Is there any systematic way to factorise expressions of the form $ n^4+pn^2 +q$

Question

Is there any systematic way to factorise expressions of the form $ n^4+pn^2 +q$

Expression like $n^4+n^2+1$ usually become helpful for telescopic type , when used in the form$(n^2+n+1)(n^2-n+1)$, but for an expression like $n^4+pn^2 +q= (n^2+an+b)(n^2+cn+d)$ systematic way to factorise it like above (if possible).

Answer 1

Yes, of course! Always there exist a way to factorize it.

If $p^2-4q\geq0,$ so we can use a way, which is a similar to the following: $$x^4+4x^2+3=x^4+3x^2+x^2+3=(x^2+1)(x^2+3).$$ But for $p^2-4q<0$ there is a way, which is a similar to the following: $$ x^4+x^2+25=x^4+10x^2+25-9x^2=(x^2-3x+5)(x^2+3x+5) $$

For $ p^2-4q<0 $ in the general case we obtain: $$ x^4+px^2+q=(x^2+\sqrt{q})^2-2x^2\sqrt{q}+px^2=(x^2+\sqrt{q})^2-(2\sqrt{q}-p)x^2=$$ $$=\left(x^2-\sqrt{2\sqrt{q}-p}x+\sqrt{q}\right)\left(x^2+\sqrt{2\sqrt{q}-p}x+\sqrt{q}\right) $$

Answer 2

Substitute $n^2=t$ and solve the quadratic which can be factorized. You can also try this here $$n^4+n^2+1=(n^2+an+b)(n^2+cn+d)$$

Answer 3

This is a quadratic in $n^2,$ so you can always factorize it over the complex numbers.

Answer 4

You can derive it by expanding the brackets and making up the system of equations: $$n^4+pn^2 +q= (n^2+an+b)(n^2+cn+d)=n^4+(a+c)n^3+(b+ac+d)n^2+(ad+bc)n+bd \Rightarrow \\ \begin{cases}a+c=0\\ b+ac+d=p\\ ad+bc=0\\ bd=q\end{cases}$$ From the first: $c=-a$.

From the third: $ad-ab=0 \Rightarrow 1) \ a=0; 2) \ a\ne 0, \ b=d$.

$1) \ a=0\Rightarrow \begin{cases}b+d=p\\ bd=q\end{cases} \Rightarrow (b,d)=\left(\frac{p\pm\sqrt{p^2-4q}}{2},\frac{p\mp\sqrt{p^2-4q}}{2}\right)$.

Hence, for $\color{red}{p^2-4q\ge 0}$: $$\color{red}{n^4+pn^2+q=\left(n^2+\frac{p+\sqrt{p^2-4q}}{2}\right)\left(n^2+\frac{p-\sqrt{p^2-4q}}{2}\right)}$$ Example: $$n^4-5n^2+6=\left(n^2+\frac{-5+\sqrt{25-24}}{2}\right)\left(n^2+\frac{-5-\sqrt{25-24}}{2}\right)=(n^2-2)(n^2-3).$$ $2) \ a\ne 0, \ b=d \Rightarrow \begin{cases}c=-a\\ a=\sqrt{2b-p}=\sqrt{2\sqrt{q}-p}\\ b=\sqrt{q}\\ \end{cases}$ Hence, for $\color{blue}{2\sqrt{q}-p\ge 0 \iff p^2-4q\le 0}$: $$\color{blue}{n^4+pn^2+q=\left(n^2+\sqrt{2\sqrt{q}-p}n+\sqrt{q}\right)\left(n^2-\sqrt{2\sqrt{q}-p}n+\sqrt{q}\right)}$$ Example: $$n^4+n^2+1=\left(n^2+\sqrt{2\sqrt1-1}n+\sqrt{1}\right)\left(n^2-\sqrt{2\sqrt{1}-1}n+\sqrt{1}\right)=(n^2+n+1)(n^2-n+1).$$