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

Question

3431 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-21T14:11:59

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

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

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

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

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

How do you factor t^3- 4t + 3t3−4t+3t^3- 4t + 3?