# 9.5: Application to Real Polynomials

• Bob Dumas and John E. McCarthy
• University of Washington and Washington University in St. Louis

If $$p$$ is a polynomial in $$\mathbb{R}[x]$$, it follows from the Fundamental Theorem of Algebra that it does have roots, but they may be complex. If it has complex roots, they must occur in complex conjugate pairs.

THEOREM 9.46. Let $$p \in \mathbb{R}[x]$$. Let $$\alpha$$ be a root of $$p$$. Then so is $$\bar{\alpha}$$.

PROOF. Let $$p(x)=\sum_{k=0}^{N} a_{k} x^{k}$$. Then $p(\alpha)=\sum_{k=0}^{N} a_{k} \alpha^{k}=0,$ SO $p(\bar{\alpha})=\sum_{k=0}^{N} a_{k} \bar{\alpha}^{k}=\overline{p(\alpha)}=0 .$ Let $$\alpha=a+i b$$. Then \begin{aligned} (x-\alpha)(x-\bar{\alpha}) &=(x-(a+i b))(x-(a-i b)) \\ &=x^{2}-2 a x+a^{2}+b^{2} \\ &=(x-a)^{2}+b^{2} . \end{aligned} So applying the Fundamental Theorem of Algebra to the real polynomial $$p$$, we first factor out the real roots, and for each pair of complex conjugate roots we get a factor as in (9.47). Thus we get:

THEOREM 9.48. Let $$p \in \mathbb{R}[x]$$ be a polynomial of degree $$N$$. Then $$p$$ can be factored into a product of linear factors $$\left(x-c_{k}\right)$$ and quadratic factors $$\left(\left(x-a_{k}\right)^{2}+b_{k}^{2}\right)$$ : $p(x)=c\left(\prod_{k=1}^{N_{1}}\left(x-c_{k}\right)\right)\left(\prod_{j=1}^{N_{2}}\left(\left(x-a_{j}\right)^{2}+b_{j}^{2}\right)\right)$ for some (not necessarily distinct) real numbers $$c, c_{j}, a_{j}, b_{j}$$. We have $$N_{1}+2 N_{2}=N$$, and the factoring is unique, up to ordering and replacing any $$b_{j} b y-b_{j}$$.

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