Let $P(x)$ be a polynomial with real coefficients such that $P(\alpha)\ge0$ for all real $\alpha$. Show that there exists polynomials with real coefficients $Q_1(x)^2, Q_2(x)^2,....Q_n(x)^2$ such that $P(x) = Q_1(x)^2 + Q_2(x)^2 + ... + Q_n(x)^2$
Asked
Active
Viewed 128 times
1
-
What have you tried? – user113102 Nov 05 '19 at 16:45
-
1If $P(\alpha)\geqslant 0$ for all $\alpha$, then $\deg P$ is even. – Luke Collins Nov 05 '19 at 16:45
1 Answers
1
Write $P(x)$ in factorised form over the reals. Note that real roots must all have even multiplicity.
The factors can be collected together as $(x-a)^2$ and as $(x-a)^2 +b^2$ for real $a$ and $b$.
Incorporating the leading coefficient, $P(x)$ is therefore a product of the form $$(f_1(x)^2+g_1(x)^2)...(f_k(x)^2+g_k(x)^2)$$ Multiplying out appropriately gives you the required sum of squared polynomials.