Question

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

Answer 1

Use distributivity and commutativity of multiplication to get:

`-2x^2y^2(3x^2-xy+4y^2) = -6x^4y^2+2x^3y^3-8x^2y^4`

Explanation:

`-2x^2y^2(3x^2-xy+4y^2)`

`=(-2x^2y^2*3x^2)+(-2x^2y^2*-xy)+(-2x^2y^2*4y^2)`

`= -6x^4y^2+2x^3y^3-8x^2y^4`

Distributivity says that `a(b+c) = ab+ac` or more generally:

`a(b_1+b_2+...+b_n) = ab_1+ab_2+...+ab_n`

Commutativity of multiplication says that:

`a*b = b*a` for any `a` and `b`.

Associativity of multiplication says that:

`a*(b*c) = (a*b)*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*a_2*...*a_n`

So for example `-2x^2y^2*3x^2 = -2*3*x^2*x^2*y^2`

Notice that when multiplying powers you add the exponents:

For example, `x^2*x^2 = x^(2+2) = x^4`