How to factor polynomials of degree larger than $2$?

Question

For instance, how do you factor a polynomial like $x^4+x^2+1$?

Secondly, is there any variation to the technique if it was a larger degree or not of the specific form I specified above?

Answer 1

$$x^4+x^2+1=x^4+x^2+1+x^2-x^2$$ $$=x^4+2x^2+1-x^2=(x^2+1)^2-x^2$$ $$=(x^2+x+1)(x^2-x+1)$$

Answer 2

$$x^4+x^2+1=x^4+2x^2+1-x^2=(x^2+1)^2-x^2=(x^2-x+1)(x^2+x+1).$$ There is also the following way. $$(x^2-1)(x^4+x^2+1)=x^6-1=(x^3-1)(x^3+1)=(x^2-1)(x^2-x+1)(x^2+x+1).$$

Answer 3

Let $z=x^2$

Then $$x^4+x^2=1$$ $$=z^2+z+1$$ Using the quadratic formula to solve when $az^2+bz+c$ equals zero $$z=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ gives $$z^2+z+1=0$$ when $$z=\frac{-1\pm \sqrt{-3}}{2}$$ $$z=\frac{-1 \pm \sqrt3i }{2}$$ Thus $$z^2+z+1=\big(z-\frac{-1+\sqrt3i}{2}\big) \big(z-\frac{-1-\sqrt3i}{2}\big)$$ $$x^4+x^2+1=\big(x^2-\frac{-1+\sqrt3i}{2}\big) \big(x^2-\frac{-1-\sqrt3i}{2}\big)$$ This can be taken further...

Answer 4

$$x^4+x^2+1=(x^2-x+1)(x^2+x+1)$$

Solving details: use method of undetermined coefficients $${{x}^{4}}+{{x}^{2}}+1=\left( {{x}^{2}}+a x+1\right) \, \left( {{x}^{2}}+b x+1\right) $$ $\Rightarrow$ $$a+b=0,\quad ab+2=1$$ $\Rightarrow$ $$a=1,\;b=-1$$ or $$ a=-1,\;b=1$$