Question

Solve for t

Answer 1

\[t^3+3t-2=0\]

Answer 2

for ax^2 +bx +c=0
the solutions for x = (-b + sqrt(b^2 -4ac))/)2a) , (-b - sqrt(b^2 -4ac))/)2a)

Answer 3

here t is x
a=1 ,b=3 and c=-2

Answer 4

THat's a cubed sign.

Answer 5

the guess solution is t=-1
putting this the l.h.s. expression is=(-1)^3-3(-1)+2=-1+3-2=0=r.h.s.
so one factor will be t+1
now lhs,
=t^3-3t-2
=t^3+t^2-t^2-t-2t-2
=(t+1)t^2-t(t+1)-2(t+1)
=(t+1)(t^2-t-2)
=(t+1)(t^2-2t+t-2)
=(t+1)(t-2)(t+1)
so the equaton becomes,
(t+1)(t-2)(t+1)=0
the solutions are t=-1,-1,2