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1.10 Sketch of proof of Theorem 1.9 (non-examinable)

Proof. As in 1.6, we factorize g(x)g(x):

g(x)=(x-a1)m1(x-as)msQ1(x)n1Qt(x)ntg(x)=(x-a_{1})^{m_{1}}\ldots(x-a_{s})^{m_{s}}Q_{1}(x)^{n_{1}}\ldots Q_{t}(x)^{% n_{t}}

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 s1s\geq 1 then we consider the polynomials (x-as)ms(x-a_{s})^{m_{s}} and

g1(x)=(x-a1)m1(x-as-1)ms-1Q1(x)n1Qt(x)nt.g_{1}(x)=(x-a_{1})^{m_{1}}\ldots(x-a_{s-1})^{m_{s-1}}Q_{1}(x)^{n_{1}}\ldots Q_% {t}(x)^{n_{t}}.

Then (x-as)msg1(x)=g(x)(x-a_{s})^{m_{s}}g_{1}(x)=g(x). If t1t\geq 1 then we set

g2(x)=(x-a1)m1(x-as)msQ1(x)n1Qt-1(x)nt-1g_{2}(x)=(x-a_{1})^{m_{1}}\ldots(x-a_{s})^{m_{s}}Q_{1}(x)^{n_{1}}\ldots Q_{t-1% }(x)^{n_{t-1}}

so that g2(x)Qt(x)nt=g(x)g_{2}(x)Q_{t}(x)^{n_{t}}=g(x).