Question

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

Answer 1

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`

Answer 2

`[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`