Factorise the following expression? So I need to factorise this expression but am a little stuck:

$x^2+3(y+z)x+(y+2z)(2y+z)=?$

Anyone?
 A: Remember that when you have an expression in factored form 
$$(x+a)(x+b)$$
When you expand it out
$$x^2+(a+b)x+ab$$
The linear term is the sum is the roots and the constant is the product.  You can see from your multi variate polynomial that your constant term (in relation to $x$) is given as a product so always see if you add the factors you get your linear term.
$$(y+2z)+(2y+z)=3y+3z=3(y+z)$$
Which is what you have.  This your factorization is
$$(x+y+2z)(x+2y+z)$$
A: There are many ways to factorize the expression based on the same idea an given by Eleven-Eleven in his/her answer.
You can write $$A=x^2+3  (y+z)x+(2 y+z) (y+2 z)=0$$ $$B=2 y^2+ (3 x+5 z)y+\left(x^2+3 x z+2 z^2\right)=0$$ $$C=2 z^2+ (3 x+5 y)z+\left(x^2+3 x y+2 y^2\right)=0$$
All these expressions are quadratic equations of the type $\alpha w^2+\beta w+\gamma=0$ and computing the roots $r_1,r_2$, you will have $$\alpha(w-r_1)(w-r_2)$$
A: Consider the more general expression $$\color{Green}{x^2+2bx+c=(x+b)^2-(b^2-c).}$$
Here $b, c$ can be any function of $y, z.$
If this expression has a proper factorization then $b^2-c$ should be a perfect square.   Why?
In your case $b^2-c$ $$=\dfrac94(y+z)^2-(y+2z)(2y+z)=\dfrac{y^2}4-\dfrac{yz}2+\dfrac{z^2}4=\color{Red}{\left(\dfrac{y}{2}-\dfrac{z}{2}\right)^2}$$
