How do you use the distributive property to factor #2k^2+4k#?

2 Answers
Apr 30, 2017

Answer:

See the solution process below:

Explanation:

You can remove or factor #2k# from each term in the expression:

#2k^2 + 4k => (2k * k) + (2k * 2) => 2k(k + 2)#

Apr 30, 2017

Answer:

#2k(k+2)#

Explanation:

You look for things that are in both parts
Demonstrated by the following example

Consider #2xx7 = 14#

I chose to partition(split) 7 into 3+4

So #2xx7# is the same as #2xx(3+4)->(3+4)#
#" "ul((3+4))larr" Add"#
#" "6+8#

Mathematical convention allows me to write #2xx(3+4)# as
#2(3+4)#

#color(brown)("Distributive - everything inside the bracket is multiplied by the 2")#
Now this is the same as #color(red)(2)color(green)((3+4))#

#color(green)((color(red)(2)xx3)+(color(red)(2)xx4)) = 6+8 = 14#

#color(red)("The "2xx" is distributed over every value inside the brackets")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using the example as a comparison to #2k^2+4k#

Working backwards from this we have:

Example#" "14->6color(white)(..)+8#
Question#" "->2k^2+4k#

Example-factor out the 2#" "color(white)(..)->2(3+4)#
Question-factor out the #2k" "->2k(k+2).....=2k^2+4k#

The #2k(k+2)# is the consequence of using the distributive property