Suppose `a ,b ,c in I` such that the greatest common divisor for `x^2+a x+b` and `x^2+bx+c` is `(x+1)` and the least common multiple of `x^2+a x+b` and `x^2+b x+c` is `(x^3-4x^2+x+6)`. Then the value of `|a+b+c|` is equal to ___________.

Answer

6

Explanation

`x^(2)+ax+b-=(x+1)(x+b)rArrb+1=a" "(1)`
also `x^(2)+bx+c-=(x+1)(x+c)rArrc+1=b`
or `b+1=c+2" "(2)`
Hence `b+1=a=c+2`
Also `(x+1)(x+b)(x+c)-=x^(3)-4x^(2)+x+6`
`rArrx^(3)+(1+b+c)x^(2)+(b+bc+c)x+bc-=x^(3)-4x^(2)+x+6`
`rArr1+b+c=-4`
`rArr 2c+2=-4`
`rArr c=-3,b=-2 and a =-1`
`rArr a+b+c=-6`