factoring to solve t^3 - 5t^2 + 3t = 0

[jshaziza]

t^3-5t^2+3t=0

I am having trouble figuring this problem out. I am used to solving similar problems that lets say have t^4+t^2+3t=0 by designating u as t^2 and u^2 as t^4; however the square of 2 doesn't go into three, so I am stuck on this problem.

Please explain how to go about solving it. Thx. for your help.

 

[axrw]

Notice that all terms have a common factor of t? Factor that out and see what you end up with.

 

[jwpaine]

Re: factoring

t^3-5t^2+3t=0

\(\displaystyle \L t(t^2 - 5t + 3) = 0\)

Is it possible to factor it anymore without having irrational roots?

John

 

[Denis]

> t^3 - 5t^2 + 3t = 0

Shaz, are you sure last term is 3t? 4t perhaps?

 

[pka]

It can be factored! See
\(\displaystyle \L t^3 - 5t^2 + 3t = t\left( {t - \left[ {\frac{{5 + \sqrt {13} }}{2}} \right]} \right)\left( {t - \left[ {\frac{{5 - \sqrt {13} }}{2}} \right]} \right)\)

 

[jshaziza]

Shaz...,Lol, yea I'm sure. The last term is 3t.