How do you factor #2x^2 +4x +6#?

1 Answer
Jun 7, 2015

#2x^2+4x+6 = 2(x^2+2x+3)#

The expression #x^2+2x+3# is of the form #ax^2+bx+c#, with #a=1#, #b=2# and #c=3#

This has discriminant given by the formula:

#Delta = b^2-4ac = 2^2-(4xx1xx3) = 4 - 12 = -8#

Since #Delta < 0# the equation #x^2+2x+3=0# has no real roots,
so there are no linear factors with real coefficients.