Question

For b>c>0, both \[\huge x^2+bx+8\] and \[\huge x^2+cx+8\] factor over the integers. Find b-c

Answer 1

Well, the two ways to factor that over the integers would be\[(x+2)(x+4) = x^2+6x+8\]and\[(x+1)(x+8)=x^2+9x+8\]So \(b=9\) and \(c=6\) so \(b-c=3\)