Question

divide the polynomial x^4 - 3x^3 - 19x^2 + 3x + 18 by the factor (x-1)

Answer 1

To divide the polynomial \( x^4 - 3x^3 - 19x^2 + 3x + 18 \) by the factor \( (x - 1) \), we will use synthetic division.

1. Write down the coefficients of the polynomial: \( 1, -3, -19, 3, 18 \).
2. Set the divisor \( x - 1 \) equal to zero to find \( x = 1 \).
3. Use synthetic division with \( 1 \):

```
1 | 1 -3 -19 3 18
| 1 -2 -21 -18
------------------------
1 -2 -21 -18 0
```

The bottom row shows the coefficients of the quotient polynomial. The division yields:

- The quotient is \( x^3 - 2x^2 - 21x - 18 \).
- The remainder is \( 0 \).

Thus, dividing \( x^4 - 3x^3 - 19x^2 + 3x + 18 \) by \( (x - 1) \) gives us:

\[
x^4 - 3x^3 - 19x^2 + 3x + 18 = (x - 1)(x^3 - 2x^2 - 21x - 18)
\]

So, the result of the division is:

\[
x^3 - 2x^2 - 21x - 18
\]