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

Question

1483 views around the world

You can reuse this answer
Creative Commons License

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) #