When the roots of x^3+alpha x^2+beta x+6=0 ar | vedclass

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(A) First,find the roots of the equation $x^4-6x^3+11x^2-6x=0$. Factoring,we get $x(x^3-6x^2+11x-6)=0$. Further factoring $x^3-6x^2+11x-6$,we get $x(x

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$\alpha-\beta+5=0$

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(A) First,find the roots of the equation $x^4-6x^3+11x^2-6x=0$. Factoring,we get $x(x^3-6x^2+11x-6)=0$. Further factoring $x^3-6x^2+11x-6$,we get $x(x-1)(x-2)(x-3)=0$. The roots are $0, 1, 2, 3$. The least root is $0$. Let the roots of $x^3+\alpha x^2+\beta x+6=0$ be $r_1, r_2, r_3$. When these roots are increased by $1$,one of the new roots is $0$. So,$r_i+1=0$ for some $i$,which means $r_i=-1$. Since $-1$ is a root of $x^3+\alpha x^2+\beta x+6=0$,we substitute $x=-1$: $(-1)^3+\alpha(-1)^2+\beta(-1)+6=0$ $-1+\alpha-\beta+6=0$ $\alpha-\beta+5=0$.

When the roots of $x^3+\alpha x^2+\beta x+6=0$ are increased by $1$,if one of the resultant values is the least root of $x^4-6 x^3+11 x^2-6 x=0$,then

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