The polynomial $$f(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$$ has the roots $$\alpha_1,\ldots,\alpha_n$$ What roots does the polynomial $$g(x)=a_nx^n+a_{n-1}bx^{n-1}+a_{n-2}b^2x^{n-2}\ldots+a_1b^{n-1}x+a_0b^n$$ has?

I tried with a second degree polynomial and obtained that the roots of $g(x)$ would be $b(\alpha_1,\ldots,\alpha_n)$.

I could use a third degree polynomial as another example but it wouldn´t be a proof for this. I also thought induction could be a way but I´m not sure it aplies for this kind of proof.

I´d appreciate your help, thank you.

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    $\begingroup$ Observe that the term $a_0b^{n}$ is the product of all roots of g(x). If you could assume or given that the roots are all the same, then the root could be easily determined. $\endgroup$ – NoChance Oct 21 '19 at 16:33


With $b \ne 0$ $$ \sum_{k=0}^n a_k b^{n-k}x^k = b^n\sum_{k=0}^n a_k\left(\frac xb\right)^k = b^n f\left(\frac xb\right) $$

  • $\begingroup$ Thank you @Cesareo $\endgroup$ – Octavio Berlanga Oct 21 '19 at 17:57
  • $\begingroup$ You are welcome! $\endgroup$ – Cesareo Oct 21 '19 at 17:59
  • $\begingroup$ While this is elegant, I am not sure how to proceed? Does this mean that $b.\alpha_{i}$ is a root for $g(x)$? $\endgroup$ – NoChance Oct 21 '19 at 20:10
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    $\begingroup$ If $f(x) = a_n\prod_{k=1}^n(x-\alpha_k)$ then $g(x) = b^na_n\prod_{k=1}^n\left(\frac xb-\alpha_k\right)$ $\endgroup$ – Cesareo Oct 21 '19 at 21:46
  • $\begingroup$ Thank you for the clarification. $\endgroup$ – NoChance Oct 22 '19 at 0:41

Hint: If $b=0$, you know that the roots of $g$ are $0$ with multiplicity $n$, so you can say that the roots are $b\alpha_1,b\alpha_2,\ldots,b\alpha_n$ in this case. If $b\neq 0$, observe that $$g(bx)=b^nf(x).$$


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