If x + 1, x + 2 are two factors of x3 + 3x2 -2αx + β, then values of

1.	If x + 1, x + 2 are two factors of x3 + 3x2 -2αx + β, then values of α + β is ……………A) – 1 B) 1 C) -2 D) 2
Answer» Correct option is (A) –1 (x+1) & (x+2) are two factors of \(p(x)=x^3+3x^2-2\alpha x+\beta\) \(\therefore\) x = -1 & x = -2 are two roots of equation p(x) = 0 \(\therefore\) \((-1)^3+3(-1)^2-2\alpha(-1)+\beta=0\) and \((-2)^3+3(-2)^2-2\alpha(-2)+\beta=0\) \(\Rightarrow\) \(-1+3+2\alpha+\beta=0\) and \(-8+12+4\alpha+\beta=0\) \(\Rightarrow\) \(2\alpha+\beta=-2\) and \(4\alpha+\beta=-4\) \(\Rightarrow\) \((4\alpha+\beta)-(2\alpha+\beta)\) = -4 - (-2) = -2 \(\Rightarrow\) \(2\alpha=-2\) \(\Rightarrow\) \(\alpha=\frac{-2}2=-1\) \(\therefore\) \(\beta=-2-2\alpha=-2-2\times-1\) = -2+2 = 0 \(\therefore\) \(\alpha+\beta\) = -1+0 = -1 Correct option is A) – 1

If x + 1, x + 2 are two factors of x3 + 3x2 -2αx + β, then values of α + β is ……………A) – 1 B) 1 C) -2 D) 2

Answer»

Correct option is (A) –1

(x+1) & (x+2) are two factors of \(p(x)=x^3+3x^2-2\alpha x+\beta\)

\(\therefore\) x = -1 & x = -2 are two roots of equation p(x) = 0

\(\therefore\) \((-1)^3+3(-1)^2-2\alpha(-1)+\beta=0\)

and \((-2)^3+3(-2)^2-2\alpha(-2)+\beta=0\)

\(\Rightarrow\) \(-1+3+2\alpha+\beta=0\) and \(-8+12+4\alpha+\beta=0\)

\(\Rightarrow\) \(2\alpha+\beta=-2\) and \(4\alpha+\beta=-4\)

\(\Rightarrow\) \((4\alpha+\beta)-(2\alpha+\beta)\) = -4 - (-2) = -2

\(\Rightarrow\) \(2\alpha=-2\)

\(\Rightarrow\) \(\alpha=\frac{-2}2=-1\)

\(\therefore\) \(\beta=-2-2\alpha=-2-2\times-1\) = -2+2 = 0

\(\therefore\) \(\alpha+\beta\) = -1+0 = -1

Correct option is A) – 1