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Let $$f(x)=x^5+ax^4+bx^3-4x^2+3x+c$$ such that it has a triple integral root. Determine $a,b,c$ and this triple root.

My attempt: there is such a number $\alpha\in\mathbb{Z}$ such that $$(x-\alpha)\vert f(x) \\ (x-\alpha)\vert f'(x) \\ (x-\alpha)\vert f''(x)\\$$ Then, if we find the root of the second derivative of the function that should be it, but this lead me to nowhere since:$$f'(x)=5x^4+4ax^3+3bx^2-8x+3 \\ f''(x)=20x^3+12ax^2+6bx-8$$ Now it seems illogical to check for all cases ($\pm1,\pm2,\pm4$). Hints would be appreciated.

Thanks in advance!

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First of all I assume that $a,b,c$ are integers, as otherwise almost any integer can be a triple root of a polynomial of the given form. Additionally if $a,b,c$ are rational numbers, then the Rational Root Theorem doesn't work.

Anyway back to the problem. Note that $\alpha$ is an integer root of both $f'(x)$ and $f''(x)$. So therefore $\alpha$ is a divisor of both $8$ and $3$, which means that $\alpha = \pm 1$

Now plugging $\alpha = 1$ gives us that:

$$f'(1) = 0 \implies 5 + 4a + 3b - 8 + 3 = 0 \implies 4a + 3b = 0$$ $$f''(1) = 0 \implies 10 + 6a + 3b - 4 = 0 \implies 6a + 3b = -6$$

Now we have that $a=-3, b = 4$. Now:

$$f(1) = 0 \implies 1 - 3 + 4 - 4 + 3 + c = 0 \implies c = -1$$

For the second case, plugging $\alpha = -1$ gives us that:

$$f'(-1) = 0 \implies 5 - 4a + 3b + 8 + 3 = 0 \implies 3b - 4a = - 16$$ $$f''(-1) = 0 \implies -10 + 6a - 3b - 4 = 0 \implies 3b - 6a = -14$$

Solving this system we have that $a = -1, b = \frac{-20}{3}$

But this violatas the condition that $b \in \mathbb{Z}$. Therefore:

$$f(x) = x^5 - 3x^4 + 4x^3 - 4x^2 + 3x - 1 = (x-1)^3(x^2+1) = 0$$

is the only possible solution.

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    Thanks a lot! Can't believe I forgot that the number also divides 3 and not only 8. – 2017-02-02