Factoring the following polynomials 
Factorize the following polynomial:

*

*$t^3 -9t +8$

*$t^6 -91t^2 +90$

 A: 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 $t$. 
$$\begin{align}t^3-9t+8&=(t-1)(t^2+t-8)\\ \\
t^6-91t^2+90&=(t^2-1)(t^4+t^2-90)\\
&=\left[(t-1)(t+1)\right]\left[(t^2-9)(t^2+10)\right]\\
&=(t-1)(t+1)(t-3)(t+3)(t^2+10)\\
&=(t-3)(t-1)(t+3)(t+3)(t^2+10)\\\end{align}$$
A: Hint:
Both have factor of $t-1$. 
How to find? You can do as instructed by Vince R. above, But another easy way is to take sum of first coefficient and third coefficient and then add with second coefficient. If it comes Zero then $t-1$ is the factor.
$t^3-9t+8$ 
$(1+8)+(-9)=0$
$t^6-91t^2+90$
$(1+91)+(-91)=0$
A: Hint: both of them have a factor $t-1$
A: 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 $$t^2=u$$
