April 8th, 2017, 12:19 PM  #11 
Senior Member Joined: Jan 2015 From: London Posts: 106 Thanks: 2  
April 8th, 2017, 12:33 PM  #12  
Math Team Joined: Jul 2011 From: Texas Posts: 2,656 Thanks: 1327  Quote:
$[(x+3)^29]2 = (x+3)^2 + 9  2 = (x+3)^2 + 7 \ne (x+3)^2 + 11$ (2) because I prefer putting everything inside $[bracket]$. if you want to leave the +2 outside the $[bracket]$ ... $2  6x  x^2 = x^2  6x + 2$ $(x^2 + 6x) + 2$ $(x^2 + 6x + 9  9) + 2$ $\left[(x^2 + 6x + 9)  9\right] + 2$ $\left[(x+3)^2  9\right] + 2$ $(x+3)^2 + 9 + 2$ $(x+3)^2 + 11$  
April 8th, 2017, 01:28 PM  #13  
Senior Member Joined: Jan 2015 From: London Posts: 106 Thanks: 2  Quote:
Thanks for all the help  

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