# For a polynomial $p(x)$ of odd degree, how many real roots does $(p(x))^2$ have?

I know that an odd degree polynomial $p(x)$ has at least one real root; what about $(p(x))^2$? It is even, so can it not possess any real roots or have two real roots? Also, let $p(x)$ be an odd degree polynomial and let $q(x) = (p(x))^2 + 2p(x) − 2$. How many real roots does $q(x)$ possess?

• $p(x_0)=0\implies p(x_0)^2=0$ – lulu Apr 19 '16 at 12:55

The polynomial $(p(x))^2$ will have all the same roots, and only the roots that $p (x)$ possesses. If you have for some $x_0$, $p (x_0)=0$, then for $(p (x _0))^2$ just substitute in what you know about $p (x_0)$ to get that it equals 0 also.
• Yes. Remember exponents distribute over multiplication. So if you had $p$ in factored form you can easily see that $p^2$ has all the same factors but with double the multiplicity. – Will Fisher May 3 '16 at 1:54
The answer here is yes, this always holds. Just think of it this way: if $p(x) =0$ for some $a$ then $p(a)^2 = 0^2 = 0$. Your second question should fall to a similar analysis!