Finding coefficients of a quadratic with roots having certain intervals If the roots of the quadratic equation $$(4p−p^2  −5)x^2  −(2p−1)x+3p=0$$ lie on either side of unity, then the number of integral values of $p$ is?

Okay so I'm having a hard time what the question means by both sides of unity. Does it mean one root is on $ x&gt;1 $ and other in $ x&lt;1$? After that, how do I adjust the coefficent so that one root is greater tha one?
First notice that leading coefficient $a=-(p^2-4p+4+1) = -(p-2)^2-1$ is negative.
If the roots of quadratic function $$f(x)=(4p−p^2  −5)x^2  −(2p−1)x+3p$$ are $x_1$ and $x_2$ (say $x_1&gt;x_2$) then we have $$x_2&lt;1&lt;x_1\implies 1-x_2&gt;0 \;\;{\rm and}\;\; 1-x_1&lt;0$$
So if we write write $f$ in factor form:$$f(x)=a(x-x_1)(x-x_2)$$
then we see that $$f(1) = a(1-x_1)(1-x_2)&gt;0$$
So you need to solve $f(1)=-(p-4)(p-1)&gt;0$ ...
If the leading coefficient is positive, then the condition $f(1)\lt 0$ suffices. If it is negative, then we want $f(1)\gt 0$. These conditions can be condensed into $$(4p-p^2-5) f(1)\lt 0 $$ Can you finish?