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
explanatation needed
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To apply the distributive law, we multiply each term in the brackets by xchdn20 wrote: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
x{x-(5x+6/x)} = 0
x^2 - (5x+6) = 0
x^2 - 5x - 6 = 0
etc