Question

Difficulty: Very hardSimplifying Expressions and Combining Like Terms

When the expression 2x(x2−3xy+2y2)−(x−2y)3−y(x2−5xy+4y2)2x(x^2 - 3xy + 2y^2) - (x - 2y)^3 - y(x^2 - 5xy + 4y^2) is fully simplified by combining like terms, what is the coefficient of the xy2xy^2 term?

Answer: -3

Answer

The coefficient of the xy2xy^2 term is −3-3.
Expanding the terms of the expression yields 2x3−6x2y+4xy22x^3 - 6x^2y + 4xy^2, −x3+6x2y−12xy2+8y3-x^3 + 6x^2y - 12xy^2 + 8y^3, and −x2y+5xy2−4y3-x^2y + 5xy^2 - 4y^3. Summing the coefficients of the xy2xy^2 terms gives 4−12+5=−34 - 12 + 5 = -3. Thus, the coefficient of xy2xy^2 is −3-3.

Step-by-Step Solution

1
Expand the first term of the expression: 2x(x2−3xy+2y2)2x(x^2 - 3xy + 2y^2)
2x3−6x2y+4xy22x^3 - 6x^2y + 4xy^2
Distribute the monomial 2x2x to each term inside the parentheses: 2x⋅x2=2x32x \cdot x^2 = 2x^3, 2x⋅(−3xy)=−6x2y2x \cdot (-3xy) = -6x^2y, and 2x⋅2y2=4xy22x \cdot 2y^2 = 4xy^2.
2
Expand the cubed binomial (x−2y)3(x - 2y)^3 and distribute the negative sign
−x3+6x2y−12xy2+8y3-x^3 + 6x^2y - 12xy^2 + 8y^3
First, expand the binomial (x−2y)3=x3−3(x2)(2y)+3(x)(2y)2−(2y)3=x3−6x2y+12xy2−8y3(x - 2y)^3 = x^3 - 3(x^2)(2y) + 3(x)(2y)^2 - (2y)^3 = x^3 - 6x^2y + 12xy^2 - 8y^3. Then, distribute the negative sign to all terms inside the parentheses.
3
Expand the third term: −y(x2−5xy+4y2)-y(x^2 - 5xy + 4y^2)
−x2y+5xy2−4y3-x^2y + 5xy^2 - 4y^3
Distribute the negative monomial −y-y to each term inside the parentheses: −y⋅x2=−x2y-y \cdot x^2 = -x^2y, −y⋅(−5xy)=5xy2-y \cdot (-5xy) = 5xy^2, and −y⋅4y2=−4y3-y \cdot 4y^2 = -4y^3.
4
Group and combine all like terms
x3−x2y−3xy2+4y3x^3 - x^2y - 3xy^2 + 4y^3
Combine the coefficients of the corresponding terms: (2x3−x3)+(−6x2y+6x2y−x2y)+(4xy2−12xy2+5xy2)+(8y3−4y3)=x3−x2y−3xy2+4y3(2x^3 - x^3) + (-6x^2y + 6x^2y - x^2y) + (4xy^2 - 12xy^2 + 5xy^2) + (8y^3 - 4y^3) = x^3 - x^2y - 3xy^2 + 4y^3.

Key Concept

Simplifying Expressions and Combining Like Terms
Estimated Time:3m 0s
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