Question 146478
Zero factor property says: if a & b are real numbers and ab=0 then, a=0 or b=0.
 1) 3x^2 + 5x -2 = 0.
    3( x^2 + 5/3x -2/3) = 0.if a & b are the zeros of above polynomial, ab = -2/3.

if ab= -2/3 , the possible values of a & b are + 1/3 ,-2 , - 1/3 ,2.
  p(x) = x^2 + 5/3 x -2/3
  p( 1/3) = (1/3)^2 + 5/3 * 1/3 -2/3
          =  1/9 + 5/9 -2/3
          =   6/9 - 2/3 = 6 -6 /9 = 0. 

p(1/3 ) = 0 i.e.(  x- 1/3 ) is a factor of p(x).
p(2) = 2^2 + 5/3 ^2 -2/3 = 4 +10/3 -2/3 = 4 +8/3 not equal to 0. i.e 2 is not a factor.
p(-2) = (-2)^ 2 + 5/3 * -2 - 2/3
      = 4 - 10/3 - 2/3 
      = 4 -12/3 = 12-12/3 =0. 
p(-2)=0  i.e. x+2=0 is a factor of p(x).
so x= -2 or 1/3.

2) 2y^2 -11y -40= 0
  2( y^2 - 11/2 y -20 )=0.
if a & b are the zeros then  ab = -20
possible values of a & b are 10,2, -10,-2,5,4,-5,-4,-20,-1,20,1,8,5/2,-8,-5/2

by trial we find that p(8) = 64 -11/2*8 -20 = 64 -44 -20 =0.
p(8)=0  i.e. x-8 =0 , x=8.

p(-5/2 ) = 25/4 +11/2* 5/2 -20 = 25/4 +55/4 -20 = 80/4 -20 = 20-20 = 0.
p(-5/2)=0 i.e. x + 5/2 =0 , x=-5/2.

solution is x= 8 or -5/2.