Simplify: -4(2x + 5)

Question

Simplify: -4(2x + 5)
A. (-4 - 2x) + (-4 - 5)
B. (-4 + 2x) multiplied by (-4 + 5)
C. (-4 divided by 2x) + (-4 divided by 5)
D. (-4 multiplied by 2x) + (-4 multiplied by 5)

GPT-4o mini · Answer

To simplify the expression $-4(2x + 5)$, we can distribute $-4$ to both terms inside the parentheses: $$ -4(2x + 5) = -4 \cdot 2x + (-4) \cdot 5 = -8x - 20 $$ Now, let's evaluate the options provided: A. $(-4 - 2x) + (-4 - 5)$ - This simplifies to $-4 - 2x - 4 - 5 = -2x - 13$ (not equivalent). B. $(-4 + 2x) \text{ multiplied by } (-4 + 5)$ - This simplifies to $(-4 + 2x) \cdot 1$, which gives $-4 + 2x$ (not equivalent). C. $(-4 \text{ divided by } 2x) + (-4 \text{ divided by } 5)$ - This simplifies to $-\frac{4}{2x} - \frac{4}{5}$ (not equivalent). D. $(-4 \text{ multiplied by } 2x) + (-4 \text{ multiplied by } 5)$ - This simplifies to $-8x - 20$, which is equivalent to the original expression. Therefore, the correct answer is: **D. $(-4 \text{ multiplied by } 2x) + (-4 \text{ multiplied by } 5)$**.

nobody · Answer

are you sure this is correct

GPT-4o mini · Answer

Yes, let's go through the simplification again to ensure clarity.

The original expression is:

\[ -4(2x + 5) \]

We need to distribute $-4$ across both terms inside the parentheses:

Multiply $-4$ by $2x$: \[ -4 \cdot 2x = -8x \]
Multiply $-4$ by $5$: \[ -4 \cdot 5 = -20 \]

Now combine these results:

\[ -8x - 20 \]

So, the simplified expression is indeed $-8x - 20$.

Now, let's evaluate the options again:

A. $(-4 - 2x) + (-4 - 5)$ simplifies to: \[ -4 - 2x - 4 - 5 = -2x - 13 \quad (\text{not equivalent}) \]
B. $(-4 + 2x) \times (-4 + 5)$ simplifies to: \[ (-4 + 2x) \times 1 = -4 + 2x \quad (\text{not equivalent}) \]
C. $(-4 \div 2x) + (-4 \div 5)$ simplifies to: \[ -\frac{4}{2x} - \frac{4}{5} \quad (\text{not equivalent}) \]
D. $(-4 \times 2x) + (-4 \times 5)$ simplifies to: \[ -8x - 20 \quad (\text{this is equivalent to } -8x - 20) \]

Therefore, the correct choice is indeed D. $(-4 \text{ multiplied by } 2x) + (-4 \text{ multiplied by } 5)$, confirming that my previous analysis was accurate.