Question 104049
Normal style of factoring would take many trials and errors, so I prefer this method:

To factor any trinomial {{{ax^2+bx+c}}},

Solve for x as if {{{ax^2+bx+c=0}}} and consequently you may opt to use the QUADRATIC FORMULA!!!

There you would find two roots: say y and z

Then,

x=y  or  x=z

x-y=0 or x-z=0

Here, {{{x-y}}} and {{{x-z}}} are the factors.

NOTE: {{{ax^2+bx+c}}} is considered PRIME or not factorable if y and/or z are irrational or imaginary.

Back to the problem of yours,

{{{-t^2+4t-3}}}

If we let {{{-t^2+4t-3=0}}},

Then,

{{{t=1 or t=3}}}

{{{t-1=0}}}  or  {{{t-3=0}}}

Therefore,

{{{-t^2+4t-3=(t-1)*(t-3)}}}