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 $t$ :

$t = -sqrt3, 0, sqrt3$

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