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

May 21, 2015

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

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

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

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

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