Question

Add. State the sum in simplest form. 2x+1/x-2 + x-3/x^2 +2x-8

Answer 1

To add the expression \(\frac{2x+1}{x-2} + \frac{x-3}{x^2} + (2x-8)\), we first need to find a common denominator and simplify where possible.

### Step 1: Rewrite the expression
We have:
\[
\frac{2x+1}{x-2} + \frac{x-3}{x^2} + (2x-8)
\]

### Step 2: Find a common denominator
The denominators are \(x-2\) and \(x^2\). The least common denominator (LCD) would be \(x^2(x-2)\).

### Step 3: Rewrite each term with the common denominator
1. For \(\frac{2x+1}{x-2}\):
\[
\frac{2x+1}{x-2} = \frac{(2x+1)x^2}{x^2(x-2)}
\]

2. For \(\frac{x-3}{x^2}\):
\[
\frac{x-3}{x^2} = \frac{(x-3)(x-2)}{x^2(x-2)}
\]

3. For \(2x-8\):
\[
2x - 8 = \frac{(2x - 8) x^2(x-2)}{x^2(x-2)} = \frac{(2x-8)x^2(x-2)}{x^2(x-2)}
\]

### Step 4: Combine all terms over the common denominator
Now we can add these fractions:
\[
\frac{(2x+1)x^2 + (x-3)(x-2) + (2x-8)x^2(x-2)}{x^2(x-2)}
\]

### Step 5: Simplify the numerator
1. Simplifying \((x-3)(x-2)\):
\[
(x-3)(x-2) = x^2 - 5x + 6
\]

2. Expanding \((2x-8)x^2(x-2)\):
\[
(2x-8)x^2(x-2) = (2x-8)(x^3 - 2x^2) = 2x^4 - 4x^3 - 8x^2 + 16x
\]

Now we combine everything in the numerator:
- From \((2x+1)x^2\):
\[
2x^3 + x^2
\]
- From \((x-3)(x-2)\):
\[
x^2 - 5x + 6
\]
- From \((2x-8)x^2(x-2)\):
\[
2x^4 - 4x^3 - 8x^2 + 16x
\]

Now the complete numerator gives us:
\[
2x^4 + (2x^3 - 4x^3) + (x^2 + x^2 - 8x^2) + (-5x + 16x + 6)
\]
Which simplifies to:
\[
2x^4 - 2x^3 - 6x^2 + 11x + 6
\]

### Step 6: Final expression
Putting this all together, we have:
\[
\frac{2x^4 - 2x^3 - 6x^2 + 11x + 6}{x^2(x-2)}
\]

### Conclusion
The simplest form of the sum \(\frac{2x+1}{x-2} + \frac{x-3}{x^2} + (2x-8)\) is:
\[
\frac{2x^4 - 2x^3 - 6x^2 + 11x + 6}{x^2(x-2)}
\]