How do you factor the trinomial #x^ 2+ 6x +27#?

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Answer 1 · 2015-12-16T05:04:06

can't be factored

D = b^2 - 4ac = 36 - 108 < 0.
Since D is not a perfect square, this trinomial can't be factored.

Answer 2 · 2015-12-16T05:15:12

#x^2+6x+27 = (x + 3 -3sqrt(2)i)(x + 3 +3sqrt(2)i)#

To factor the trinomial, we can use the quadratic formula:

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

where:
#a=1#
#b=6#
#c=27#

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

#x=(-(6)+-sqrt((6)^2-4(1)(27)))/(2(1))#

#x=(-6+-sqrt(36-108))/2#

#x=(-6+-sqrt(-72))/2#

#x=(-6+-6sqrt(-2))/2#

#x=(-6+-6sqrt(2)sqrt(-1))/2#

#x=(-6+-6sqrt(2)i)/2lArrsqrt(-1)=i#

#x=(2(-3+-3sqrt(2)i))/((2)1)#

#x=(color(red)cancelcolor(black)2(-3+-3sqrt(2)i))/((color(red)cancelcolor(black)2)1)#

#x=-3+-3sqrt(2)i#

Hence factors:

#(x + 3 -3sqrt(2)i)(x + 3+3sqrt(2)i)#