2016/105) Solve $x^4+4x^3+3x^2-14x+26=0$

As it is found that this does not have a rational factor and as we know that complex roots appear as conjugate pairs
so it can be factored to quadratic with real coefficient
so Let $x^4+4x^3+3x^2-14x+26$
$ = (x^2+px + q) (x^2 + sx + t)$ as coeffiefint of $x^4$  = 1
$= x^4 + x^3(s + p) + x^2(ps + q + t) + x(pt + qs) + qt$
Comparing coefficients
$s+ p = 4$
$ps +q + t = 3$
$pt+qs = - 14$
$qt = 26 => ( 1,26),(-1,-26),(2,13), (-2, - 13)$
it can be easily solved if we have rational coefficient (above 4 combinations)
by taking $q = 1, t = 26$ or $q = -1 t = -26$ you shall not find any solution(I have tried) but
if we take $q = 2, t = 13$ we get
$s+ p = 4 \cdots(1)$
$sp + 15 = 3$ OR  $ps = -12 \cdots(2)$
$13p + 2s = - 14 \cdots(3)$
From (1) and (3) you get p = - 2 and s = 6 and it satisfies (2)
So we get quadratic factor as $(x^2-2x+2)(x^2+6x+13) = 0$
So we are lucky to get factors with rational coefficients
$(x^2-2x+2) = 0$ gives 2 roots $1\pm i$
And $(x^2+6x+13) = 0$  gives 2 roots $-3\pm 2i$