Question 89056



  factor completely: 
y3-12y^2-36y          y^3-12y^2-36y
                      = y(y^2-12y-36)
    use the std formula y = -b+-sq rtb^2-4ac/2a   where a = 1  b = -12 c = -36  
                        
                       y = -(-12)+-sq rt(-12)^2-4.1(-36)/2.1

                         = 12+-sq rt144+144/2

                         = 12+-sq rt2.144/2
                         = 12+-12.sq rt2/2
                         = 12(1+-sq rt2)/2
                         = 6(1+-sq rt2)

     therefore  the value of y = 0 or value of y = 6(1+-sq rt2)