Proof. As in 1.6, we factorize g(x)g(x):
where a1,…,asa_{1},\ldots,a_{s} are distinct real numbers and Q1(x),…,Qt(x)Q_{1}(x),\ldots,Q_{t}(x) are distinct irreducible quadratic polynomials. We will prove the theorem by induction on s+ts+t.
Assuming s≥1s\geq 1 then we consider the polynomials (x-as)ms(x-a_{s})^{m_{s}} and
Then (x-as)msg1(x)=g(x)(x-a_{s})^{m_{s}}g_{1}(x)=g(x). If t≥1t\geq 1 then we set
so that g2(x)Qt(x)nt=g(x)g_{2}(x)Q_{t}(x)^{n_{t}}=g(x).