How to find an alternate form of this polynomial (factorize?)

Question

I am trying to find the limit of the function

$$\lim_{t \to 1} {{t^3-2t+1}\over{t^3+t^2-2}}$$

And it obviously evaluates to ${0\over0}$ so at first glance it is indetermined.

But I have these two polynomials:

$${t^3-2t+1}$$

And:

$${t^3+t^2-2}$$

So I used Wolfram Alpha to find alternate forms for both, being:

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

And:

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

And I have read through many sites describing how to factor polynomials but I just can't find an answer; I think this is the most well explained article I've read: Simple Polynomial Factoring.

How do I get to those alternate forms? Is factorization the right way to go?

Answer 1

To factorise cubics and higher degree polynomials, firstly, note that the constant term of the polynomial must be the product of the roots. For example, for $ax^2+bx+c$, $c$ must be the product of $m$ and $n$ if the quadratic factorises to $(x-n)(x-m)$. With this in mind, the roots must be the plus or minus factors of that constant term. Substitute in these factors as values of your $t$ and check if they result in $0$, if they do it is obviously a root and one can use long division/synthetic division to take out the factor. Continue until you get a quadratic and then you can use the quadratic formula to get the other roots if necessary.

Answer 2

Hint: Cancel out the common factor $t-1$ when you take the limit. Note that $t \neq 1$.

Observe that $t = 1$ is a zero of both numerator and denominator, hence a well known "theorem" states that $t-1$ is a factor of both of them and you can use Synthetic division or long division to factorise....