How do you factor #7k(k-3)+4(k-3)#?

1 Answer
Jul 21, 2015

Answer:

#7k(k-3) + 4(k-3) = (7k+4)(k-3)#

Explanation:

The factor #color(blue)((k-3))# occurs in both terms:
#color(white)("XXXX")##color(red)(7k)color(blue)((k-3))#
#color(white)("XXXX")#and
#color(white)("XXXX")##color(red)(+4)color(blue)((k-3))#

In general, the distributive property tells us that
#color(white)("XXXX")##color(red)(a)color(blue)((c))color(red)(+b)color(blue)((c)) = color(red)((a+b))color(blue)((c))#

So, in this case:
#color(white)("XXXX")##color(red)(7k)color(blue)((k-3))color(red)(+4)color(blue)((k-3)) = color(red)((7k+4))color(blue)((k-3))#