How do you factor #-x^{2}-12x-18=0#?

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Answer 1 · 2018-04-22T14:29:37

The solution is #x=-6±3sqrt2#

and hence the factorization is
#(x+6-3sqrt2)(x+6+3sqrt2))=0#

#-x^2-12x-18=0#
#x^2+12x+18=0#

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

#x= (-12+-sqrt(144-72))/(2 times1)#

#x=(-12+-sqrt72)/2#

#x= (-12+-6sqrt2)/2#

#x=-6±3sqrt2#

These are the solutions when #-x^2-12x-18=0#

The factorization can be written in the form #(x-a)(x-b)# where #a# and #b# are the solutions, also known as zeros.

#(x-(-6+3sqrt2))(x-(-6-3sqrt2))=0#

#(x+6-3sqrt2)(x+6+3sqrt2))=0#