Questions: 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(y-z+8) b. (x+n)(x+n+g) c. x(n+1)-x a. 7(y-z+8)= b. (x+n)(x+n+g)= c. x(n+1)-x=

Solution Steps

To solve these problems, we will apply the distributive property of multiplication over addition. This property states that \( a(b + c) = ab + ac \). We will distribute the multiplication over each term inside the parentheses and simplify the expressions where possible.

For the expression \(7(y-z+8)\), we apply the distributive property:

\[ 7(y-z+8) = 7 \cdot y - 7 \cdot z + 7 \cdot 8 \]

Simplifying each term:

\[ = 7y - 7z + 56 \]

For the expression \((x+n)(x+n+g)\), we expand using the distributive property (also known as the FOIL method for binomials):

\[ (x+n)(x+n+g) = (x+n)x + (x+n)n + (x+n)g \]

Expanding each term:

\[ = x^2 + nx + nx + n^2 + gx + ng \]

Combine like terms:

\[ = x^2 + 2nx + n^2 + gx + ng \]

For the expression \(x(n+1)-x\), we distribute \(x\) in the first term:

\[ x(n+1) - x = x \cdot n + x \cdot 1 - x \]

Simplifying:

\[ = xn + x - x \]

Combine like terms:

\[ = xn \]

Final Answer