If alpha is a root of multiplicity 3 of the e | vedclass

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(A) Let $f(x) = x^5-8x^4+25x^3-38x^2+28x-8$. Since $\alpha$ is a root of multiplicity $3$,we have $f(\alpha) = 0$,$f'(\alpha) = 0$,and $f''(\alpha) =

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(A) Let $f(x) = x^5-8x^4+25x^3-38x^2+28x-8$. Since $\alpha$ is a root of multiplicity $3$,we have $f(\alpha) = 0$,$f'(\alpha) = 0$,and $f''(\alpha) = 0$. Calculating the derivatives: $f'(x) = 5x^4-32x^3+75x^2-76x+28$ $f''(x) = 20x^3-96x^2+150x-76$ Testing $x = 2$: $f(2) = 32-128+200-152+56-8 = 0$ $f'(2) = 80-256+300-152+28 = 0$ $f''(2) = 160-384+300-76 = 0$ $f'''(2) = 60(4)-192(2)+150 = 240-384+150 = 6 
eq 0$. Thus,$\alpha = 2$ is a root of multiplicity $3$. Therefore,$\alpha^2-5\alpha+6 = (2)^2-5(2)+6 = 4-10+6 = 0$.

If $\alpha$ is a root of multiplicity $3$ of the equation $x^5-8x^4+25x^3-38x^2+28x-8=0$,then $\alpha^2-5\alpha+6=$

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