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

by DivMit   Last Updated August 14, 2019 07:20 AM

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).

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#### Answers 3

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

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

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: $$x^4+px^2+q=(x^2+\sqrt{q})^2-2x^2\sqrt{q}+px^2=(x^2+\sqrt{q})^2-(2\sqrt{q}-p)x^2=$$ $$=\left(x^2-\sqrt{2\sqrt{q}-p}x+\sqrt{q}\right)\left(x^2+\sqrt{2\sqrt{q}-p}x+\sqrt{q}\right)$$

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