Question

1. $3x^{2}y^{4} cdot 4x^{3}y^{2} cdot -3x^{2}$
2. $v^{3} cdot -2vu^{2}$

-36x³y⁶

1. $4a^{2} cdot -4ab^{3}$
2. $y^{2} cdot yx^{4}$

Explanation:

Problem 13: \( 3x^{2}y^{4} \cdot 4x^{3}y^{2} \cdot (-3x^{2}) \)

Step 1: Multiply the coefficients

First, multiply the numerical coefficients: \( 3 \times 4 \times (-3) \).
\( 3 \times 4 = 12 \), then \( 12 \times (-3) = -36 \).

Step 2: Multiply the \( x \)-terms using exponent rule

For the \( x \)-terms, use the rule \( a^{m} \cdot a^{n} = a^{m + n} \). So \( x^{2} \cdot x^{3} \cdot x^{2} = x^{2 + 3 + 2} = x^{7} \).

Step 3: Multiply the \( y \)-terms using exponent rule

For the \( y \)-terms, \( y^{4} \cdot y^{2} = y^{4 + 2} = y^{6} \).

Step 4: Combine all parts

Combine the coefficient, \( x \)-terms, and \( y \)-terms: \( -36x^{7}y^{6} \).

Step 1: Multiply the coefficients

The coefficient of the first term is \( 1 \) (for \( v^{3} \)) and the second is \( -2 \). So \( 1 \times (-2) = -2 \).

Step 2: Multiply the \( v \)-terms using exponent rule

For \( v \)-terms: \( v^{3} \cdot v = v^{3 + 1} = v^{4} \).

Step 3: Include the \( w \)-term

The \( w \)-term is \( w^{2} \) (from the second factor), so we include it as is.

Step 4: Combine all parts

Combine the coefficient, \( v \)-terms, and \( w \)-terms: \( -2v^{4}w^{2} \).

Step 1: Multiply the coefficients

Multiply \( 4 \) and \( -4 \): \( 4 \times (-4) = -16 \).

Step 2: Multiply the \( a \)-terms using exponent rule

For \( a \)-terms: \( a^{2} \cdot a = a^{2 + 1} = a^{3} \).

Step 3: Include the \( b \)-term

The \( b \)-term is \( b^{3} \) (from the second factor), so we include it as is.

Step 4: Combine all parts

Combine the coefficient, \( a \)-terms, and \( b \)-terms: \( -16a^{3}b^{3} \).

Answer:

\( -36x^{7}y^{6} \)

Problem 14: \( v^{3} \cdot (-2vw^{2}) \)