2009/024) Resolve into factors X^4-3x+20

if this can be factored with real coefficients then it can be as product of 2 quadratic
we shall get
(x^2+ax+b)(x^2+cx+d)

now coeffcient of x^3 = 0 so a +c = 0 or c = -a

so we get(x^2+ax+b)(x^2-ax+d)
= x^4 + (d + b - a^2) x^2 + x(ad-ab) + bd

comparing coefficients

a^2 = b + d
bd = 20
so ad -ab = - 3
by trial and error we see that a = 3 , b= 5 and d= 4 (this is using 20 = 1 * 20 = 2(*10 - 4 * 5 outof which only 4 +5 = 9 we get perfect square) satisfies
this can be factored with rational coefficient as
(x^2+3x+5)(x^2-3x+4)
and not further this can be checked by quadaric factorisation

hence X^4-3x+20 = (x^2+3x+5)(x^2-3x+4)