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

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Answer 1 · 2015-10-04T10:16:32

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)#

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)#