Help with problem

Sep 2015
2
0
Los Angeles
Hey guys my names Joey I'm new and there is something that I'm stuck on in math. How do I find the roots of the equation t(to the 3rd power) + 2t(to the 2nd power) – 4t – 8=0
 

romsek

MHF Helper
Nov 2013
6,767
3,043
California
$p(t)=t^3+2t^2 - 4t -8=0$

well...

If it can't be factored there is a formula for the roots of a cubic equation but it's quite involved.

If it can be factored by the rational root theorem you know that any factors $(t+a)$ will be such that $a$ divides 8, the constant term of the polynomial.

that limits the values of $a$ to

$\pm 1, \pm 2, \pm 4, \pm 8$

so start trying to divide $p(t)$ by $(t \pm1), (t\pm 2), \dots$

or you could head over to Wolframalpha.com and type

Factor[t^3+2t^2 - 4t -8]

and it will do the grunt work for you.
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
$t^3+2t^2-4t-8=0$

$t^2(t+2) -4(t+2)=0$

$(t+2)(t^2-4)=0$

Can you finish from here?

Next time, post your problem in the proper forum ... this one could go in the algebra forum. Thanks.