To simplify the expression \(\frac{2x^2 + x - 1}{x} \div \frac{2x - 1}{2x + 1}\), we first rewrite the division of fractions as multiplication by the reciprocal:
\[ \frac{2x^2 + x - 1}{x} \times \frac{2x + 1}{2x - 1} \]
Next, we need to simplify the expression step-by-step.
- Factor the numerator \(2x^2 + x - 1\).
To factor \(2x^2 + x - 1\), we look for two numbers that multiply to \(2 \cdot (-1) = -2\) and add to \(1\). The numbers \(2\) and \(-1\) work since \(2 \cdot (-1) = -2\) and \(2 + (-1) = 1\).
Now we can rewrite the polynomial:
\[ 2x^2 + 2x - x - 1 = (2x^2 + 2x) + (-x - 1) = 2x(x + 1) - 1(x + 1) = (2x - 1)(x + 1) \]
Now substituting this back into the original expression, we have:
\[ \frac{(2x - 1)(x + 1)}{x} \times \frac{2x + 1}{2x - 1} \]
- Cancel the common terms \((2x - 1)\):
In the expression now, we can cancel out the \((2x - 1)\) in the numerator and denominator:
\[ \frac{(x + 1)}{x} \times (2x + 1) \]
- Now multiply:
\[ \frac{(x + 1)(2x + 1)}{x} \]
- Simplify:
Now we need to multiply \((x + 1)\) and \((2x + 1)\):
\[ (x + 1)(2x + 1) = x(2x) + x(1) + 1(2x) + 1(1) = 2x^2 + x + 2x + 1 = 2x^2 + 3x + 1 \]
So, we substitute this back into the fraction:
\[ \frac{2x^2 + 3x + 1}{x} \]
- Finally, we can write this as:
\[ 2x + 3 + \frac{1}{x} \]
Thus, the simplified form of the expression is:
\[ 2x + 3 + \frac{1}{x} \]