# How do you list all possible roots and find all factors of #3x^2+2x+2#?

##### 1 Answer

This quadratic has zeros:

#3x^2+2x+2#

#= 3(x+1/3-sqrt(5)/3i)(x+1/3+sqrt(5)/3i)#

#= 1/3(3x+1-sqrt(5)i)(3x+1+sqrt(5)i)#

#### Explanation:

Given:

#3x^2+2x+2#

Note that this is in standard quadratic form

This has discriminant

#Delta = b^2-4ac = 2^2-(4*3*2) = 4-24 = -20 = -2^2*5#

Since

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

#=(-b+-sqrt(Delta))/(2a)#

#=(-2+-sqrt(-20))/6#

#=(-2+-2sqrt(5)i)/6#

#=1/3(-1+-sqrt(5)i)#

That is:

#x_1 = 1/3(-1+sqrt(5)i)#

#x_2 = 1/3(-1-sqrt(5)i)#

Both

#3(x-x_1) = (3x+1-sqrt(5)i)#

#3(x-x_2) = (3x+1+sqrt(5)i)#

If we multiply these together we will get a leading term

#3x^2+2x+2 = 1/3(3x+1-sqrt(5)i)(3x+1+sqrt(5)i)#

Alternatively, if we prefer monic factors, we can write:

#3x^2+2x+2 = 3(x+1/3-sqrt(5)/3i)(x+1/3+sqrt(5)/3i)#