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

Apr 30, 2017

See the solution process below:

#### Explanation:

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

$2 {k}^{2} + 4 k \implies \left(2 k \cdot k\right) + \left(2 k \cdot 2\right) \implies 2 k \left(k + 2\right)$

Apr 30, 2017

$2 k \left(k + 2\right)$

#### Explanation:

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

Consider $2 \times 7 = 14$

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

So $2 \times 7$ is the same as $2 \times \left(3 + 4\right) \to \left(3 + 4\right)$
$\text{ "ul((3+4))larr" Add}$
$\text{ } 6 + 8$

Mathematical convention allows me to write $2 \times \left(3 + 4\right)$ as
$2 \left(3 + 4\right)$

$\textcolor{b r o w n}{\text{Distributive - everything inside the bracket is multiplied by the 2}}$
Now this is the same as $\textcolor{red}{2} \textcolor{g r e e n}{\left(3 + 4\right)}$

$\textcolor{g r e e n}{\left(\textcolor{red}{2} \times 3\right) + \left(\textcolor{red}{2} \times 4\right)} = 6 + 8 = 14$

$\textcolor{red}{\text{The "2xx" is distributed over every value inside the brackets}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using the example as a comparison to $2 {k}^{2} + 4 k$

Working backwards from this we have:

Example$\text{ } 14 \to 6 \textcolor{w h i t e}{. .} + 8$
Question$\text{ } \to 2 {k}^{2} + 4 k$

Example-factor out the 2$\text{ } \textcolor{w h i t e}{. .} \to 2 \left(3 + 4\right)$
Question-factor out the $2 k \text{ } \to 2 k \left(k + 2\right) \ldots . . = 2 {k}^{2} + 4 k$

The $2 k \left(k + 2\right)$ is the consequence of using the distributive property