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

Mar 5, 2017

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

#### Explanation:

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

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

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

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

Now we have

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

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

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

so

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

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

${t}^{2} = 3$

$t = \pm \sqrt{3}$

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

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