Long Division of Polynomials

[allyonline]

1+8z^2-2z/(1+4z)

Will someone please work this out for me? I'm not getting the answer I'm suppose to .

 

[mmm4444bot]

That's not how we do things here.

Please show us your work; it will save everybody time.

Cheers

PS: Please read the post titled, "Read Before Posting."

 

[BigGlenntheHeavy]

\(\displaystyle f(z) \ = \ \frac{1+8z^2-2z}{1+4z} \ = \ \frac{8z^2-2z+1}{4z+1} \ = \ 2z-1+\frac{2}{4z+1}\)

\(\displaystyle Observe \ that \ y \ = \ 2z-1 \ is \ an \ oblique \ asymptote \ and \ z \ = \ -\frac{1}{4} \ is \ a \ vertical \ asymptote\)

\(\displaystyle of \ f(z). \ See \ graph \ below.\)

[attachment=0:13xr2967]ddd.jpg[/attachment:13xr2967]

 

[allyonline]

\(\displaystyle f(z) \ = \ \frac{1+8z^2-2z}{1+4z} \ = \ \frac{8z^2-2z+1}{4z+1} \ = \ 2z-1+\frac{2}{4z+1}\)

\(\displaystyle Observe \ that \ y \ = \ 2z-1 \ is \ an \ oblique \ asymptote \ and \ z \ = \ -\frac{1}{4} \ is \ a \ vertical \ asymptote\)

\(\displaystyle of \ f(z). \ See \ graph \ below.\)

[attachment=0bt5yjv2]ddd.jpg[/attachmentbt5yjv2]

Oh Okay ! thanks , I see now that I'm suppose to put the divisor in order !