## Precalculus (6th Edition) Blitzer

In order to factor by grouping, we arrange the terms in pairs and factor out the common factor for each pair, hoping to obtain an expression of the form =(term 1)[expression 1]+(term 2)[expression 1], that is, the brackets hold $\underline{the~same~expression}$. If that is the case, we factor it out: =[expression 1][(term 1) + (term 2) ] --- For example, $3x^{2}+3y^{2}x+2y^{2}x+2y^{4}=[3x^{2}+3y^{2}x]+[2y^{2}x+2y^{4}]$ $=3x(x+y^{2})+2y^{2}(x+y^{2})\quad$...same expression! =$(x+y^{2})(3x+2y^{2})$