An Interesting Rational Expression

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[dwm]

I was taught long division of polynomials many years ago, when I was about 13, so I did it that way, assuming it would divide, or the question would have been pointless, I suppose. Horrible, with so many opportunities to make errors! Your method 2 is so much more elegant, and fun...as you said.

[yaser]

If we multiply nominator and denominator by (x-1) and then we have in nominator (x^9-1) + (x^5-x^8) + (x - x^4), and x^3-1 in denominator. That can be cancelled simply

[davidellis]

Expanded synthetic division is a streamlined alternative to polynomial long division.

[tygrataps]

Polynomial division always makes me unease. Its so easy to get things wrong. To help keep things straight, I will add in 0 values for the missing exponents. So x^8+0x^7+0x^6 and so on. At least it helps me keep the columns straight. I hope this helps someone!

[Qermaq]

My first impulse was to multiply top and bottom by (x - 1) to get a difference of cubes at the bottom. Boy was that out of the way! Wound up dividing like you did.

[lesnyk]

I never learned polynomial division while in school - it just wasn't offered. I sort of taught myself several years ago, but never used it, and have since forgotten the detailed "how". So I just assumed a 6th order polynomial (monic form with "1" as the constant), multiplied it by the denominator, & equated resulting coefficients. Lots of algebra, lots of mistakes, but eventually made my way to the correct solution. I really liked method 2 - I've got to get better at factoring, or at least in anticipating how completing the square (or cube) might help...

[davidtaran]

we divide the polynomial (x^8+x^4+1) by the polynomial (x^2+x+1)
we get (x^6-x^5+x^3-x+1)

[benjaminvatovez]

We could use remarkable products for x²+x+1 = (x^3-1)/(x-1) and the same for
Then we develop a^4-1=(a-1)(a+1)(a²+1) for a being x and x^3.
Then the fraction is reduced to From there, let apply b^3+1=(b+1)(b^2-b+1) for b being x and x^2 and we get the answer.

[dan-florinchereches]

My guess was to bring denominator to anicer form
Either (x-1)(x^2+x+1)=x^3-1 or
(X^+x+1)(X^2-x+1)=X^4+x^2+1
The second expression looks very promising so the expected result is
(X^4-x^2+1)(X^2-x+1) For the division

I was lsitening to the video and you said x^4-x^2+1 is not factorable. Euler showed any quartic is factorable and in fact i am pretty sure the factora will have the form of
(X^2+ax+1)(X^2-ax+1)= X^4+(-a^2+2)x^2+1=x^4-x^2+1 equating coefficients
a^2=3 so it factors but not into anything very useful

[bkkboy-cmeb]

=(x²-x+1)(x⁴-x²+1)

[mystychief]

In fact the expression is equal to f(x)/f(x^4) with f(x)=x²+x+1, but I could not use this to find a solution. :-)