How do you solve $t^{3} - 13 t - 12 = 0$ $t^3-13t-12=0$ ?

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Answer 1 · 2016-12-22T20:25:44

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

Given:

$f (t) = t^{3} - 13 t - 12$ $f(t) = t^3-13t-12$

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

We find:

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

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

$t^{3} - 13 t - 12 = (t + 1) (t^{2} - t - 12)$ $t^3-13t-12 = (t+1)(t^2-t-12)$

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

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

$t^{2} - t - 12 = (t - 4) (t + 3)$ $t^2-t-12 = (t-4)(t+3)$

So the other two zeros are:

$t = 4$ $t = 4" "$ and $t = - 3$ $" "t = -3$

How do you solve t^3-13t-12=0t3−13t−12=0t^3-13t-12=0 ?