Question

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

Answer 1

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)`

Answer 2

`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