Ruffini's rule

Do the division $\dfrac{p(x)}{q(x)}$, where $p(x)=-x^4+ax^3-3x^2+2x-3$ and $q(x)=x-2$, and impose the value of the parameter $a$ so that the division has a remainder equal to $3$.

We apply Ruffini's rule:

  $-1$ $+a$ $-3$ $+2$ $-3$
$2$   $-2$ $2(a-2)$ $2(2(a-2)-3)$ $2(2(2(a-2)-3)+2)$
  $-1$ $a-2$ $2(a-2)-3$ $2(2(a-2)-3)+2$ $2(2(2(a-2)-3)+2)-3$

Therefore, now we have to solve the following equation:

$$2(2(2(a-2)-3)+2)-3=3$$

So:

$$2(2(2(a-2)-3)+2)-3=3 \Leftrightarrow 2(2(2(a-2)-3)+2)=0 \Leftrightarrow$$

$$2(2(a-2)-3)+2=0 \Leftrightarrow 2(2(a-2)-3)=-2 \Leftrightarrow$$

$$2(a-2)-3=-1 \Leftrightarrow 2(a-2)=2 \Leftrightarrow a=3$$

With the value of $a=3$, the result of the division has a remainder equal to $3$.

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