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:

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

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