How do you factor completely #27g ^ { 2} + 39g c + 14c ^ { 2}#?

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Answer 1 · 2017-01-02T06:04:26

#(3g + 2c)(9g + 7c)#

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

#27g^2 + 39gc + 14c^2#

#a = 27#
#b =39#
#c = 14#

Let's use #x# in the meantime

#x = (-39 +- sqrt(39^2 - 4(27)(14)))/(2*27)#

#x = (-39 +- sqrt(1521 - 1512))/54#

#x = (-39 +- sqrt9)/54#

#x = (-39 +- 3)/54#

#x = -36/54 = -2/3, x= -42/54 = -7/9#

#(3g + 2c)(9g + 7c)#