# How do you simplify -2x^2y^2(3x^2-xy+4y^2)?

Jun 26, 2015

Use distributivity and commutativity of multiplication to get:

$- 2 {x}^{2} {y}^{2} \left(3 {x}^{2} - x y + 4 {y}^{2}\right) = - 6 {x}^{4} {y}^{2} + 2 {x}^{3} {y}^{3} - 8 {x}^{2} {y}^{4}$

#### Explanation:

$- 2 {x}^{2} {y}^{2} \left(3 {x}^{2} - x y + 4 {y}^{2}\right)$

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

$= - 6 {x}^{4} {y}^{2} + 2 {x}^{3} {y}^{3} - 8 {x}^{2} {y}^{4}$

Distributivity says that $a \left(b + c\right) = a b + a c$ or more generally:

$a \left({b}_{1} + {b}_{2} + \ldots + {b}_{n}\right) = a {b}_{1} + a {b}_{2} + \ldots + a {b}_{n}$

Commutativity of multiplication says that:

$a \cdot b = b \cdot a$ for any $a$ and $b$.

Associativity of multiplication says that:

$a \cdot \left(b \cdot c\right) = \left(a \cdot b\right) \cdot c$ for any $a$, $b$ and $c$

More generally, we can miss out the brackets and not worry what order multiplication occurs in in a product like:

${a}_{1} \cdot {a}_{2} \cdot \ldots \cdot {a}_{n}$

So for example $- 2 {x}^{2} {y}^{2} \cdot 3 {x}^{2} = - 2 \cdot 3 \cdot {x}^{2} \cdot {x}^{2} \cdot {y}^{2}$

Notice that when multiplying powers you add the exponents:

For example, ${x}^{2} \cdot {x}^{2} = {x}^{2 + 2} = {x}^{4}$