is when I add it in what's left after eliminating the x6 and x5 I get .. 7x^4-8x^3+3x^2+0x +5

dont know what to do from there.. do I have to multiply the 3x^4 by some sort of fraction to get it to equal to 7 to cancel out..
 
please help! not sure what to do .. I get stuck at the next step .. not sure what to multiply by, don't cancel the x to the fourth's out
 
The second partial division and subtraction yielded a remainder. The process needed inclusion of the last both two terms in order to perform a sensible second subtraction; I needed to use all FIVE terms of the divisor, which is why the last two expanded form terms of the dividend were needed.

I have no easy way to typeset or express the numeric process in this forum. Let me just say, my final result after simplification was
-3x^2+(7/3)+(-8x^3 + 3x^2-(14/3)x+4)/(3x^4 -2x+1)

See if you can duplicate that! So many symbols and slightly lengthy, maybe I made an error.
 
thats pretty much what I came up with , not sure though with it all being a remainder like that.. that's half the reason I thought my answer was wrong
 
acuraintegra9 said:
thats pretty much what I came up with , not sure though with it all being a remainder like that.. that's half the reason I thought my answer was wrong

What course are you in? How does your textbook treat this topic? Did your teacher mention or give any example like the one you posted?

You could try checking my result using multiplying the result by the divisor; it should be found equivalent to the dividend.
 
Back
Top