Factoring the following polynomials Factorize the following polynomial:

$t^3 -9t +8$
$t^6 -91t^2 +90$
It's not so bad to be baffled... :) 

Putting $t=1$ and and $t^2=1$ into the first and second expressions respectively will make them zero, so $(t-1)$ and $(t^2-1)$ are factors for the first and second expressions respectively. 

You can work out the other factor(s) by equating coefficients of powers of $t$. 

$$\begin{align}t^3-9t+8&amp;=(t-1)(t^2+t-8)\\ \\
t^6-91t^2+90&amp;=(t^2-1)(t^4+t^2-90)\\
&amp;=\left[(t-1)(t+1)\right]\left[(t^2-9)(t^2+10)\right]\\
&amp;=(t-1)(t+1)(t-3)(t+3)(t^2+10)\\
&amp;=(t-3)(t-1)(t+3)(t+3)(t^2+10)\\\end{align}$$
To solve $t^3 -9t +8$ try to find one root by checking different vaule for t. 

From the rational root theorem follows that a rational solution of a polynomial with integer coefficients and leading coefficient $1$ must be an integer dividing the constant term of the polynomial. So if the polynomial $t^3 -9t +8$ has a rational solution it must be among the numbers $\{\pm 8, \pm4, \pm2, \pm1\}$. By substituting $1$ for $t$ in the polynomial you see that $1$ is a zero of the polynomial.

From the he factor theorem follows: if $1$ is a zero of $t^3 -9t +8$ then $t-1$ is a divisor of $t^3 -9t +8$. So we divide $t^3 -9t +8$ by $t-1$  using th polynomial division  algorithm to get the quotient $t^2+t-8$. The remainder is $0$ because $t-1$ is a divisor. 

So we now know that $$t^3 -9t +8=(t-1)(t^2+t-8)$$. The remaining solutions can be found by solving
$$t^2+t-8=0$$
(Why?)

To solve $t^6 -91t^2 +90$ we start with substituting $u$ for $t^2$ in the equation to get 

$$u^3 -91u +90$$
now we have  a third degree polynomial as we had in the former case. We solve it using the samce technic.
After we have found solutions $u$ we can find solution $t$ of the sixth degree polynomial by solving the equation $$t^2=u$$