Suppose a, b, c ∈ I such that the greatest common divisor of x2 + ax + b and x2 + bx + c is (x + 1) and the least common multiple of x2 + ax + b and x2 + bx + c is (x3 - 4x2 + x + 6). Then the value of |a + b + c| is equal to __________.

x2+ax+b≡(x+1)(x+b)⇒b+1=a …… (1)also x2+bx+c≡(x+1)(x+c)⇒c+1=bor b+1=c+2 ….. (2)Hence b+1=a=c+2Also (x+1)(x+b)(x+c)≡x3−4x2+x+6⇒ x3+(1+b+c)x2+(b+bc+c)x+bc≡x3−4x2+x+6⇒ 1+b+c=−4⇒ 2c+2=−4⇒ c=−3;b=−2 and a=−1⇒ a+b+c=−6

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Suppose a, b, c $\in$ I such that the greatest common divisor of x² + ax + b and x² + bx + c is (x + 1) and the least common multiple of x² + ax + b and x² + bx + c is (x³ - 4x² + x + 6). Then the value of |a + b + c| is equal to __________.

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answer is -6.

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Detailed Solution

$x^{2} + ax + b \equiv (x + 1) (x + b) \Rightarrow b + 1 = a$ …… (1)

also $x^{2} + bx + c \equiv (x + 1) (x + c) \Rightarrow c + 1 = b$

or $b + 1 = c + 2$ ….. (2)

Hence $b + 1 = a = c + 2$

Also $(x + 1) (x + b) (x + c) \equiv x^{3} - 4 x^{2} + x + 6$

$\begin{array}{l} \Rightarrow x^{3} + (1 + b + c) x^{2} + (b + bc + c) x + bc \equiv x^{3} - 4 x^{2} + x + 6 \\ \Rightarrow 1 + b + c = - 4 \\ \Rightarrow 2 c + 2 = - 4 \\ \Rightarrow c = - 3; b = - 2 and a = - 1 \\ \Rightarrow a + b + c = - 6 \end{array}$