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 5 of 5: Take the maximum and finish
a≤(n+1)c⟹ac≤n+1a\le(n+1)c\Longrightarrow \frac{a}{c}\le n+1
Detailed analysis

The preceding cases all establish a≤(n+1)ca\le(n+1)c. Since c>0c>0 for the non-zero polynomial gg, dividing by cc gives ac≤n+1\frac{a}{c}\le n+1, equivalently ac≤n+1\frac{a}{c}\le n+1 under the source's normalized coefficient notation.