Find the values of b and c.

Use the Factor Theorem.

I think Viete's Formulae might come in handy:
If the roots of the polynomial below are x_1, x_2, x_3 then by the fundamental theorem of algebra:
$ax^3+bx^2+cx+d=a(x-x_1)(x-x_2)(x-x_3)$ (*)
and therefore
$x_1+x_2+x_3=-\frac ba$
$x_1x_2+x_2x_3+x_3x_1=\frac c a$
$x_1x_2x_3=-\frac d a$

If you haven't seen this before, check it out! it's completely straightforward algebraic manipulation of (*).

You then only need to denote the third root (which you know it's real, but it actually doesn't matter) with say y and substitute in the above to get:
-1-2+y=-b/2
2-y-2y=-5/2
2y=-c/2

so, from the second one y=3/2 then put this in the first to get b=3 and finally in the last to get c=-6.

I will try to lead you through this problem anyway. I'm not well aware what your level of understanding is, so I enclosed by [] ideas that might be hard for you to comprehend. Feel free to ignore whatever i wrote between square brackets. I will start with a few definitions:

def1.a polynomial is an function of the form
$f(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+x_0$ [with coefficients $a_i$ in a ring R which might be $\mathbb{Z}$, or $\mathbb{Q}$, or $\mathbb{R}$, or ultimately $\mathbb{C}$]
def2.a root of a polynomial is a [complex] number c such that $f(c)=0$
def3.A factor g of the polynomial f is a polynomial such that g|f, i.e. there exists a third polynomial h such that gh=f (or if you want g(x)h(x)=f(x) for all x.

Now onto the theorem you have trouble with:

thm1. c is a root of f IF AND ONLY IF (x-c) (which is itself a polynomial) is a factor of f.

A proof of this might be complicated and quite long-winded . However I will try to help you understand the its mechanics.

In the expression for f in def1. take

e.g.1.n=1 and assume $a_1\neq0$. Then
$f(x)=a_1x+a_0=a_1(x-(-\frac {a_0}{a_1}))$ (trivial algebraic manipulations)
so $(x-(-\frac{a_0}{a_1}))$ is a factor of f (take h=a_1 in def.3)
On the other hand,
$f(c)=0\Leftrightarrow a_1c+a_0=0 \Leftrightarrow a_1c=-a_0 \Leftrightarrow c=-\frac{a_0}{a_1}$.
We can summarize the above as
c is a root of f $\Leftrightarrow c=-\frac{a_0}{a_1} \Leftrightarrow$ (x-c) is a factor of f.

e.g.2. n=2. I will not work out the general case, but rather a simple example. The general case might be a good exercise for you, if you are indeed interested in learning maths.
Set $f(x)=x^2-3x+2$. Then f has roots
$\frac{b\pm\sqrt{b^2-4ac}}{2a}$= 1 or 2. [i trust you know how to solve quadratics. if not I might help you with that too].
Now to check thm1., note that:
$f(x)=x^2-x-2x+2=x(x-1)-2(x-1)=(x-1)(x-2)$.
Therefore (x-1) is a factor of f [take], and (x-2) is a factor of f [take]. As claimed!

Now onto the final bit, the problem we have to solve:

We are given that x+1 is a factor of $f(x):=2x^3+bx^2-5x+c$. Then, according to thm1., -1 is a root of f, and equivalently,
$f(-1)=0\Leftrightarrow -2+b+5+c=0 \Leftrightarrow b+c=-3$ (1)
Likewise, x+2 is a factor of f, so -2 is a root of f, i.e.
$f(-2)=0 \Leftrightarrow -16+4b+10+c=0 \Leftrightarrow 4b+c=6$ (2)
So you end up with a system of 2 simultaneous linear equations with two unknowns, which you can solve, for instance by subtracting (1) from (2) and get 3b=9, so b=3 and substituting this back in (1) to get c=-3-b, so c=-6, as before!

I think this is a much more appropriate approach for you and I hope the above will be of help. Concerning my previous solution, I think it rather defeats the purpose of the question (for which I apologise). However, if you are interested in learning more about The Fundamental Theorem of Algebra and Viete's Formulae, you can look here:
http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra
http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formulas

If x+1 is a factor then putting x= -1 in the equation should give the answer 0
If x+2 is a factor then putting x= -2 in the equation should also give the answer 0

Do this and you have a pair of simultaneous equations:-

-2+b+5+c = 0
-16+4b+10+c = 0

Solve for b and c

OK?

MP