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

• Correct answer
  A

  -1
• B

  25
• C

  -36
• D

  48

Solution

Given that $\alpha, \beta, \gamma, 2$ and $\varepsilon$ are roots of the equation $x^5+4 x^4-13 x^3-52 x^2+36 x+144=0$ Put $x=-2$ $-32+64+104-208-72+144=0$ So, $x=-2$ is also roots Put $x=3$ $243+324-351-468+108+144=0$ So, $x=3$ is also roots $\begin{aligned} & \therefore(x-2),(x+2),(x-3) \text { are factors } \\ & \therefore(x-2)(x+2)(x-3)=x^3-3 x^2-4 x+12 \end{aligned}$

Now, $x^2+7 x+12=0$ $\begin{aligned} & \Rightarrow x^2+3 x+4 x+12=0 \Rightarrow(x+3)(x+4)=0 \\ & \Rightarrow x=-3,-4 \end{aligned}$ Hence roots are $-2,2,3,-3,-4$ Given that $\alpha \lt \beta \lt \gamma \lt 2 \lt \varepsilon$ $\therefore \alpha=-4, \beta=-3, \gamma=-2, \varepsilon=3$ Now, $\alpha+2 \beta+3 \gamma+5 \varepsilon=-4-6-6+15=-1$.

Hence, the correct answer is option: 'A'.