Homework Help?

1.
When $x^4 + 6x^3 - ax^2 - 45x -15$ is divided by $x^2 - x - 6$ the remainder is $3$.
What is $a$?

$\text{Let $x^2 - x - 6 = (x-3)(x+2)$}$

$\begin{array}{|lrcll|} \hline & \mathbf{\dfrac{x^4 + 6x^3 - ax^2 - 45x -15}{(x-3)(x+2)}} &=& \mathbf{q(x) + \dfrac{3}{(x-3)(x+2)}} \quad | \quad \times (x-3)(x+2) \\\\ x=3: & x^4 + 6x^3 - ax^2 - 45x -15 &=& q(x)(x-3)(x+2) + 3 \\ & 3^4+6(3^3)-a(3^2)-45(3)-15 &=& q(x)(0)(5) + 3 \\ & 3^4+6(3^3)-a(3^2)-45(3)-15 &=& 3 \\ & 81+162-9a-135-15 &=& 3 \\ & 93-9a &=& 3 \quad | \quad :3 \\ & 31-3a &=& 1 \\ & 3a &=& 31-1 \\ & 3a &=& 30 \quad | \quad :3 \\ & \mathbf{a} &=& \mathbf{10} \\ \hline \end{array}$

2.

Find the remainder when r^14 is divided by r - 1.

The remainder is 1