How do you factor #6xy+15x#?

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Answer 1 · 2017-04-09T19:43:56

#3x(2y+5)#

Both terms have the variable #x#,

#x(6y+15)#

Both terms can be factored by #3#,

#3x(2y+5)#

You can check if you factored correctly by multiplying through the brakctets,

#(3x*2y)+(3x+5)#

#=6xy+15x#

Answer 2 · 2017-04-09T19:44:54

#3x(2y+5)#

Let's expand everything we can and then factor out common factors:
#6xy+15x#
#2*color(blue)(3)*color(green)(x)*y+color(blue)(3)*5*color(green)(x)#

There are #x#s in both values and a #3# for both. Let's factor those out:
#3*x(2*y+5)#
#3x(2y+5)#.

Just to make sure the new expression is still equal tothe old one, let's distribute the #3x# int #2y+5#. We should get #6xy+15x#

#3x(2y+5)#
#2y*3x+5*3x#
#6xy+15x#
Yep. it's still the same! Good job!