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

Answer 1

Avery Alexander

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#

#-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#

Multiplication's commutativity states that:

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

According to associativity of multiplication,

#a*(b*c) = (a*b)*c# for any #a#, #b# and #c#

In general, we don't need to worry about the order of multiplication in a product like this if we omit the brackets.

#a_1*a_2*...*a_n#

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

Observe that the exponents are added when multiplying powers:

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

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Answer 2

Eva Adamson

To simplify the expression -2x^2y^2(3x^2-xy+4y^2), distribute -2x^2y^2 into each term within the parentheses:

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

Combine the terms:

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

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.