Prove that any polynomial of odd degree has atleast one real root
Solution:
We want to show that if "P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\\cdots+a_{1} x+a_{0}" is a polynomial with n odd and "a_{\\mathrm{n}} \\neq 0" , then there is a real number c, such that P(c)=0
Firstly we know that every polynomial is continuous on the real line. We also know that
"\\lim _{n \\rightarrow \\infty} \\frac{P(x)}{a_{n} x^{n}}=1 \\text { and } \\lim _{n \\rightarrow-\\infty} \\frac{P(x)}{a_{n} x^{n}}=1"
Consequently for "|x|" large enough, "P(x)" and "a_nx^n" have the same sign. But "a_nx^n" has opposite signs for positive "x" and negative "x". Thus it follows that if "a_n>0", there are real numbers "x_0<x_1" such that "P(x_0)<0" and "P(x_1)>0". Similarly, if "a_n<0", we can find "x_0<x_1" such that "P(x_0)>0" and "P(x_1)<0". In each case, it now follows directly from the Intermediate Value Theorem that (for d = 0) there is a real number "c\\in\\left[x_{0}, x_{1}\\right]\\ with\\ P(c)=0"
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