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Zorluk: Çok zorSimplifying Expressions and Combining Like Terms

When the expression −13x(x−2y)2−12x(xy−3y2)−(x3−2x2y)-\frac{1}{3}x(x - 2y)^2 - \frac{1}{2}x(xy - 3y^2) - (x^3 - 2x^2y) is fully simplified, what is the coefficient of the x2yx^2y term?

  1. A
    103\frac{10}{3}
  2. B
    56\frac{5}{6}
  3. C
    −76-\frac{7}{6}
  4. 176\frac{17}{6}Cevap
  5. E
    32\frac{3}{2}

Cevap

The coefficient of the x2yx^2y term is 176\frac{17}{6}.
By expanding all components of the expression: the first term yields −13x3+43x2y−43xy2-\frac{1}{3}x^3 + \frac{4}{3}x^2y - \frac{4}{3}xy^2, the second term yields −12x2y+32xy2-\frac{1}{2}x^2y + \frac{3}{2}xy^2, and the third term yields −x3+2x2y-x^3 + 2x^2y. Summing the coefficients of the x2yx^2y term gives 43−12+2=176\frac{4}{3} - \frac{1}{2} + 2 = \frac{17}{6}.

Adım Adım Çözüm

1
Expand the first term of the expression.
−13x(x−2y)2=−13x(x2−4xy+4y2)=−13x3+43x2y−43xy2-\frac{1}{3}x(x - 2y)^2 = -\frac{1}{3}x(x^2 - 4xy + 4y^2) = -\frac{1}{3}x^3 + \frac{4}{3}x^2y - \frac{4}{3}xy^2
Applying binomial expansion to (x−2y)2(x-2y)^2 and distributing −13x-\frac{1}{3}x.
2
Expand the second term of the expression.
−12x(xy−3y2)=−12x2y+32xy2-\frac{1}{2}x(xy - 3y^2) = -\frac{1}{2}x^2y + \frac{3}{2}xy^2
Distributing the term −12x-\frac{1}{2}x over the parenthetical terms.
3
Distribute the negative sign in the third term.
−(x3−2x2y)=−x3+2x2y-(x^3 - 2x^2y) = -x^3 + 2x^2y
Distributing the negative sign across all terms inside the parentheses.
4
Combine the coefficients of the x2yx^2y terms.
43−12+2=86−36+126=176\frac{4}{3} - \frac{1}{2} + 2 = \frac{8}{6} - \frac{3}{6} + \frac{12}{6} = \frac{17}{6}
Finding a common denominator of 6 to sum the coefficients of the x2yx^2y term.

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Simplifying expressions by distributing terms, expanding binomials, and combining like terms with fractional coefficients.
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Simplifying Expressions and Combining Like Terms

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