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 $$3$$.
See development and solution
Development:
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:
$$2(2(2(a-2)-3)+2)-3=3 \Leftrightarrow 2(2(2(a-2)-3)+2)=0 \Leftrightarrow$$
$$2(2(a-2)-3)+2=0 \Leftrightarrow 2(2(a-2)-3)=-2 \Leftrightarrow$$
$$2(a-2)-3=-1 \Leftrightarrow 2(a-2)=2 \Leftrightarrow a=3$$
Solution:
With the value of $$a=3$$, the result of the division has a remainder equal to $$3$$.