Question

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

Answer 1

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)`