Question

How do you factor `3x^2+9x+28=0`?

Answer 1

`3x^2+9x+28=(x-(-2/3 +sqrt(255)/6 i))(x-(-2/3 -sqrt(255)/6 i))`

This can be written in other ways if you prefer.

Method
You will not factor this by trial and error. The discriminant (from the qudratic formula) is

`9^2-4(3)(28)=81-336=-255` So the solutions are imaginary and have irrational imaginary parts.

Find the solutions first and then factor.

`x=(-9 +- sqrt(9^2-4(3)(28)))/(2(3))=(-9 +- sqrt(-255))/(6)=-2/3 +- sqrt(255)/6 i`

The zeros of the quadratic are:
`z_1=-2/3 +sqrt(255)/6 i` and `z_2=-2/3 - sqrt(255)/6 i`

So the factors are `(x-z_1)` and `(x-z_2)`

`3x^2+9x+28=(x-(-2/3 +sqrt(255)/6 i))(x-(-2/3 -sqrt(255)/6 i))`

This can be written in other ways if you prefer.