Find the upper bounds for the real roots of the equation $2x^3-5x^2-7x+4 = 0$ Problem: Find the upper bounds for the real roots of the equation $2x^3-5x^2-7x+4 = 0$

Reading the definition in the book, If upon dividing f(x) by (x-r), r is an upper bound if it is greater than or equal to 0 and the coefficients of the quotients of the quotient as well as the remainder are positive.

Trial: The question gave an equation but no given (x-r) or r. What I did is get the roots of the cubic equation.

Root 1 = 3.36374665 

Root 2 = 0.4519194497

Root 3 = -1.3256661

It says that for r to be upper bound, first it should be greater than zero, so I ignored the root -1.32

@root 1; f(r) = $2(3.36374665)^3-5(3.36374665)^2-7(3.36374665)+4 = 1.0014 x 10^-9$

And with that I know what I'm doing is wrong so I stopped. So, how to answer these kinds of problem? By the way, the choices are : 4,2,3 and 5.

Any help would be appreciated.
4 is a upper bound.
By rational root theorem, the possible zeros are $\frac{4}{1},-\frac{4}{1},\frac{4}{2}, -\frac{4}{2}$ or $4, -4, 2, -2$

By the rules of bounds, you divide the polynomial by $4$, $(x-4)$, by using synthetic division, and the numbers of the last row are all positive or zero. Therefore, $4$ is a upper bound.
Check you definition of an upper bound.  An upper bound of a set of reals is any number greater than or equal to all members of the set.  When the set is finite the set of upper bounds is all values from the greatest to $\infty$.