Which of the following is equivalent to $18$ more than the product of $9-6p$ and $p-2$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $3(7p+6)$ $\newline$ (B) $-3p(2p-7)$ $\newline$ (C) $(27-6p)(p-2)$ $\newline$ (D) $(9-6p)(p+16)$

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Grade 6

Factor numerical expressions using the distributive property

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Q. Which of the following is equivalent to

18

more than the product of

9-6p

and

p-2

\newline

Choose

1

answer:

\newline

(A)

3(7p+6)

\newline

(B)

-3p(2p-7)

\newline

(C)

(27-6p)(p-2)

\newline

(D)

(9-6p)(p+16)

Express Algebraically: First, let's express the given problem algebraically. The expression for ＂ $18$ more than the product of $9-6p$ and $p-2$ ＂ can be written as: $(9-6p)(p-2) + 18$ .
Distribute Multiplication: Next, we need to distribute the multiplication across the terms inside the parentheses. This means we multiply each term in the first parentheses by each term in the second parentheses: $(9 \times p) + (9 \times -2) + (-6p \times p) + (-6p \times -2)$ .
Perform Multiplication: Perform the multiplication: $9p - 18 - 6p^2 + 12p$ .
Combine Like Terms: Combine like terms: $-6p^2 + (9p + 12p) - 18 = -6p^2 + 21p - 18$ .
Add $18$ : Now, add $18$ to both sides to account for the ＂ $18$ more than＂ part of the expression: $-6p^2 + 21p - 18 + 18$ .
Simplify Expression: Simplify the expression by combining like terms, which in this case are the constants $-18$ and $+18$ , resulting in: $-6p^2 + 21p$ .
Factor Out $-3$ : We notice that all terms in the expression $-6p^2 + 21p$ can be divided by $-3$ , so let's factor $-3$ out: $-3(2p^2 - 7p)$ .
Correct Mistake: However, there seems to be a mistake in the previous step regarding the sign and coefficient in front of the $p^2$ term when factoring out $-3$ . The correct factoring should maintain the original terms' signs and coefficients, so the correct step should have been to recognize the mistake and correctly factor the expression if necessary. Let's correct this: The correct factoring should actually be $-3(2p^2 - 7p)$ , but we need to ensure the original expression matches this form correctly.