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(A) Given the polynomial equation $x^5+4 x^4-13 x^3-52 x^2+36 x+144=0$. We are given that $2$ is one of the roots. By performing polynomial division o

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(A) Given the polynomial equation $x^5+4 x^4-13 x^3-52 x^2+36 x+144=0$. We are given that $2$ is one of the roots. By performing polynomial division or testing factors,we can factor the expression. Dividing $x^5+4 x^4-13 x^3-52 x^2+36 x+144$ by $(x-2)$,we get the quotient $x^4+6x^3-x^2-54x-72$. Further factoring,we find the roots of the equation are $-4, -3, -2, 2, 3$. Given the condition $\alpha $\alpha = -4, \beta = -3, \gamma = -2, \varepsilon = 3$. Now,calculate the expression: $\alpha+2 \beta+3 \gamma+5 \varepsilon = (-4) + 2(-3) + 3(-2) + 5(3)$ $= -4 - 6 - 6 + 15$ $= -16 + 15 = -1$.

$\alpha, \beta, \gamma, 2, \varepsilon$ are the roots of the equation $x^5+4 x^4-13 x^3-52 x^2+36 x+144=0$. If $\alpha < \beta < \gamma < 2 < \varepsilon$,then find the value of $\alpha+2 \beta+3 \gamma+5 \varepsilon$.

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