Question

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

Answer 1

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)`

Explanation:

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)`