Question

x^2-5x+k, k>0
find all the values for k so that the trinomial can be factored using integers

Answer 1

let x^2-5x+k=(x+m)(x+n)
this gives m+n=-5 and mn=k;
as k>0. so both m and n must be either +ve or -ve.
but +ve m,n is not possible as m+n=-5;
so m,n is -ve,
(m,n)=(-1,-4) or( -2,-3) or (-3,-2) or (-4,-1);
this gives k={4,6}