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

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Answer 1 · 2017-03-05T00:54:22

$t = - \sqrt{3}, 0, \sqrt{3}$ $t = -sqrt3, 0, sqrt3$

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

$t^{5} - 6 t^{3} + 9 t = t (t^{4} - 6 t^{2} + 9)$ $t^5 - 6t^3 + 9t = t(t^4 - 6t^2 + 9)$

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

${(t^{2})}^{2} - 6 (t^{2}) + 9 = (t^{2} - 3) (t^{2} - 3) = {(t^{2} - 3)}^{2}$ $(t^2)^2 - 6(t^2) + 9 = (t^2-3)(t^2-3) = (t^2-3)^2$

Now we have

$g (t) = t {(t^{2} - 3)}^{2} = 0$ $g(t) = t(t^2-3)^2 = 0$

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

$t = 0$ $t = 0$

which gives us a solution already, or

${(t^{2} - 3)}^{2} = 0$ $(t^2-3)^2 = 0$

so

${(t^{2} - 3)}^{2} = 0$ $(t^2-3)^2=0$

$t^{2} - 3 = 0$ $t^2-3 = 0$

$t^{2} = 3$ $t^2 = 3$

$t = \pm \sqrt{3}$ $t = +-sqrt3$

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

$t = - \sqrt{3}, 0, \sqrt{3}$ $t = -sqrt3, 0, sqrt3$

How do you find all zeros of g(t)=t5−6t3+9tg(t)=t^5-6t^3+9t?