how to find out the following statements are true or false? Let $p(x)$ be an odd degree polynomial and let $q(x)=(p(x))^2+ 2p(x)-2$ 
a) The equation $q(x)=p(x)$ admits atleast two distinct real solutions.
b) The equation $q(x)=0$ admits atleast two distinct real solutions.
c) The equation $p(x)q(x)=4$ admits atleast two distinct real solutions.
which of the following are true?
i know that all the three are true but donot know how to prove them
 A: $$q=p\iff p^2+p-2=(p-1)(p+2)=0$$
If $\;p(x)\;$ is given and fixed the above can easily be false, but if you meant $\;p(x)\;$ can variate then it is true.
Again, as before, this depends on whether $\;p(x)\;$ is fixed or if it is asked whether there exists an odd degree polynomial such that etc. Anyway, if 
$$\;\deg p=1\implies \deg qp=3\implies qp-4=0\;\;\text{can have at most three different solutions}$$
With (b) is the same, but $\;x^3-1\;$ is an odd degree polynomial with one single real root.
A: a) Let $q(x)=p(x)$, then we have $(p(x))^2+p(x)-2 = (p(x)-1)(p(x)+2) = 0$. From here $p(x) = 1$ or $p(x) = -2$. Since $p(x)$ is an odd degree polynomial, it has at least one solution in each case, and the overall equation $q(x)=p(x)$ has at least two real and distinct solutions.
b) I do not see how $p(x) = 0$ can have at least two distinct solutions. However, $q(x) = 0$ does. $(p(x))^2+2p(x)-2 = 0$ has two distinct roots for $p(x)$ and, thus, following the same logic as in a), we conclude that $q(x) = 0$ has at least two distinct roots.
c) $q(x)p(x) = (p(x))^3+2(p(x))^2-2p(x)-4 = 0$, which can be written as $(p(x)+2)((p(x))^2-2) = 0$. This equation has three distinct roots for $p(x)$ and thus $q(x)p(x)=0$ has at least three distinct solutions.
Hope this helps.
