If one of the roots of the equation 6x^3-25x^ | vedclass

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(D) Let $f(x) = 6x^3-25x^2+2x+8$. By the Rational Root Theorem,possible integer roots are factors of $8$ divided by factors of $6$. Testing $x=2$: $f(

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$4$

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(D) Let $f(x) = 6x^3-25x^2+2x+8$. By the Rational Root Theorem,possible integer roots are factors of $8$ divided by factors of $6$. Testing $x=2$: $f(2) = 6(8)-25(4)+2(2)+8 = 48-100+4+8 = -40 
eq 0$. Testing $x=4$: $f(4) = 6(64)-25(16)+2(4)+8 = 384-400+8+8 = 0$. So,$x=4$ is a root. Dividing $6x^3-25x^2+2x+8$ by $(x-4)$,we get $6x^2-x-2=0$. Factoring this quadratic: $(2x+1)(3x-2)=0$. The roots are $x = -1/2$ and $x = 2/3$. Given $\alpha > 0$ and $\beta < 0$,we have $\alpha = 2/3$ and $\beta = -1/2$. Then,$\frac{4}{\alpha} + \frac{1}{\beta} = \frac{4}{2/3} + \frac{1}{-1/2} = 6 - 2 = 4$.

If one of the roots of the equation $6x^3-25x^2+2x+8=0$ is an integer and $\alpha > 0$,$\beta < 0$ are the other two roots,then $\frac{4}{\alpha}+\frac{1}{\beta}=$

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