All terms share #t# as a common factor, let's factor out #t:#
#g(t)=t(t^4-6t^2+9)#
Now, let's focus on the trinomial #t^4-6t^2+9#. This somewhat resembles a quadratic form; however, the fourth degree appears somewhat confusing. So, let's say #u=t^2#, meaning #u^2=(t^2)^2=t^4,# and rewrite the trinomial in terms of #u:#
#t^4-6t^2+9=u^2-6u+9#
We can factor this easily. Let's think of two numbers which add together to give #-6# and multiply together to give #9.# That would be #-3# and #-3.# Thus,
#u^2-6u+9=(u-3)(u-3)#
Rewrite in terms of t, as we know #u=t^2#
#(u-3)(u-3)=(t^2-3)(t^2-3)#
We can't really factor this further. Our fully factored form is then
#t(t^2-3)(t^2-3)#
Let's set this equal to #0# and solve for #t:#
#t(t^2-3)(t^2-3)=0#
#t=0# is a solution.
#t^2-3=0#
#t^2=3#
#t=+-sqrt(3)#
#sqrt(3),-sqrt(3)# are solutions.
