Question

"Express z^5 -1 as the product of 2 factors, one of which is linear." We then have to find the zeroes of that expression in polar form, and finally express z^4 + z^3 + z^2 + z + 1 as a product of 2 real quadratic factors. I sort of know how to do it, but I don't have a good understanding of WHY i am doing it! Can you recommend a good resource to explain this please?"Express z^5 -1 as the product of 2 factors, one of which is linear." We then have to find the zeroes of that expression in polar form, and finally express z^4 + z^3 + z^2 + z + 1 as a product of 2 real quadratic factors. I sort of know how to do it, but I don't have a good understanding of WHY i am doing it! Can you recommend a good resource to explain this please?@Mathematics

Answer 1

If you want a linear factor, that implies something of the form
\[(z-a)(z^4+bz^3+cz^2+dz+e)\]
Which, when multiplied out, gives
\[z^5+bz^4+cz^3+dz^2+ez-az^4-abz^3-acz^2-adz-ae\]
\[=z^5+(b-a)z^4+(c-ab)z^3+(d-ac)z^2+(e-ad)z-ae\]
which implies that, since the coefficients in front of everything but z^5 and 1 be zero,
\[b=a\]
\[c=ab=a^2\]
\[d=ac=a^3\]
\[e=ad=a^4\]
and, since we need the constant terms to match,
\[ae=a^5=1\]
It would appear, then, that
\[a=1^{\frac{1}{5}} =1\]
and so our factor becomes
\[(z-1)(z^4+z^3+z^2+z+1)\]

Answer 2

Is this along the lines of what you're asking, or are you trying to find some place in life where you'd actually use something like this? :)

Answer 3

No that's lovely thanks! You explained it a lot better than my textbook did ;)

Answer 4

The thing is, I know how to find the quadratic factors too, by first finding the linear factors and then using sum/product of pairs of those to get a quadratic factor. I just wonder if you could show me your way of solving it if you have time please, as I definitely learned from your other demo. Thanks!

Answer 5

Sorry about the delay. Made an algebra boo-boo, as I have an unfortunate tendency to do sometimes. Correct answer is forthcoming... haha