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

1 answer

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: \[ \frac{2x^4 - 2x^3 - 6x^2 + 11x + 6}{x^2(x-2)} \]