3x^2+9x+28=(x-(-2/3 +sqrt(255)/6 i))(x-(-2/3 -sqrt(255)/6 i))3x2+9x+28=(x−(−23+√2556i))(x−(−23−√2556i))
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=-25592−4(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 ix=−9±√92−4(3)(28)2(3)=−9±√−2556=−23±√2556i
The zeros of the quadratic are:
z_1=-2/3 +sqrt(255)/6 iz1=−23+√2556i and z_2=-2/3 - sqrt(255)/6 iz2=−23−√2556i
So the factors are (x-z_1)(x−z1) and (x-z_2)(x−z2)
3x^2+9x+28=(x-(-2/3 +sqrt(255)/6 i))(x-(-2/3 -sqrt(255)/6 i))3x2+9x+28=(x−(−23+√2556i))(x−(−23−√2556i))
This can be written in other ways if you prefer.
