Root and factorization of a polynomial

We will try to modify the polynomial to have an expression similar to the square of a sum:

$$x^2+10x+25=x^2+2\cdot5x+5^2=(x+5)^2$$

We effect a change of variable $x^2=t$:

$$t^2-625$$

And now we can apply the difference of two squares:

$$t^2-625=(t+25)\cdot(t-25)$$

The solutions are $t=25$ and $t=-25$. We undo the change:

$$x^2=25 \Rightarrow \left\{\begin{array}{c} x=5 \\ x=-5 \end{array} \right.$$

The other polynomial $x^2+25$ is irreducible.

We look for candidates that are roots of the polynomial, specifically the divisors of the independent term (in this case $6$):

values: $1,-1,2,-2,3,-3$

Therefore:

$$p(1)=1^2+1-6=-4$$

$$p(-1)=(-1)^2+(-1)-6=-6$$

$$p(2)=2^2+2-6=0$$

$$p(-2)=(-2)^2+(-2)-6=-4$$

$$p(3)=3^2+3-6=6$$

$$p(-3)=(-3)^2+(-3)-6=0$$