How do you find all zeros of #g(t)=t^5-6t^3+9t#?

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Answer 1 · 2017-03-05T00:54:22

#t = -sqrt3, 0, sqrt3#

Since all terms have a #t# in them, you can factor that out:

#t^5 - 6t^3 + 9t = t(t^4 - 6t^2 + 9)#

You can factor that, too, treating #t^2# as the subject, a bit like #x# in a standard quadratic.

#(t^2)^2 - 6(t^2) + 9 = (t^2-3)(t^2-3) = (t^2-3)^2#

Now we have

#g(t) = t(t^2-3)^2 = 0#

For this to equal to #0#, at least one of the factors has to equal #0#, so either

#t = 0#

which gives us a solution already, or

#(t^2-3)^2 = 0#

so

#(t^2-3)^2=0#

#t^2-3 = 0#

#t^2 = 3#

#t = +-sqrt3#

Therefore, we have three total solutions for #t#:

#t = -sqrt3, 0, sqrt3#