x^3+6x-7=0 To factor a quadratic of the form ax^2+bx+c=y you need to find two values, j and k, which satisfy two conditions...
ac=jk=-7 and b=j+k=6 so j and k must be 7 and -1
Now you replace the linear coefficient, bx, with jx and kx to get:
x^2-x+7x-7=0 now you can factor the 1st and 2nd pair of terms...
x(x-1)+7(x-1)=0
(x+7)(x-1)=0 So there are two solutions when x=-7 and when x=1.
Or in (x,y) notation the points (-7, 0) and (1, 0)