Let $Δ (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 x^{3} - 3 x^{2} & 5 x + 7 & 2 4 x^{3} - 7 x & 3 x + 2 & 1 7 x^{3} - 8 x^{2} & x - 1 & 3 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$= a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4}$

Column I Column II

A $a_{0}$ equals top p -73

B $a_{1}$ equals to q 46

C $a_{3}$ equals to r -43

D $a_{4}$ equals to t 0

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Solution

$Δ (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 x^{3} - 3 x^{2} & 5 x + 7 & 2 4 x^{3} - 7 x & 3 x + 2 & 1 7 x^{3} - 8 x^{2} & x - 1 & 3 \end{matrix} ∣ ∣ ∣ ∣ ∣ = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4}$

$R_{3} \to R_{3} - 3 R_{2}, R_{1} \to R_{1} - 2 R_{2}$

= $∣ ∣ ∣ ∣ ∣ \begin{matrix} - 6 x^{3} - 3 x^{2} + 14 x & 3 - x & 0 4 x^{3} - 7 x & 3 x + 2 & 1 - 5 x^{3} - 8 x^{2} + 21 x & - 8 x - 7 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$= - 1 [(6 x^{3} + 3 x^{2} - 14 x) (8 x + 7) - (5 x^{3} + 8 x^{2} - 21 x) (x - 3)]$

By comparing the coefficient method
$a_{0} = 0, a_{1} = 161, a_{3} = - 73, a_{4} = - 43, a_{2} = 46$

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