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

1 Answer
Jun 26, 2015

Answer:

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#