explanatation needed

[chdn20]

How is the x distributed the answer from the book is:

x{x-(5x+6/x)}=0

x^2-(5x+6)=0
x^2-5x-6=0
(x-6)(x+1)=0

Why is the x^2 on the second portion?


The question is from Manhattan GMAT prep Guide 1 page 129.

Thanks

[Brent@GMATPrepNow]

How is the x distributed the answer from the book is:

x{x-(5x+6/x)}=0

x^2-(5x+6)=0
x^2-5x-6=0
(x-6)(x+1)=0

Why is the x^2 on the second portion?
The question is from Manhattan GMAT prep Guide 1 page 129.
Thanks

To apply the distributive law, we multiply each term in the brackets by x
x{x-(5x+6/x)} = 0
x^2 - (5x+6) = 0
x^2 - 5x - 6 = 0
etc