Find a and b of $x^3+ax^2+bx−26=0$ I am doing practise papers and one of the questions is:


  The cubic equation $x^3+ax^2+bx−26=0$ has $3$ positive, distinct,
  integer roots.
  
  Find the values of $a$ and $b$


The mark scheme says:


  $3$ roots are $1, 2, 13$. Equation is $x^3-16x^2+41x-26=0$


I'm guessing this has something to do with the factor theorem thing where $x-a$ is a factor if $f(a)=0$ but with 3 unknowns I don't get how you find any of these.

I tried just putting $x$ as $1$:

$$
1^3+a+b-26=0
$$
$$
a+b=25
$$
then doing the same with $2$:
$$
2^3+4a+2b-26=0
$$
$$
4a+2b=18
$$
and after subsituting the two equations I got the answer of a $a=-16$ and $b=41$. However, I realised this was just by chance that these numbers were factors (it didn't work with $x=3$ for example).

What is the proper way to solve this?
If $p$, $q$, $r$ are distinct positive integer roots of $P(x) = x^3 + ax^2 + bx - 26$, then we must have $$(x-p)(x-q)(x-r) =  x^3 - (p+q+r)x^2 + (pq+qr+rp)x - pqr = P(x),$$ hence equating the constant coefficient, $$pqr = 26 = 2 \cdot 13.$$  Since the prime factorization of $26$ contains only $2$ and $13$, it immediately follows that there is only one unique solution (up to permutation) for this equation that satisfies the given conditions, namely $\{p, q, r\} = \{1, 2, 13\}$ in some order.  The rest follows easily.
Here's a hint: every cubic can be factored in the following way: 
$$p(x) = (x-r_1)(x-r_2)(x-r_3)$$
You know, from the given, that $r_1$, $r_2$, $r_3$ are distinct positive integers, and you know that $r_1r_2r_3 = 26$.  It turns out that there is only one possible unordered triple $(r_1, r_2, r_3)$ that satisfies these properties – can you show this?

Once you find these roots, you can use Vieta's, as others have mentioned, to find the values of $a$ and $b$.