\( ( x ^ { 3 } + 2 x + 1 ) \div ( x - 1 ) = \square + \frac { \square } { x - 1 } \)

To perform the division \( (x^3 + 2x + 1) \div (x - 1) \), we can use polynomial long division. 1. **Divide the leading term**: Divide the leading term of the dividend \( x^3 \) by the leading term of the divisor \( x \): \[ x^3 \div x = x^2 \] 2. **Multiply and subtract**: Multiply \( x^2 \) by the entire divisor \( (x - 1) \): \[ x^2(x - 1) = x^3 - x^2 \] Now subtract this from the original polynomial: \[ (x^3 + 2x + 1) - (x^3 - x^2) = x^2 + 2x + 1 \] 3. **Repeat the process**: Now divide the leading term \( x^2 \) by \( x \): \[ x^2 \div x = x \] Multiply \( x \) by the divisor: \[ x(x - 1) = x^2 - x \] Subtract this from the current polynomial: \[ (x^2 + 2x + 1) - (x^2 - x) = 3x + 1 \] 4. **Repeat again**: Now divide \( 3x \) by \( x \): \[ 3x \div x = 3 \] Multiply \( 3 \) by the divisor: \[ 3(x - 1) = 3x - 3 \] Subtract this from the current polynomial: \[ (3x + 1) - (3x - 3) = 4 \] 5. **Final result**: Now we have a remainder of \( 4 \). Therefore, we can write the result of the division as: \[ x^2 + x + 3 + \frac{4}{x - 1} \] Putting it all together, we have: \[ (x^3 + 2x + 1) \div (x - 1) = x^2 + x + 3 + \frac{4}{x - 1} \] Thus, filling in the squares: \[ \boxed{x^2 + x + 3} + \frac{\boxed{4}}{x - 1} \]