How do you factor #9t ^ { 2} + 16t + 4#?

1 Answer
Aymeric D.
May 14, 2017

#1/9(9t+8-2sqrt7)(9t+8+2sqrt7)#

Explanation:

#9t^2+16t+4 = 3^2t^2 + 2*3*(8/3)t+(8/3)^2-(8/3)^2+4 #
#= (3^2t^2 + 2*3*(8/3)t+(8/3)^2)-(8/3)^2+4 #
#= (3t+8/3)^2-4(4/3)^2+4 #
#=(3t+8/3)^2+4(1-16/9)#
#=(3t+8/3)^2+4((9-16)/9)#
#=(3t+8/3)^2-4/9*7#
#=(3t+8/3)^2-((2sqrt7)/3)^2#
#=(3t+8/3-(2sqrt7)/3)(3t+8/3+(2sqrt7)/3)#

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