Let f(x) = x^10 −3x^5 +2. Find all α ∈ C satisfying f(α) = 0 and use your results to express f(x) as a product of linear factors (factorize f(x) completely)?

Teacher gave hint :Let z = x5 and consider the roots of g(z) = f(x). Write the roots of g(z) in exponential (or polar) form.

1 Answer
Ananda Dasgupta
Apr 14, 2018

#{alpha|f(alpha)=0}={1,e^{2ipi/5},e^{4ipi/5},e^{6ipi/5},e^{8ipi/5}, root{5}{2},root{5}{2}e^{2ipi/5},root{5}{2}e^{4ipi/5},root{5}{2}e^{6ipi/5},root{5}{2}e^{8ipi/5}}#

#f(x) = (x-1)(x-e^{2ipi/5})(x-e^{4ipi/5})(x-e^{6ipi/5})(x-e^{8ipi/5}) #
#quad (x-root{5}{2})(x-root{5}{2}e^{2ipi/5})(x-root{5}{2}e^{4ipi/5})(x-root{5}{2}e^{6ipi/5})(x-root{5}{2}e^{8ipi/5})#

Explanation:

Denote #x^5 # by #z#. Then the equation becomes

#z^2-3z+2 = 0 implies (z-2)(z-1) = 0#

So, either #z=2# or #z=1#

If #z=1#, we have #x^5=1# - so #x# is one of the five fifth roots of unity.

Note that we can write #1 = e^{2i pi n},quad nin ZZ# So, a fifth root of unity can be written as

#root{5}{e^{2i pi n}} = exp(2i pi n/5),qquad n in ZZ#

Note that putting #n=m+5# and #n=m# , respectively, in this expression returns the same value, so we get distinct roots only for #n =0,1,2,3,4#

Similarly, if #z = 2# we have

#x = root{5}{2e^{2i pi n}} = root{5}{2} exp(2i pi n/5),qquad n=0,1,2,3,4#

Thus the set of all #alpha# satisfying #f(alpha)=0# is

#{1,e^{2ipi/5},e^{4ipi/5},e^{6ipi/5},e^{8ipi/5}, root{5}{2},root{5}{2}e^{2ipi/5},root{5}{2}e^{4ipi/5},root{5}{2}e^{6ipi/5},root{5}{2}e^{8ipi/5}}#

and a complete factorization of #f(x)# is

#f(x) = (x-1)(x-e^{2ipi/5})(x-e^{4ipi/5})(x-e^{6ipi/5})(x-e^{8ipi/5}) #
#quad (x-root{5}{2})(x-root{5}{2}e^{2ipi/5})(x-root{5}{2}e^{4ipi/5})(x-root{5}{2}e^{6ipi/5})(x-root{5}{2}e^{8ipi/5})#

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