Question

18x^2+9x+3 divided by -3x

Answer 1

To divide the expression \( 18x^2 + 9x + 3 \) by \( -3x \), we can rewrite the division as follows:

\[
\frac{18x^2 + 9x + 3}{-3x}
\]

Now, we can separate the terms in the numerator:

\[
\frac{18x^2}{-3x} + \frac{9x}{-3x} + \frac{3}{-3x}
\]

Now, we can simplify each term:

1. For the first term:
\[
\frac{18x^2}{-3x} = -6x \quad (\text{since } 18 \div -3 = -6 \text{ and } x^2 \div x = x)
\]

2. For the second term:
\[
\frac{9x}{-3x} = -3 \quad (\text{since } 9 \div -3 = -3 \text{ and } x \div x = 1)
\]

3. For the third term:
\[
\frac{3}{-3x} = -\frac{1}{x} \quad (\text{since } 3 \div -3 = -1)
\]

Now, putting it all together, we have:

\[
-6x - 3 - \frac{1}{x}
\]

So the final answer is:

\[
-6x - 3 - \frac{1}{x}
\]