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.

How do you factor 2c^2 + 4cd + 3c + 3d2c^2 + 4cd + 3c + 3d?