Question
\( \left. \begin{array} { l } { \left. \begin{array} { l l } { ( 8 x ^ { 3 } ) ( 2 x ^ { 2 } ) = } & { ( - 3 x y ) ( - 2 x ^ { 2 } ) ( - 4 x ^ { 2 } y ) = } \\ { ( - 5 a ) ( 3 b ) = } & { ( 3 x ) ( 3 x ) ( 3 x ) ( 3 x ) = } \\ { ( - 6 x y ) ( - 6 x y ) = } & { ( 2 a ) ( 3 b ) ( 5 c ) = } \\ { ( x ^ { 2 } y z ^ { 3 } ) ( x y ^ { 4 } ) = } & { ( x y ^ { 2 } ) ( x y z ) ( x y z ^ { 3 } ) = } \end{array} \right. } \end{array} \right. \)
Ask by Gibbs Reid. in the United States
Jan 24,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
\[
\begin{array}{ll}
(8x^3)(2x^2) = 16x^5 &
(-3xy)(-2x^2)(-4x^2 y) = -24x^5 y^2 \\
(-5a)(3b) = -15ab &
(3x)(3x)(3x)(3x) = 81x^4 \\
(-6xy)(-6xy) = 36x^2 y^2 &
(2a)(3b)(5c) = 30abc \\
(x^2 y z^3)(x y^4) = x^3 y^5 z^3 &
(x y^2)(x y z)(x y z^3) = x^3 y^4 z^4 \\
\end{array}
\]
Solution
Certainly! Let's evaluate each of the given algebraic expressions by multiplying the coefficients and combining like variables.
\[
\begin{array}{ll}
(8x^3)(2x^2) = 16x^5 &
(-3xy)(-2x^2)(-4x^2 y) = -24x^5 y^2 \\
(-5a)(3b) = -15ab &
(3x)(3x)(3x)(3x) = 81x^4 \\
(-6xy)(-6xy) = 36x^2 y^2 &
(2a)(3b)(5c) = 30abc \\
(x^2 y z^3)(x y^4) = x^3 y^5 z^3 &
(x y^2)(x y z)(x y z^3) = x^3 y^4 z^4 \\
\end{array}
\]
### Detailed Breakdown:
1. **\((8x^3)(2x^2)\)**
- **Coefficients:** \(8 \times 2 = 16\)
- **Variables:** \(x^{3+2} = x^5\)
- **Result:** \(16x^5\)
2. **\((-3xy)(-2x^2)(-4x^2 y)\)**
- **Coefficients:** \((-3) \times (-2) \times (-4) = -24\)
- **Variables:** \(x^{1+2+2} = x^5\) and \(y^{1+1} = y^2\)
- **Result:** \(-24x^5 y^2\)
3. **\((-5a)(3b)\)**
- **Coefficients:** \(-5 \times 3 = -15\)
- **Variables:** \(a \times b = ab\)
- **Result:** \(-15ab\)
4. **\((3x)(3x)(3x)(3x)\)**
- **Coefficients:** \(3 \times 3 \times 3 \times 3 = 81\)
- **Variables:** \(x^{1+1+1+1} = x^4\)
- **Result:** \(81x^4\)
5. **\((-6xy)(-6xy)\)**
- **Coefficients:** \((-6) \times (-6) = 36\)
- **Variables:** \(x^{1+1} = x^2\) and \(y^{1+1} = y^2\)
- **Result:** \(36x^2 y^2\)
6. **\((2a)(3b)(5c)\)**
- **Coefficients:** \(2 \times 3 \times 5 = 30\)
- **Variables:** \(a \times b \times c = abc\)
- **Result:** \(30abc\)
7. **\((x^2 y z^3)(x y^4)\)**
- **Coefficients:** \(1 \times 1 = 1\) (implicit)
- **Variables:**
- \(x^{2+1} = x^3\)
- \(y^{1+4} = y^5\)
- \(z^3\) (unchanged)
- **Result:** \(x^3 y^5 z^3\)
8. **\((x y^2)(x y z)(x y z^3)\)**
- **Coefficients:** \(1 \times 1 \times 1 = 1\) (implicit)
- **Variables:**
- \(x^{1+1+1} = x^3\)
- \(y^{2+1+1} = y^4\)
- \(z^{0+1+3} = z^4\) (Note: \(z\) in the first term has an exponent of 0)
- **Result:** \(x^3 y^4 z^4\)
Feel free to ask if you need further explanations on any of these steps!
Reviewed and approved by the UpStudy tutoring team
Like
error msg
Explain 
Simplify this solution Mind Expander
Let’s dive into the world of multiplying algebraic expressions! When you multiply terms, remember to multiply the coefficients (numbers in front) and then combine the variables by adding their exponents. For example, in \( (8x^3)(2x^2) \), you would calculate \( 8 \times 2 = 16 \) and \( x^3 \times x^2 = x^{3+2} = x^5 \), giving you \( 16x^5 \). A common pitfall in these operations is forgetting to keep track of the signs. When multiplying negative coefficients, like in \( (-3xy)(-2x^2) \), both negatives will result in a positive coefficient; so it’s \( 6x^{1+2}y = 6x^3y \). Remembering these simple rules will transform you into a multiplication maestro in no time!
