`(14x^(2)yz-28x^(2)y^(2)z^(3)+32y^(2)z^(2))div(-4xy)` is equal to

Options

A. `(7)/(2)yz+7xyz^(2)+8xyz`
B. `-(7)/(2)xz+7xyz^(3)-(8yz^(2))/(x)`
C. `-(7)/(2)xz-7xyz^(3)+(8yz^(2))/(x)`
D. `(7)/(2)xz-7xyz^(2)-(8yz^(2))/(x)`

Answer

To solve the expression \((14x^{2}yz - 28x^{2}y^{2}z^{3} + 32y^{2}z^{2}) \div (-4xy)\), we will divide each term in the numerator by \(-4xy\).

### Step 1: Divide the first term
The first term is \(14x^{2}yz\).

\[ \frac{14x^{2}yz}{-4xy} = \frac{14}{-4} \cdot \frac{x^{2}}{x} \cdot \frac{y}{y} \cdot z \]

Calculating each part:
- \(\frac{14}{-4} = -\frac{7}{2}\)
- \(\frac{x^{2}}{x} = x^{1} = x\)
- \(\frac{y}{y} = 1\)

So, the first term simplifies to:
\[ -\frac{7}{2}xz \]

### Step 2: Divide the second term
The second term is \(-28x^{2}y^{2}z^{3}\).

\[ \frac{-28x^{2}y^{2}z^{3}}{-4xy} = \frac{-28}{-4} \cdot \frac{x^{2}}{x} \cdot \frac{y^{2}}{y} \cdot z^{3} \]

Calculating each part:
- \(\frac{-28}{-4} = 7\)
- \(\frac{x^{2}}{x} = x^{1} = x\)
- \(\frac{y^{2}}{y} = y^{1} = y\)

So, the second term simplifies to:
\[ 7xyz^{2} \]

### Step 3: Divide the third term
The third term is \(32y^{2}z^{2}\).

\[ \frac{32y^{2}z^{2}}{-4xy} = \frac{32}{-4} \cdot \frac{y^{2}}{y} \cdot z^{2} \]

Calculating each part:
- \(\frac{32}{-4} = -8\)
- \(\frac{y^{2}}{y} = y^{1} = y\)

So, the third term simplifies to:
\[ -8yz \]

### Final Result
Now, we combine all the simplified terms:

\[ -\frac{7}{2}xz + 7xyz^{2} - 8yz \]

Thus, the expression \((14x^{2}yz - 28x^{2}y^{2}z^{3} + 32y^{2}z^{2}) \div (-4xy)\) is equal to:

\[ -\frac{7}{2}xz + 7xyz^{2} - 8yz \]