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Given polynomials^{3}+10 s^{2}+50 s+500=0This criterion is based on arranging the coefficients of characteristic equation into an array called "Routh array".q(s)=a_{0} s^{n}+a_{1} s^{n-1}+a_{2} s^{n-2}+\cdots+a_{n}=0 .Routh array: \quad a_{0}=1 ; a_{1}=10 ; a_{2}=50 ; a_{3}=500s^{n} \quad a_{0} \quad a_{2} \quad a_{4} \quad a_{6} \ldotss^{n-1} \quad a_{1} \quad a_{3} \quad a_{5} \quad a_{7} \ldots where;\begin{array}{l}s^{n-2} \quad b_{1} b_{2} b_{3} \ldots \ldots \quad b_{1}=\frac{a_{1} a_{2}-a_{0} a_{3}}{a_{1}} \\s^{n-3} c_{1} \quad c_{2} \quad c_{3} \ldots b_{2}=\frac{a_{1} a_{4}-a_{0} a_{5}}{a_{1}} \\c_{1}=\frac{b_{1} a_{3}-a_{1} b_{2}}{b_{1}} \\s^{0} \quad a_{n} \\c_{2}=\frac{b_{1} a_{5}-a_{1} b_{3}}{b_{1}} \\5^{3} 150 \\s^{2} \quad 10 \quad 500 \\s^{\prime} \quad 0 \quad 0 \\b_{1}=\frac{10 \times 50-500}{10} \\b_{1}=0 \\\end{array}No sign change in the first column of Routh table, we got zero, so the system is marginally stable.\rightarrow Answer is option (C) ...