Determine polynomial of $3$ degree real coefficients $f(x)$ such that $f(x) \vdots (x-2)$ and $f(x)$ dividing by $x^2-1$ have remainder $2x$

Question

I need help in this problem.

Problem: Determine polynomial of $3$ degree real coefficients $f(x)$ such that $f(x) \vdots (x-2)$ and $f(x)$ divided by $x^2-1$ has remainder $2x$ and suppose $f(x)=x^3+ax^2+bx+c$.

Answer 1

Let $f(x) = ax^3 + bx^2 + cx + d$

and also $f(x) = (px+q)(x^2-1) + 2x$

Comparing the coefficients of $f(x)$, we have $a = p$, $b = q$, $c = 2 - p$, $d = -q$

Also $f(2) = 0 \implies 8a + 4b + 2c + d = 0$

Substituting the values of $a, b, c$ in this equation and simplifying

$6p + 3q + 4 = 0$

You can find infinite number of solutions to the above equation.

For example, if you choose $p = 1$ then $q = -\frac{10}{3}$

and $f(x) = x^3 -\frac{10}{3}x^2 + x + \frac{10}{3}$

Answer 2

Hint:

$f(x)=(x^2-1)(x-a)+2x$ and $f(2)=0$.

Solve for $a$.

$f(x)=x^3-ax^2+x+a$.