# Roots of a polynomial.

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.

• 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. – NoChance Oct 21 '19 at 16:33

With $$b \ne 0$$ $$\sum_{k=0}^n a_k b^{n-k}x^k = b^n\sum_{k=0}^n a_k\left(\frac xb\right)^k = b^n f\left(\frac xb\right)$$
• While this is elegant, I am not sure how to proceed? Does this mean that $b.\alpha_{i}$ is a root for $g(x)$? – NoChance Oct 21 '19 at 20:10
• If $f(x) = a_n\prod_{k=1}^n(x-\alpha_k)$ then $g(x) = b^na_n\prod_{k=1}^n\left(\frac xb-\alpha_k\right)$ – Cesareo Oct 21 '19 at 21:46
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 $$g(bx)=b^nf(x).$$