How do you solve #t^3-13t-12=0# ?

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Answer 1 · 2016-12-22T20:25:44

The solutions are #t=-1#, #t=4# and #t=-3#.

Given:

#f(t) = t^3-13t-12#

Notice that #12+1=13#, so is there some value like #t=+-1# which will give #f(t) = 0# ?

We find:

#f(-1) = -1+13-12 = 0#

So #t=-1# is a zero and #(t+1)# a factor:

#t^3-13t-12 = (t+1)(t^2-t-12)#

To factor #t^2-t-12# find a pair of factors of #12# which differ by #1#.

The pair #4, 3# works. Hence we find:

#t^2-t-12 = (t-4)(t+3)#

So the other two zeros are:

#t = 4" "# and #" "t = -3#