Factorization of polynomials with degree higher than 2 I need help to factorize $x^4-x^2+16$. I have tried to take $x^4$ as $(x^2)^2$ and factorize it in the typical way of factorizing a quadratic expression but that did not help. Can someone help me to factor this and also introduce me to the procedure that i need to follow to factorize expressions with degree higher than two?
 A: $$x^4-x^2+16$$
$$=[(x^2)^2+2\cdot x^2\cdot 4+4^2]-9x^2$$
$$=(x^2+4)^2-(3x)^2$$
$$=(x^2+4-3x)(x^2+4+3x)$$
by using $a^2-b^2=(a+b)(a-b)$.
A: $$x^4+8x^2+16-9x^2=(x^2+4)^2-(3x)^2=$$
$$=(x^2-3x+4)(x^2+3x+4)$$
A: Can be done by making perfect squares
$$
Let\ ax^2 +bx + c=0
\\Try\ to\ make\ the\ equation\ look\ in\ the\ form
\\ax^2 +2\sqrt {ca}x +c-(2\sqrt{ca}-b)x=0
\\You\ will\ see\ that\ ax^2 +2\sqrt {ca}x +c\ makes\ a \ perfect\ square
\\\therefore \quad  ax^2 +2\sqrt {ca}x +c=(\sqrt ax + \sqrt c )^2
\\Thus\ the\ equation\ will\ convert\ to\ (\sqrt ax + \sqrt c )^2-(2\sqrt{ca}-b)x=0
\\Consider \ the\ term\ (2\sqrt{ca}-b)x
\\We\ can\ write\ it\ as\ {[\sqrt{(2\sqrt{ca}-b)x}]}^2
\\Thus\ the\ equation\ is\ converted\ to\ form\ p^2-q^2=0
\\where\ p=\sqrt ax + \sqrt c \quad and\quad q=\sqrt{(2\sqrt{ca}-b)x}
\\Equation\implies [\sqrt ax + \sqrt c]^2 - {[\sqrt{(2\sqrt{ca}-b)x}]}^2=0
$$
Now you can apply $p^2-q^2=(p+q)(p-q)$ to get the required factorization.

Regarding your question Consider
$$ a= 1,b=-1 ,c=16 \ and\ usual \ substitution\ x=x^2\ (both\ x\ are\ different)
\\ \therefore p=x^2+4 \quad q=3\sqrt {x^2}=3x 
$$
After this its easy mathematics,you have to factorize 2 more easy equations thus giving you 4 roots.I hope , I did well to explain you :)
A: Can be done by a simple completion of squares. The presence of $x^4$ and 16 hints towards a possibility of a $(x^2+4)^2$ so just calculate that and compare it with your question function. $(x^2+4)^2 = x^4+8x^2+16$ which leaves us with a difference only in the $x^2$ term. This difference turns out to be $9x^2$. So our function reduces to $$(x^2+4)^2-(3x)^2$$
It's easy to see after this, a simple use of $a^2-b^2 = (a+b)(a-b)$ gives the final answer to be $$(x^2-3x+4)(x^2+3x+4)$$
