Respuesta :
The GCD is [tex]x+1[/tex], so [tex]x+1[/tex] divides both [tex]x^2+ax+b[/tex] and [tex]x^2+bx+c[/tex]. For some [tex]\alpha[/tex] and [tex]\beta[/tex] we can write
[tex]x^2 + ax + b = (x + 1) (x - \alpha)[/tex]
[tex]x^2 + bx + c = (x + 1) (x - \beta)[/tex]
Expanding the right sides, we get
[tex]x^2 + ax + b = x^2 + (1 - \alpha) x - \alpha[/tex]
[tex]x^2 + bx + c = x^2 + (1 - \beta) x - \beta[/tex]
[tex]x+1[/tex] also divides the LCM, so
[tex]x^3 - 4x^2 + x + 6 = (x + 1) (x - \alpha) (x - \beta)[/tex]
and expanding gives
[tex]x^3 - 4x^2 + x + 6 = x^3 + (1 - \alpha - \beta) x^2 + (\alpha \beta - \alpha - \beta) x + \alpha \beta[/tex]
Matching up all the coefficients, we have
[tex]\begin{cases} a = 1 - \alpha \\ b = -\alpha \\ b = 1 - \beta \\ c = -\beta \\ -4 = 1 - \alpha - \beta \\ 1 = \alpha \beta - \alpha - \beta \\ 6 = \alpha \beta \end{cases} \implies \alpha + \beta = 5[/tex]
All the unknowns are integers, and we have the prime factorization 6 = 2×3 that also satisfies 5 = 2 + 3.
• If [tex]\alpha=2[/tex] and [tex]\beta=3[/tex], then [tex]a=-1[/tex], [tex]b=-2[/tex], and [tex]c=-3[/tex].
• If [tex]\alpha=3[/tex] and [tex]\beta=2[/tex], then there is no solution for [tex]a,b,c[/tex] since
[tex]\begin{cases}b = -\alpha \\ b = 1-\beta \end{cases} \implies \alpha = \beta - 1[/tex]
but 3 ≠ 2 - 1 = 1.
It follows that [tex]a + b + c = -1 - 2 - 3 = \boxed{-6}[/tex].
