How do you factor #x^2+3x+12=0#?

2 Answers
Mar 5, 2018

Answer:

There are no Real factors of #x^2+3x+12#

Explanation:

If you are willing to include Complex values in your solution
then you can use the quadratic formula to get
#(-b+-sqrt(b^2-4ac))/(2a) = (-3+-sqrt(3^2-(4 * 1 * 12)))/(2 *1) = (-3+-sqrt(39)i)/2#
for the factoring:
#(x+(3+sqrt(39)i)/2)(x+(3-sqrt(39)i)/2)=0#

Mar 5, 2018

Answer:

#[x + 3/2 +(sqrt(39)color(white)(..)i)/2] color(white)(..)[x+3/2 -(sqrt(39)color(white)(..)i)/2] = 0#

Explanation:

First of all there are no whole number factors of 12 that can be use to sum to 3 from the #3x#. Consequently we are looking at fractional factors.

Lets check to see if the plot actually crosses the x-axis. If it does then the determinate part of #x=(-b+-sqrt(b^2-4ac))/(2a)# will be positive.

Determinate #->b^2-4ac color(white)("d") ->color(white)("d") 3^2-4(1)(12) = 9-48 <0#

Thus the roots do not belong to the set of real numbers ( #RR# )

They belong to the set of numbers called complex ( #CC# )
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#x=(-3+-sqrt(-39))/(2) = -3/2 +-((sqrt(39))color(white)(..)i)/2#

#x=(-3/2 -((sqrt(39))color(white)(..)i)/2)#

#x=(-3/2 +((sqrt(39))color(white)(..)i)/2)#

#[x + 3/2 +((sqrt(39))color(white)(..)i)/2] color(white)(..)[x+3/2 -((sqrt(39))color(white)(..)i)/2] = 0#

Tony B