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

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

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Answer 1 · 2015-06-26T23:28:02

Use distributivity and commutativity of multiplication to get:

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

Explanation:

$- 2 x^{2} y^{2} (3 x^{2} - x y + 4 y^{2})$ $-2x^2y^2(3x^2-xy+4y^2)$

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

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

Distributivity says that $a (b + c) = a b + a c$ $a(b+c) = ab+ac$ 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, $x^2*x^2 = x^(2+2) = x^4$

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

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Explanation:

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How do you simplify -2x^2y^2(3x^2-xy+4y^2)−2x2y2(3x2−xy+4y2)-2x^2y^2(3x^2-xy+4y^2)?

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Explanation:

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