# Polynomial Factorisation

Consider that we have a polynomial like $$x^3- (a + b +c ) x^2+abx-abc+s$$ Which is multiplication of $$(x-a)(x-b)(x-c)+s$$ Is it possible to reach value= $abc$ knowing the Coefficients and exponents of the upper polynomial or factorizing of the polynomial is difficult like integer factorization ?

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You can find $abc$ by taking the constant coefficient of $(x-a)(x-b)(x-c)$ and negating it. Is that your question or do I misunderstand? – anon Sep 24 '13 at 3:01
sorry I edited my question – sima Sep 24 '13 at 3:04
If you know the coefficients of the expansion of $(x-a)(x-b)(x-c)+s$ but you don't know $a,b,c,s$ then you cannot determine $abc$. – anon Sep 24 '13 at 3:05
Is it a hard problem like integer factorization? would you plaese introduce me a source – sima Sep 24 '13 at 3:08
It is an impossible problem, like guessing a random number I wrote down without letting you see it. This is because you can vary the numbers $a,b,c,s$ in such a way to get any configuration of coefficients in the expansion and value of $abc$ you want simultaneously, so knowing the three coefficients tells us nothing about the product's value. – anon Sep 24 '13 at 3:10

So you said you are given a polynomial, and you wish to find $abc$ and $s$ which satisfies $$x^3 - (a+b+c)x^2 + abx - abc + s = (x-a)(x-b)(x-c)+s$$

By Vieta's formula, when we compare the coefficient of $x$, we know \begin{align}ab =& ab+bc+ca\\0=&c(a+b) \end{align}

If $a+b = 0$, the coefficient of $x^2$ gives $-c$. Multiply this with the negative of the coefficient of $x$ to get $abc$. Otherwise, if $c = 0$, we get $abc = 0$ definitely.

Therefore there can be two solutions for $abc$ from such polynomial $x^3 + Ax^2 + Bx + C$: either $abc = -AB$, or $abc = 0$.

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Based on the first draft of the question I suspect the OP simply has a typo in the expansion of $(x-a)(x-b)(x-c)$. I could be wrong though. – anon Sep 24 '13 at 3:54
sorry i can't understand or maybe there is a mistake – sima Sep 24 '13 at 5:03
we have a product of a polynomial+secret is it possible to determine s – sima Sep 24 '13 at 5:04
Notice that when you say a polynomial $x^3 + Ax^2 + Bx + C$ can be written as both $x^3 - (a+b+c)x^2 + abx - abc + s = (x-a)(x-b)(x-c)+s$, when you compare the coefficient of $x$, we know $ab = ab+bc+ca$. This limits the possibility of the value $abc$, which can be either $-AB$ or $0$. Your "secret" $s$ is either $C-AB$ or $C$. – peterwhy Sep 24 '13 at 16:06