Find a polynomial f(x) of degree 5 such that 2 properties hold. I have been trying to find a polynomial $f(x)$ such that these $2$ properties hold:


*

*$f(x)-1$ is divisible by $(x-1)^3$

*$f(x)$ is divisible by $x^3$
To start, I set $f(x) =ax^5 + bx^4 + cx^3 + dx^2 + ex + f$.
This is divisible by $x^3$, so $d, e, f $  must be $0$. So polynomial $f(x) = ax^5 + bx^4 + cx^3$
This minus $1$ is divisible by $(x-1)^3$. So I used synthetic division and got that the remainder of $ax^5 + bx^4 + cx^3$ divided by $(x-1)$ is $a+b+c-1$. But it should divide out evenly, so $a+b+c-1 = 0$.
From here, I can't find any other equations involving the three variables. Could someone have any suggestions for how to continue? Thanks in advance. :)
 A: When a repeated factor $(x-a)^3$ is present in polynomial $P(x)$, then $P(a) = P'(a) = P''(a) = 0$
So set $f(1) - 1 = 0, f'(1) = 0, f''(1) = 0$
$f(x) - 1 = ax^5 + bx^4 + cx^3 - 1$
$f'(x) = 5ax^4 + 4bx^3 + 3cx^2$
$f''(x) = 20ax^3 + 12bx^2 + 6cx$
Hence your three equations are:
$a + b + c - 1 = 0 $  
$5a + 4b + 3c = 0$    
$20a + 12b + 6c = 0$   
and you should be able to solve them to get $(a,b,c) = (6,-15, 10)$ so $f(x) = 6x^5 - 15x^4 + 10x^3$ = $x^3(6x^2-15x + 10)$.
A: Because $f$ is divisible by $x^3$, we must have $f(x)=r_3x^3+r_4x^4+r_5x^5$.
The second condition gives, comparing coefficients of powers of $x$, that
$$
f(x)=6x^5 - 15x^4 + 10x^3.
$$
Edit: Comparing coefficients gives me the linear equations $r_4 + 3r_5 - 3=0, r_3 - 3r_5 + 8$ and $r_5=6$.
A: Since $x^2$ and $(x-1)^2$ divide $p'(x)$ we have $$p'(x) =ax^2(x-1)^2 =ax^4-2ax^3+ax^2,$$ thus $$p(x) = {a\over 5}x^5-{a\over 2}x^4+{a\over 3}x^3+c$$
Since $p(0)=0$ we have $c=0$ and since $p(1)=1$ we have ${a\over 5}-{a\over 2}+{a\over 3}=1$ so $a=30$.
