Divide (2x3+4x2+5x−6)÷(x+1)

To perform synthetic division of the polynomial \( 2x^3 + 4x^2 + 5x - 6 \) by \( x + 1 \), we first need to find the root of \( x + 1 = 0 \), which is \( x = -1 \).

Next, we write down the coefficients of the polynomial we are dividing:

- The coefficients for \( 2x^3 + 4x^2 + 5x - 6 \) are:
- \( 2 \) (for \( x^3 \))
- \( 4 \) (for \( x^2 \))
- \( 5 \) (for \( x^1 \))
- \( -6 \) (for the constant term)

So, the coefficients are \( 2, 4, 5, -6 \).

Now we set up synthetic division using \( -1 \) (the value we found) and the coefficients:

```
-1 | 2 4 5 -6
| -2 -2 -3
---------------------
2 2 3 -9
```

1. Bring down the first coefficient \( 2 \).
2. Multiply \( -1 \) by \( 2 \) and write it under the next coefficient \( 4 \): \(-1 \times 2 = -2\).
3. Add \( 4 + (-2) = 2\).
4. Repeat for the next coefficient: \( -1 \times 2 = -2\), so \( 5 + (-2) = 3\).
5. For the last part, \(-1 \times 3 = -3\), so \(-6 + (-3) = -9\).

The bottom row \( 2, 2, 3 \) represents the coefficients of the quotient polynomial \( 2x^2 + 2x + 3 \), and the remainder is \(-9\).

Thus, the quotient is \( 2x^2 + 2x + 3 \) with a remainder of \(-9\).

The missing term in the quotient (the coefficient of \( x \) term, which is already filled) is:

**2**.