MathLabs

Problem 3

Let f(x)=anxn+an−1xn−1+⋯+a0f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 and g(x)=cn+1xn+1+cnxn+⋯+c0g(x)=c_{n+1}x^{n+1}+c_nx^n+\cdots+c_0 be non-zero real polynomials such that g(x)=(x+r)f(x)g(x)=(x+r)f(x) for some real rr. If a=max⁡(∣an∣,…,∣a0∣)a=\max(|a_n|,\ldots,|a_0|) and c=max⁡(∣cn+1∣,…,∣c0∣)c=\max(|c_{n+1}|,\ldots,|c_0|), prove that ac≤n+1\frac{a}{c}\le n+1.
Step 2 of 5: Handle the case r=0
r=0⟹a=c⟹ac=1≤n+1r=0\Longrightarrow a=c\Longrightarrow \frac{a}{c}=1\le n+1
Detailed analysis

If r=0r=0, then c0=0c_0=0 and ck=ak−1c_k=a_{k-1} for 1≤k≤n1\le k\le n, while cn+1=anc_{n+1}=a_n. Thus the maxima agree: a=ca=c, and because ff is non-zero, a=c>0a=c>0. Hence ac=1≤n+1\frac{a}{c}=1\le n+1.