The sum of all non-integer roots of the equat | vedclass

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(D) Given the equation: $x^5-6x^4+11x^3-5x^2-3x+2=0$. By testing integer roots using the Rational Root Theorem,we find that $x=1$ and $x=2$ are roots.

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$3$

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(D) Given the equation: $x^5-6x^4+11x^3-5x^2-3x+2=0$. By testing integer roots using the Rational Root Theorem,we find that $x=1$ and $x=2$ are roots. Dividing the polynomial by $(x-1)(x-2) = x^2-3x+2$,we get: $(x-1)(x-2)(x^3-3x^2+1)=0$. The non-integer roots are the roots of the cubic equation $x^3-3x^2+1=0$. Let the roots of this cubic equation be $\alpha, \beta, \gamma$. By Vieta's formulas,the sum of the roots is given by $-\frac{b}{a} = -\frac{-3}{1} = 3$. Thus,the sum of all non-integer roots is $3$.

List-$I$	List-$II$
$A$. All the roots are negative	$I$. $(b - 3)^2 = 36 + P^2$ for $P \in R$
$B$. Two roots are complex	$II$. $-3 < b < 9$
$C$. Two roots are positive	$III$. $b \in (-\infty, -3) \cup (9, \infty)$
$D$. All roots are real and distinct	$IV$. $b = 9$
	$V$. $b = -3$

The sum of all non-integer roots of the equation $x^5-6x^4+11x^3-5x^2-3x+2=0$ is

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