How do you factor #12+ 16x - 3x ^ { 2}#?

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Answer 1 · 2017-06-15T10:44:52

#=(6-x)(2+3x)#

#12+16x-3x^2#

#=12-2x+18x-3x^2#

#=2(6-x)+3x(6-x)#

#=(6-x)(2+3x)#

Answer 2 · 2017-06-15T10:50:00

The equation can be factorised to #(-3x - 2)(x - 6)#

Use the general formula
#x = (-b +- sqrt(b^2 -4ac))/(2a)#

Substitute in the given values
#x = (-16 +- sqrt(16^2 -4*-3*12))/(-6)#

simplify
#x = (-16 +- 20)/(-6)#

#x_"1" = 6#
#x_"2" = -2/3#

Put those values back into #a(x-x_"1")(x-x_"2")#

gives

#-3(x + 2/3)(x - 6)#

rearranging
#(-3x - 2)(x - 6)#