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:

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

2. $f(x)$ 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. :)

Solution 1:

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 $r_5=6$.

Solution 2:

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 $f(x) = 6x^5 - 15x^4 + 10x^3$ = $x^3(6x^2-15x + 10)$.