# How do you multiply 4x(2x^2 + x)?

Jun 24, 2015

$4 x \left(2 {x}^{2} + x\right) = 8 {x}^{3} + 4 {x}^{2}$

#### Explanation:

You would do this by distributing $4 x$ into the quantities in the parenthesis.

For example,

$a \left(b + c\right) = a b + a c$

The $a$ was distributed into the parenthesis and multiplied by each unit inside of it. The same concept applies here, we just have expressions instead of constant numbers.

First, distribute the $4 x$,

$4 x \left(2 {x}^{2} + x\right) = \left(4 x \cdot 2 {x}^{2}\right) + \left(4 x \cdot x\right)$

You don't need to add parenthesis, I just added them to clarify how the $4 x$ was distributed.

Now you can simplify. We can use exponent rules here.

${x}^{a} \cdot {x}^{b} = {x}^{a + b}$

Using this, we can simplify further. Note that the constant coefficients just get multiplied normally.

$4 x \left(2 {x}^{2} + x\right) = \left(4 {x}^{1} \cdot 2 {x}^{2}\right) + \left(4 {x}^{1} \cdot {x}^{1}\right) = 8 {x}^{1 + 2} + 4 {x}^{1 + 1} = 8 {x}^{3} + 4 {x}^{2}$

Jun 24, 2015

 = color(blue)( 8x^3 + 4x^2

#### Explanation:

$\textcolor{b l u e}{4 x} \left(2 {x}^{2} + x\right)$
Here $\textcolor{b l u e}{4 x}$ needs to be multiplied with each term within bracket:

$= \textcolor{b l u e}{4 x} . \left(2 {x}^{2}\right) + \textcolor{b l u e}{4 x} . x$
 = color(blue)( 8x^3 + 4x^2