I wanted to put the answer in a formatted way first. Now I'll talk about how I did this problem.

First, I started by looking at 2x^{2}. If I divide this first term by 2x, what do I get? That answer is x. So when I multiply x by 2x+3, I get the quantity 2x^{2} + 3x. Just like in long division, you subtract this product out, leaving -12x - 20. -12x goes into 2x -6 times. Multiply 2x+3 by -6 gives you -12x - 18, which you subtract out leaving -2. You can't go any further so the remainder is written in fractional form with the divisor as the denominator: -2/(2x+3). So you final answer is x - 6 - 2/(2x+3). This satisfies the question, as R has x to the 0th power and D has x to the 1st power, and has been expressed as Q + R/D... quotient plus remainder over divisor.

## Comments

I wanted to put the answer in a formatted way first. Now I'll talk about how I did this problem.

First, I started by looking at 2x

^{2}. If I divide this first term by 2x, what do I get? That answer is x. So when I multiply x by 2x+3, I get the quantity 2x^{2}+ 3x. Just like in long division, you subtract this product out, leaving -12x - 20. -12x goes into 2x -6 times. Multiply 2x+3 by -6 gives you -12x - 18, which you subtract out leaving -2. You can't go any further so the remainder is written in fractional form with the divisor as the denominator: -2/(2x+3). So you final answer is x - 6 - 2/(2x+3). This satisfies the question, as R has x to the 0th power and D has x to the 1st power, and has been expressed as Q + R/D... quotient plus remainder over divisor.And I apologize for the grammatical errors. I can't edit my comment to go back and fix them, only the original posting. I hope this helps you.