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Rename each of the following using the distributive property of multiplication over addition so that there are no parentheses in the final answer. Simplify when possible. a. 7(f+g-2) b. (z+x)(z+x+c) c. z(x+1)-Z a. 7(f+g-2)= b. (z+x)(z+x+c) = c. z(x+1)-z=

Rename each of the following using the distributive property of multiplication over addition so that there are no parentheses in the final answer. Simplify when possible. a. 7(f+g-2) b. (z+x)(z+x+c) c. z(x+1)-Z a. 7(f+g-2)= b. (z+x)(z+x+c) = c. z(x+1)-z=

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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### Applying the Distributive Property The task is to rename each expression using the distributive property of multiplication over addition so that there are no parentheses in the final answer. Simplify when possible. 1. **Expression: 7(f + g – 2)** - Original: \( 7(f + g – 2) \) - Apply the distributive property: \( 7 \cdot f + 7 \cdot g - 7 \cdot 2 \) - Simplified: \( 7f + 7g - 14 \) 2. **Expression: (z + x)(z + x + c)** - Original: \( (z + x)(z + x + c) \) - Apply the distributive property: - First, distribute \( (z + x) \) to each term inside the parentheses: - \( (z + x)z + (z + x)x + (z + x)c \) - Apply distributive property inside each term: - \( z \cdot z + z \cdot x \) - \( x \cdot z + x \cdot x \) - \( z \cdot c + x \cdot c \) - Combine these: - \( z^2 + zx + xz + x^2 + zc + xc \) - Notice \( zx \) and \( xz \) are the same: - Final Simplified: \( z^2 + 2zx + x^2 + zc + xc \) 3. **Expression: z(x + 1) - z** - Original: \( z(x + 1) - z \) - Apply the distributive property: - \( zx + z \cdot 1 - z \) - Simplified: \( zx + z - z \) - Remove like terms: - Final Simplified: \( zx \) ### Final Expressions a. \( 7(f + g – 2) = 7f + 7g - 14 \) b. \( (z + x)(z + x + c) = z^2 + 2zx + x^2 + zc + xc \) c. \( z(x + 1) - z = zx \) These steps help in
Transcribed Image Text:### Applying the Distributive Property The task is to rename each expression using the distributive property of multiplication over addition so that there are no parentheses in the final answer. Simplify when possible. 1. **Expression: 7(f + g – 2)** - Original: \( 7(f + g – 2) \) - Apply the distributive property: \( 7 \cdot f + 7 \cdot g - 7 \cdot 2 \) - Simplified: \( 7f + 7g - 14 \) 2. **Expression: (z + x)(z + x + c)** - Original: \( (z + x)(z + x + c) \) - Apply the distributive property: - First, distribute \( (z + x) \) to each term inside the parentheses: - \( (z + x)z + (z + x)x + (z + x)c \) - Apply distributive property inside each term: - \( z \cdot z + z \cdot x \) - \( x \cdot z + x \cdot x \) - \( z \cdot c + x \cdot c \) - Combine these: - \( z^2 + zx + xz + x^2 + zc + xc \) - Notice \( zx \) and \( xz \) are the same: - Final Simplified: \( z^2 + 2zx + x^2 + zc + xc \) 3. **Expression: z(x + 1) - z** - Original: \( z(x + 1) - z \) - Apply the distributive property: - \( zx + z \cdot 1 - z \) - Simplified: \( zx + z - z \) - Remove like terms: - Final Simplified: \( zx \) ### Final Expressions a. \( 7(f + g – 2) = 7f + 7g - 14 \) b. \( (z + x)(z + x + c) = z^2 + 2zx + x^2 + zc + xc \) c. \( z(x + 1) - z = zx \) These steps help in
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