One more confusing factoring question. The question is:

$x^6 + 5x^3 + 8$

Please can someone help me in factorising this. I saw some solutions but they are not meant for a IX grade student. 
Thanks for the help.
 A: Replace $ x^3 $ by say $ t $ and then you have simple quadratic expression $$ t^2 + 5t + 8 $$
You can factorise it by using basic rules.
OR


*

*If you want to equate it to some constant.
$$ (x^3 + 5/2)^2 + \frac{7}{4}$$ 

A: We can rearrange the polynomial as follows and use a difference of cubes:
\begin{align*}
x^6+5x^3+8&=(x^2+2)^3-6x^4+5x^3-12x^2\\
&=(x^2+2)^3-x^3-x^2(6x^2-6x+12)\\
&=\left[(x^2+2)-x\right]\left[(x^2+2)^2+x(x^2+2)+x^2\right]-6x^2(x^2-x+2)\\
&=(x^2-x+2)\left[(x^2+2)^2+x(x^2+2)-5x^2\right]\\
&=(x^2-x+2)\left(x^4+x^3-x^2+2x+4\right)\\
\end{align*}
A: You can basically extend the logic of the rational root technique to solve this.
For the rational root technique, we'd be looking for a root of the form $ax+b$, and so by expanding, we confirm that it must have $a$ dividing the leading coefficient, and $b$ dividing the trailing coefficient.
The same logic can be applied here. It's just a little longer.
We have
$$
x^6+5x^3+8
$$
Now, if we have a product of the form
$$
(ax^2+bx+c)(dx^4+ex^3+fx^2+gx+h)
$$
then the leading coefficient is $ad$ and the trailing coefficient is $ch$. So we know that $a$ divides 1 and $c$ divides 8. We also know that $d$ divides 1, so without loss of generality, we'll take $a=d=1$.
Now, we have to match the rest of the terms up. This is a little more complicated. We have to determine possible values of $b$ by starting with a given value of $c$. So we'd probably start with $c=1$.
From this, we get
$$
x^6+(b+e)x^5+(1+be+f)x^4+(e+bf+g)x^3+(f+bg+8)x^2+(8b+g)x+8
$$
And so we immediately have $e=-b$ and $g=-8b$ (to make the $x^5$ and $x$ coefficients zero). $x^4$ gives $b^2=1+f$, while $x^2$ gives $8b^2=f+8$. Matching these two gives us $b^2=1$ and $f=0$. However, this would make our $x^3$ coefficient either $-9$ or $9$, and thus we arrive at a contradiction.
The same process can be taken for each possible $c$. It's certainly more complicated than the simple rational root case, but we don't have much choice, if we want to approach the solution process in this manner.
When we get to $c=2$, we get
$$
x^6+(b+e)x^5+(2+be+f)x^4+(2e+bf+g)x^3+(2f+bg+4)x^2+(4b+2g)x+8
$$
and so we have $e=-b$, $g=-2b$, $b^2=2+f$, $b^2=2+f$ (that's a repeat), and $b(f-4)=5$.
What you'll notice is that, this time, we get an extra degree of freedom - the $x^2$ and $x^4$ coefficients are zero for the same conditions. This is convenient, and means that we can always factor the equation in this form for some $b$ (which may not be an integer). But since we want an integer solution, we need only test for that case. Combining our two equations in $b$ and $f$ will produce a cubic in $b$, which is then susceptible to the rational root technique.
As it turns out, we get $b=-1$ as a solution, and thus our factor is $x^2-x+2$.
