Question

How do you factor `3r^2+ r + 1`?

Answer 1

This has no linear factors with Real coefficients.

You can use the quadratic formula to find:

`3r^2+r+1 = 3(r+(1-i sqrt(11))/6)(r+(1+i sqrt(11))/6)`

Explanation:

Let `f(r) = 3r^2+r+1`. This is of the form `ar^2+br+c` with `a = 3`, `b=1` and `c=1`.

This quadratic expression has discriminant `Delta` given by the formula:

`Delta = b^2 - 4ac = 1^2 - (4xx3xx1) = 1-12 = -11`

Since this is negative, the equation `f(r) = 0` has no Real solutions. It has a conjugate pair of Complex roots given by the quadratic formula:

`r = (-b+-sqrt(Delta))/(2a) = (-1+-sqrt(-11))/6 = (-1+-i sqrt(11))/6`

Hence:

`f(r) = 3(r - (-1+i sqrt(11))/6)(r - (-1-i sqrt(11))/6)`

`=3(r+(1-i sqrt(11))/6)(r+(1+i sqrt(11))/6)`