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

Nov 17, 2015

This does not factor into linear factors.

#### Explanation:

Suppose:

$2 {c}^{2} + 4 c d + 3 c + 3 d = \left(p c + q d + r\right) \left(s c + t d + u\right)$

for some $p , q , r , s , t , u \in \mathbb{R}$

Then $r u = 0$, so without loss of generality $u = 0$

$2 {c}^{2} + 4 c d + 3 c + 3 d = \left(p c + q d + r\right) \left(s c + t d\right)$

Next looking at the coefficient of ${d}^{2}$, we have $q t = 0$, but $t \ne 0$ since $3 d$ has no factor $c$. So we must have $q = 0$

$2 {c}^{2} + 4 c d + 3 c + 3 d = \left(p c + r\right) \left(s c + t d\right)$

Looking at the coefficients of $3 c$ and $3 d$ we have $r s = r t = 3$.
So $s = t$

$2 {c}^{2} + 4 c d + 3 c + 3 d = s \left(p c + r\right) \left(c + d\right)$

Then looking at the coefficients of ${c}^{2}$ and $c d$ we have:

$2 = s p = 4$

which is false.

So there is no such linear factorisation.