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

Question

1535 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-29T07:48:40

This quadratic has zeros: #-1/3+-sqrt(5)/3i# and hence can be factored as:

#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)#

Given:

#3x^2+2x+2#

Note that this is in standard quadratic form #ax^2+bx+c# with #a=3#, #b=2# and #c=2#.

This has discriminant #Delta# given by the formula:

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

Since #Delta < 0#, this quadratic has no Real zeros, but we can find the Complex zeros using the quadratic formula:

#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 #(x-x_1)# and #(x-x_2)# are factors of our quadratic, so also are:

#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 #9x^2#, so we need to divide by #3# somewhere. So the factorisation can be expressed:

#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)#