Take all polynomials we're working with to be monic. Clearly, there is no loss of generality. A product of polynomials of the form $A^{2}+xB^{2}\ (*)$ is again of this form, so it suffices to write $f$ as a product of first or second degree such polynomials. Its second degree irreducible factors are of this form, its first degree factors of the type $x+t, t\ge 0$ are again of this form, and the factors of the type $x-t, t\ge 0$ appear with even exponents, as can easily be seen, so we can pair them up into factors of the type $(x-t)^{2}$, which are also of the type $(*)$.

First we prove the statement for polynomials of degree 1 and 2. Consider p(x)=ax+b of degree 1 such that p(x)>=0 for all x>=0. Then a>=0 and b>=0 because of this. So we can write p(x) to this form : p(x) = (b^.5)^2 + x(a^.5)^2 Now consider p(x) = ax^2 + bx + c , and p(x)>=0 for all x>=0. Then c>=0 and a>=0 and b^2-4ac >=0. So we can divide all coefficients by positive real number a!=0 , and consider p(x) = x^2 + bx + c . If p(x) have real roots we can write it to form (x-r1)(x-r2) and using previous expression. So we can consider that p(x) has complex (unreal) roots a+bi and a-bi (because p(x) is real polynomial) , so the polynomial p(x) is to the form p(x) = x^2 - 2ax + (a^2+b^2). So we can write it to the form: p(x) = (x - (a^2+b^2)^.5)^2 + x((2(a^2+b^2)^.5 - 2a)^.5)^2 (notice: we used that (a^2+b^2)^.5-a>=0 ) Now we claim that if p(x) and q(x) be real polynomials and we can write them to this form, then we can write p(x)q(x) to this form. Proof. p(x)=A(x)^2 + xB(x)^2 q(x)=C(x)^2 + xD(x)^2 so p(x)q(x) = (AC)^2 + x(AD)^2 + x(BC)^2 + (xBD)^2 = (xBD-AC)^2 + x(AD+BC)^2 this completes the proof of claim. Finaly, Consider the real polynomial p(x) be reduced in R[x], by multiplication of some degree 1 and degree 2 (for complex unreal roots) polynomials. p(x)=a(x^2-2a1.x+(a1^2+b1^2))^l1...(x^2-2ar.x+(ar^2+br^2))^lr(x-c1)^k1...(x-cs)^ks where r,s are index of ai 's and bi 's and ci 's.and li and ki are occurence of roots. The coefficent a is don't care because we can enter it to the expression. (a is possitive) ap(x)=(a^.5 A)^2 + x(a^.5 B)^2 and ci>=0 or li=0(mod 2) else we have sign variance in x=ci. and square polynomial can be written to this form: A^2 = (A)^2+x(0) So we can write p(x) to this form, because each of it factors can be written to this form.