1. You have, and can be written as .
From this you get, .
2. There are infinitely many numbers. Some of them are :- 2, 3, 5, 7, 13, 17, 113......
3. There are three different ways which I know to solve this problem.
→ 1. Find local points of local extrema, sort them up, substitute in the given expression and find how many times the expression changes its sign from -∞ to +∞.(generally used for polynomials of lower degree)
→ 2. Descartes's method. (Simplified form of the method mentioned above, usually used when the expression has simplified terms)
→ 3. We have something that says, " For a polynomial f(x) if f(a).f(b)<0 then there exists a real root between 'a' and 'b' ". ( I am gonna apply this. I admit it's time-consuming but you are 13 and might not probably understand the above two methods )
Now, start from -∞, you can see that expression is positive.
Take -1, even now f(x)>0.
Take 0, f(x)<0, observe that sign of f(x) has changed when moving from -1 to 0, this means they're is a root between -1 & 0.
Take 1, f(x)<0, no root between 0 & 1.
Take 3, f(x)>0, there's another root.
You may verify that expression remains positive from here on, indicating there no root between 3 and +∞.
Hence, there are only two real roots possible and the other two are complex.
4. You have, , first p was noted to be 17 and the roots were found to be -2 & -15.
You get, q=(-2)(-15)=30. Then p was corrected as 13. New equation you get is .
Roots of this equation are -10 & -3.