Finding polynomial. Let $f(x)$ be a fifth degree polynomial. Such that$(x-1)^3$ divides $(f(x)+2)$
&$(x+1)^3$ divides $(f(x)-2)$.
Find $f(x)$.
It was to be find without using calculus.but l don't know how to do this .
I was surprised that how it can be done using calculus.
Please tell me how to solve such questions
I would be thankful.
 A: There's "mild" calculus in this solutions, basically the fact that if $r$ is a triple root of $P(x)$ then $P'(r)=P''(r)=0$. This can be proved by pure algebra.
Since $(x-1)^3$ divides $f(x)+2$ you can write
$$f(x)+2=(x-1)^3\,P(x)$$
where $P(x)$ is a second degree polynomial. Then
$$f(x)-2=(x-1)^3\,P(x)-4$$
Since $(x+1)^3$ divides that polynomial, we get $0$ if we plug in $x=-1$ which gives
$$0=-8P(-1)-4\Leftrightarrow P(-1)=-\dfrac{1}{2}$$
We also know that $x=-1$ is a root of the derivative of $f(x)-2$ which is
$$f'(x)=3(x-1)^2\,P(x)+(x-1)^3P'(x)$$
so that
$$0=12P(-1)-8P'(-1)$$
since we know $P(-1)$, we find easily $P'(-1)$. Another application of the same idea gives us $P''(-1)$. Now $P$ is a second degree polynomial, let's write it as
$$P(x)=a(x+1)^2+b(x+1)+c$$
(sort of a Taylor series near $-1$). Then $c=P(1)$, $b=P'(1)$, $a=\dfrac{P''(1)}{2}$ which are all known numbers at this stage. Thus we obtain $P(x)$, and then we can use
$$f(x)+2=(x-1)^3\,P(x)$$
to get $f(x)$.
A: Hint: As $(x-1)^3$ divide $f(x)+2$, we have that $(x-1)^2$ divide $f^{\prime}(x)$. Also, $(x+1)^2$ divide $f^{\prime}(x)$. As $f$ is of degree $5$, there exists a constant $c$ such that $f^{\prime}(x)=c(x-1)^2(x+1)^2=c(x^2-1)^2$. I think you can take it from here...  
A: Hint:
From the given,
$$f(x)=(ax^2+bx+c)(x-1)^3-2=(a'x^2+b'x+c')(x+1)^3+2.$$
But expanding and identifying the coefficients, you will form a system of six equations in six unknowns. You can also write the equality for six distinct values of $x$.

From $x=\infty$, $a'=a$ and from $x=0$, $c'+2=-c-2$. Then $x=1$ gives $8(a'+b'+c')+2=-2$ and $x=-1$, $-8(a-b+c)-2=2$. Just two unknowns remain...
