Determine a and b in a polynomial long division so that the result is a polynomial

Question

I'm stuck with this, I've been trying factoring, rooting and I just don't understand what to do.

Determine a&b so that the result of the long division is polynomial $$ \frac{x^3+ax^2+bx-4a}{x^2-2x-3}$$

Answer 1

@dxiv provided the answer. So take the following as a step-by-step elaboration, and I don't know... Upvote one of @dxiv's answers, or whatever...

What he/she is saying is that the polynomial remainder theorem states that the remainder of the division of a polynomial $p(x)$ by a linear form $x-n$ is equal to $p(n),$ i.e. the polynomial evaluated at $n:$

$$\frac{p(x)}{x-n} =q(x) + \frac{p(n)}{x-n}.$$

In here, $q(x)$ is the quotient polynomial (the answer of the division when the remainder is zero). This is the case in your problem: the result of the long division is a polynomial.

The expression above can be re-formulated as

$$p(x)=(x-n)\;q(x)\;+\;p(n).$$

We know that we can decompose the expression in your original question as:

$$\frac{x^3+ax^2+bx-4a}{x^2-2x-3}=\frac{x^3+ax^2+bx-4a}{(x+1)(x-3)}$$

Since the remainder is zero, i.e. $p(n)=0$, you can arrive in a couple of steps to

$$x^3+ax^2+bx-4a=(x+1)(x-3) \;q'(x)\tag 1$$

We don't know what $q(x)$ is right away, but it doesn't matter, because we know that when $x=-1,$ the expression on the RHS will be zero, and if we calculate the LHS of Eq. 1, we arrive at @dxiv equation under comments:

$$p(-1)=-1+a-b-4a=0$$

and the parallel equation for $x=3:$

$$p(3)= 27 +9a+3b -4a=0$$

Now we have a system with two unknowns and two equations.

Answer 2

Hint

If you use long division, $$\frac{x^3+ax^2+bx-4a}{x^2-2x-3}=x+a+2+\frac{x (2 a+b+7)-a+6}{x^2-2x-3}$$ Then $a-6=0$ and $2a+b+7=0$ then $a$ and $b$.