What is the set of values of $p$ for which $$p(x^2+2)<2x^2+6x+1,$$ for all real values of $x$?
So I found this one pretty tricky. Feel free to have a look yourself before I outline my thinking.
So I started by taking everything to the RHS forming a quadratic inequality $>0$. Then as long as $2-p$ is positive we know the discriminant must be negative yielding $p<-1$ or $p>\frac{7}{2}$ but we specified $2-p$ is positive so $p<-1$ is the only range of p satisfying the original inequality.
The answer given in the textbook is $-1<p<\frac{7}{2}$. Is this an error or is there something I’ve missed?
Your approach is entirely correct, and seems like the best approach to me as well. I'll write out the details here, if only for myself as I don't have any pen and paper at hand.
Bringing all terms to one side, for each value of $p$ we get a quadratic polynomial in $x$, and we want $$(2-p)x^2+6x+(1-2p)>0,$$ for all $x$. This happens if and only if $2-p>0$ and the discriminant of the quadratic is negative, i.e. $$(-6)^2-4(2-p)(1-2p)<0.$$ The latter is a quadratic in $p$, and expanding the products yields $$-4(2p^2-5p-7)<0.$$ By the quadratic formula we see that this inequality holds if and only if $$p<\frac{5-\sqrt{25+56}}{4}=-1\qquad\text{ or }\qquad p>\frac{5+\sqrt{25+56}}{4}=\frac72.$$ We already found the necessary condition that $2-p>0$, so together this shows that $$p(x^2+2)<2x^2+6x+1,$$ if and only if $p<-1$. It seems that you are correct, and your textbook is wrong.
Just as a sanity check; plugging in $p=0$ yields the inequality $$0<2x^2+6x+1,$$ which should hold for all $x$ according to the solution in your textbook. But this inequality fails for $x=-1$.