Question

1. what is the quotient of the expression

\\(\frac{6x^{3}y + 3x^{2}y^{2} - 24x}{3x}\\)?
a. \\(3x^{2}y + xy^{2} - 8x\\)
b. \\(3x^{2} - 24x\\)
c. \\(2x^{2}y + xy - 8\\)
d. \\(2x^{2}y + xy^{2} - 8\\)

Explanation:

Step1: Split the fraction

We can split the given fraction \(\frac{6x^{3}y + 3x^{2}y^{2}-24x}{3x}\) into three separate fractions: \(\frac{6x^{3}y}{3x}+\frac{3x^{2}y^{2}}{3x}-\frac{24x}{3x}\).

Step2: Simplify each fraction

- For the first fraction \(\frac{6x^{3}y}{3x}\), divide the coefficients and subtract the exponents of like bases. The coefficient \(6\div3 = 2\), and for \(x\), \(x^{3}\div x=x^{3 - 1}=x^{2}\), so it simplifies to \(2x^{2}y\).
- For the second fraction \(\frac{3x^{2}y^{2}}{3x}\), the coefficient \(3\div3 = 1\), for \(x\), \(x^{2}\div x=x^{2 - 1}=x\), and \(y^{2}\) remains, so it simplifies to \(xy^{2}\)? Wait, no, wait. Wait, the original second term in the numerator is \(3x^{2}y^{2}\)? Wait, no, looking back, the original numerator is \(6x^{3}y+3x^{2}y^{2}-24x\). Wait, when we divide \(\frac{3x^{2}y^{2}}{3x}\), the \(y^{2}\) is in the numerator, but wait, no, wait, maybe I made a mistake. Wait, no, let's re - check. Wait, the second term is \(3x^{2}y^{2}\)? Wait, no, the original problem: \(\frac{6x^{3}y + 3x^{2}y^{2}-24x}{3x}\). Wait, no, wait, maybe the second term is \(3x^{2}y\)? No, the user's problem shows \(3x^{2}y^{2}\). Wait, no, let's do the division again. \(\frac{3x^{2}y^{2}}{3x}=\frac{3}{3}\times\frac{x^{2}}{x}\times y^{2}=1\times x^{2 - 1}\times y^{2}=xy^{2}\)? But wait, the option D has \(xy^{2}\), but let's check the third term. \(\frac{- 24x}{3x}=\frac{-24}{3}\times\frac{x}{x}=- 8\). Wait, but the first term: \(\frac{6x^{3}y}{3x}=\frac{6}{3}\times x^{3-1}y = 2x^{2}y\). So putting it together: \(2x^{2}y+xy^{2}-8\)? But wait, option D is \(2x^{2}y + xy^{2}-8\), and option C is \(2x^{2}y+xy - 8\). Wait, maybe I misread the numerator. Wait, maybe the second term in the numerator is \(3x^{2}y\) instead of \(3x^{2}y^{2}\)? Wait, let's check the original problem again. The user's problem: " \(\frac{6x^{3}y + 3x^{2}y^{2}-24x}{3x}\)". Wait, maybe it's a typo, but assuming the problem is as written. Wait, no, wait, maybe I made a mistake in the second term. Wait, \(\frac{3x^{2}y^{2}}{3x}=xy^{2}\), \(\frac{6x^{3}y}{3x}=2x^{2}y\), \(\frac{-24x}{3x}=-8\). So the result is \(2x^{2}y + xy^{2}-8\), which is option D? Wait, no, wait, let's check the options again. Option D: \(2x^{2}y+xy^{2}-8\), option C: \(2x^{2}y + xy-8\). Wait, maybe the second term in the numerator is \(3x^{2}y\) instead of \(3x^{2}y^{2}\). Let's assume that maybe there was a typo, and the second term is \(3x^{2}y\). Then \(\frac{3x^{2}y}{3x}=xy\). Then the result would be \(2x^{2}y+xy - 8\), which is option C. But according to the given problem, the second term is \(3x^{2}y^{2}\). Wait, maybe I misread the problem. Let me check again. The problem says: \(\frac{6x^{3}y + 3x^{2}y^{2}-24x}{3x}\). So let's do the division correctly.

First term: \(\frac{6x^{3}y}{3x}=2x^{2}y\) (since \(6\div3 = 2\), \(x^{3}\div x=x^{2}\), and \(y\) remains).

Second term: \(\frac{3x^{2}y^{2}}{3x}=xy^{2}\) (since \(3\div3 = 1\), \(x^{2}\div x=x\), and \(y^{2}\) remains).

Third term: \(\frac{-24x}{3x}=-8\) (since \(-24\div3=-8\) and \(x\div x = 1\)).

So combining these, we get \(2x^{2}y+xy^{2}-8\), which is option D. Wait, but let's check the options again. Option D: \(2x^{2}y+xy^{2}-8\), option C: \(2x^{2}y + xy-8\). So if the second term in the numerator is \(3x^{2}y\) (a typo), then it's option C, but as per the given problem, it's \(3x^{2}y^{2}\), so it's option D. Wait, maybe the original problem has a typo, but let's go with the given problem.

Wait, no, wait, maybe I made a mistake in the second term. Let's re - express the original fraction:

\(\frac{6x^{3}y+3x^{2}y^{2}-24x}{3x}=\frac{6x^{3}y}{3x}+\frac{3x^{2}y^{2}}{3x}+\frac{-24x}{3x}\)

\(=2x^{2}y + xy^{2}-8\)

Which is option D.

Answer:

D. \(2x^{2}y + xy^{2}-8\)