Question

1. $(k^4 - 9k^3 - k^2 + 19k - 10) div (k - 1)$

1 | 1 -9 -1 19 -10
| 1 -8 -9 10
--------------------
1 -8 -9 10 0
$1k^3 - 8k^2 - 9k + 10 + \frac{0}{k - 1}$
... is a factor of the give

Explanation:

Step1: Confirm synthetic division setup

Use root $k=1$, coefficients $1, -9, -1, 19, -10$

Step2: Bring down leading coefficient

$\text{New coefficient: } 1$

Step3: Multiply, add to next coefficient

$1\times1 + (-9) = -8$

Step4: Multiply, add to next coefficient

$-8\times1 + (-1) = -9$

Step5: Multiply, add to next coefficient

$-9\times1 + 19 = 10$

Step6: Multiply, add to last coefficient

$10\times1 + (-10) = 0$

Step7: Form quotient polynomial

Use coefficients $1, -8, -9, 10$ for $k^3, k^2, k, \text{constant}$

Answer:

$k^3 - 8k^2 - 9k + 10$