Multiply `3x^(2)y` and `-2xy^(2)`. Verify the product for x = 1 and y = 2.

Answer

To solve the problem of multiplying \(3x^{2}y\) and \(-2xy^{2}\), we will follow these steps:

### Step 1: Write down the expressions to be multiplied
We have:
\[ 3x^{2}y \quad \text{and} \quad -2xy^{2} \]

### Step 2: Multiply the coefficients
The coefficients are \(3\) and \(-2\):
\[ 3 \times -2 = -6 \]

### Step 3: Multiply the variables
Now we will multiply the variables. We have:
- For \(x\): \(x^{2}\) and \(x\)
- For \(y\): \(y\) and \(y^{2}\)

Using the property of exponents \(a^{m} \times a^{n} = a^{m+n}\), we can calculate:
- For \(x\):
\[ x^{2} \times x = x^{2+1} = x^{3} \]
- For \(y\):
\[ y \times y^{2} = y^{1+2} = y^{3} \]

### Step 4: Combine the results
Now, we combine the coefficient and the variables:
\[ 3x^{2}y \times -2xy^{2} = -6x^{3}y^{3} \]

### Step 5: Verify the product for \(x = 1\) and \(y = 2\)
Now we will substitute \(x = 1\) and \(y = 2\) into the expression \(-6x^{3}y^{3}\):
\[ -6(1)^{3}(2)^{3} \]
Calculating \(1^{3}\) and \(2^{3}\):
\[ 1^{3} = 1 \quad \text{and} \quad 2^{3} = 8 \]
So we have:
\[ -6 \times 1 \times 8 = -6 \times 8 = -48 \]

### Final Result
The product of \(3x^{2}y\) and \(-2xy^{2}\) is \(-6x^{3}y^{3}\), and when verified for \(x = 1\) and \(y = 2\), the result is \(-48\).

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