Find the roots of $6z^5 + 15z^4 + 20z^3 + 15z^2 + 6z + 1 = 0.$
I know how to do this without the coefficients, but I do not know what to do in this problem.
Thanks
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asked Jul 1, 2015 at 0:17
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2$\begingroup$ Apropos of Andre's answer, you should train yourself to instantly recognise the lower order binomial coefficients (using Pascal's triangle, for instance). $\endgroup$– DeepakCommented Jul 1, 2015 at 0:53
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2$\begingroup$ What's the story behind the title? $\endgroup$– JiKCommented Jul 1, 2015 at 4:20
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1 Answer
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Hint: We are solving $(z+1)^6-z^6=0$, or equivalently $\left(1+\frac{1}{z}\right)^6=1$.
answered Jul 1, 2015 at 0:20
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$\begingroup$ I got $(z+1)^6 -z^6 = 0$, but how did you get $(1+1/z)^6=1$? $\endgroup$ Commented Jul 1, 2015 at 16:41
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$\begingroup$ Rewrite as $(z+1)^6=z^6$. Any solution of this is non-zero, so we can divide both sides by $z^6$. obtaining $\left(\frac{z+1}{z}\right)^6=1$. To make further processing easier later, I rewrote $\frac{z+1}{z}$ as $1+\frac{1}{z}$. $\endgroup$ Commented Jul 1, 2015 at 16:51