The cubic equation x^3+ax^2+bx-26=0 has 3 positive,distinct integer roots.
Find the values a and b.
(considering that a and b could be any value i cant understand how to find out what the integer roots are)

Question

The cubic equation x^3+ax^2+bx-26=0 has 3 positive,distinct integer roots.
Find the values a and b.
(considering that a and b could be any value i cant understand how to find out what the integer roots are)

kainui · Answer

Hmm, well I'm trying some stuff out right now, this is an interesting question.

ganeshie8 · Answer

descartes rule of signs tells you $a$  must be negative, and $b$ must be positive

ganeshie8 · Answer

rational root theorem gives you the possible roots  :  $\pm 1, \pm 2, \pm 13$

parthkohli · Answer

The product of roots is 26.  By inspection, there are only three positive integral factors.

ganeshie8 · Answer

$$x^3+ax^2+bx-26 = (x-1)(x-2)(x-13)$$

compare coefficients

parthkohli · Answer

Alternative: exploit the sum of roots and cyclic sum of roots.  :P

epoweritheta · Answer

x1x2x3=26 ; x1+x2+x3=-a;x1x2+x2x3+x1x3=b ;
26=1*2*13 only possible positive dinstinct integer combination , implies a=-16 ,b=41

note the problem is possible to solve only when these 3 conditions are given : positive dinstinct integer