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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?

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    $\begingroup$ $p(x_0)=0\implies p(x_0)^2=0$ $\endgroup$ – lulu Apr 19 '16 at 12:55
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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.
The second part of your question is much more difficult and there's nothing that I know of that can generally be said about that (but I don't know everything; just know it's a lot more complicated :) ).

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  • $\begingroup$ so is it like the roots of p(x)^2 has all roots of p(x) but with multiplicity 2 $\endgroup$ – sindhu May 3 '16 at 1:51
  • $\begingroup$ 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. $\endgroup$ – Will Fisher May 3 '16 at 1:54
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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!

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