How do you factor #2c^2 + 4cd + 3c + 3d#?

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Answer 1 · 2015-11-17T21:17:29

This does not factor into linear factors.

Suppose:

#2c^2+4cd+3c+3d = (pc+qd+r)(sc+td+u)#

for some #p, q, r, s, t, u in RR#

Then #ru = 0#, so without loss of generality #u = 0#

#2c^2+4cd+3c+3d = (pc+qd+r)(sc+td)#

Next looking at the coefficient of #d^2#, we have #qt=0#, but #t != 0# since #3d# has no factor #c#. So we must have #q = 0#

#2c^2+4cd+3c+3d = (pc+r)(sc+td)#

Looking at the coefficients of #3c# and #3d# we have #rs = rt = 3#.
So #s = t#

#2c^2+4cd+3c+3d = s(pc+r)(c+d)#

Then looking at the coefficients of #c^2# and #cd# we have:

#2 = sp = 4#

which is false.

So there is no such linear factorisation.