How do you factor $y^2+3y+yx+3x$ by grouping?

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Answer 1 · 2016-07-03T22:19:37

$y^2+3y+yx+3x =(y+3) (x + y)$

Proposing

$y^2+3y+yx+3x =(a x + b y + c) (d x + y)$

and equating same power coefficients

${ (3 - c = 0), (1 - b = 0), (3 - c d = 0), (1 - a - b d = 0),( -a d = 0) :}$

Solving for $a,b,c,d$ we obtain

${a = 0, b = 1, c = 3, d = 1}$

giving

$y^2+3y+yx+3x =(y+3) (x + y)$

How do you factor y^2+3y+yx+3xy^2+3y+yx+3x by grouping?