How do you factor $t^3- 4t + 3$ ?

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Answer 1 · 2015-05-21T14:11:59

Consider the related equation $t^3-4t+3 = 0$
An obvious solution to this is $t=1$

Therefore
$(t-1)$ is a factor of $t^3-4t+3$

Applying synthetic division we can obtain:
$(t^3-4t+3) div (t-1) = t^2+t-3$

Therefore
$t^3-4t+3) = (t-1)(t^2+t-3)$

and factoring the quadratic is fairly simple:
$t^3-4t+3 = (t-1)^2(t+2)$

How do you factor t^3- 4t + 3t^3- 4t + 3?