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

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Question

(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.

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(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 $3$.

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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