The funny thing about this problem is, that in order
to factor the binomial that results, you have think
of two numbers whose product is 21 and whose sum is 10,
which means you have to solve the problem in order to
solve it!!!!


   (x-3)(x-7) = 0
 

        xy = 21
       x+y = 10

Solve the second equation for y

       x+y = 10
         y = 10-x

Substitute 10-x for y in

        xy = 21
   x(10-x) = 21 
       10x-x² = 21
   -x²+10x-21 = 0

Multiply through by -1

    x²-10x+21 = 0
   (x-3)(x-7) = 0   <--- here is where you have to know the solution
                         to the problem already in order to do this factoring!!!
x-3 = 0     x-7 = 0
  x = 3       x = 7 

Substituting each in y = 10-x

y = 10-3   y = 10-7  
y = 7      x = 3

It looks like there's two solutions but
it's really only one pair of numbers, {3,7}

Edwin