factor x^2 + 3x + 6 over the complex numbers

[NEHA]

factor x^2 + 3x + 6 over the set of complex numbers.
ok i got stuck at the end..can you help me out

x^2 + 3x + 6 = 0
quadratic formula

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

x = -3 +- sqrt(3^2 - 4 (1*6) ) / 2(1)

x= -3 +- sqrt( 9 - 4(6) ) / 2

x = -3 +- qrt(9 - 24) / 2

x = -3 +- sqrt(-15)/ 2

now what?

 

[steve_b]

factor x^2 + 3x + 6 over the set of complex numbers.
ok i got stuck at the end..can you help me out

x^2 + 3x + 6 = 0
quadratic formula

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

x = -3 +- sqrt(3^2 - 4 (1*6) ) / 2(1)

x= -3 +- sqrt( 9 - 4(6) ) / 2

x = -3 +- qrt(9 - 24) / 2

x = -3 +- sqrt(-15)/ 2

===============

First, you should be putting parentheses around the first part of your expression, because the 2 at the end divides all of the first part:

x = (-3 +- sqrt(-15)) / 2

Now that we've gotten that (very important) technicality out of the way, you have to rewrite the sqrt(-15) as: i*sqrt(15), where "i" is the sqrt(-1). This will complete the answer:

x = (-3 +- i sqrt(15))/ 2

Steve

 

[NEHA]

This will complete the answer:

x = (-3 +- i sqrt(15))/ 2

ok thanks so that is the answer?
i don't have to divide the top parts by 2 right?

 

[steve_b]

x = (-3 +- i sqrt(15))/ 2

============

You can remove the parentheses I added by dividing both terms by 2:

x = -3/2 +- (i sqrt(15))/2

or

x = -1.5 +- .5 i sqrt(15)

Steve