Factoring a polynomial over real numbers (no real roots)

Question

Polynomial is: $$4x^4+2x+\frac{15}{16}$$ I know that the degree of the highest irreducible polynomial over reals is 2, so it should be possible to factor this polynomial into two second degree polynomials?

Answer 1

Using the Lagrange resolvent method, the characteristic polynomial is

$$s^6-\frac{15}{16}s^2-\frac14,$$ having the real root

$$s=\sqrt{\sqrt[3]{\frac{\sqrt{61}i}{64}+\frac18}+\sqrt[3]{-\frac{\sqrt{61}i}{64}+\frac18}}.$$

Hence the factorization

$$4(x^2+sx+\frac{s^2}2-\frac1{4s})(x^2-sx+\frac{s^2}2+\frac1{4s}).$$

Answer 2

WOLG, let $4x^4+2x+15/16=(2x^2+ax+b)(2x^2+cx+d)$ be the disired factoring, then $$ a+c=0,\quad 2d+2b+ac=0,\quad bc+ad=2,\quad bd=15/16. $$ The rest is yours (you need to solve a degree 3 equation of $a^2$, if there is no obvious rational root, then the problem is not interesting.).