Prove that one root of the equation is x=2 and hence find the remaining roots
$∣ ∣ ∣ ∣ \begin{matrix} x 2 - 3 \end{matrix} \begin{matrix} - 6 - 3 x 2 x \end{matrix} \begin{matrix} - 1 x - 3 x + 2 \end{matrix} ∣ ∣ ∣ ∣ = 0$

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Solution

$∣ ∣ ∣ ∣ \begin{matrix} x & - 6 & - 1 2 & - 3 x & x - 3 - 3 & 2 x & x + 2 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow x [- 3 x (x + 2) - 2 x (x - 3)] + 6 [2 (x + 2) + 3 (x - 3)] - 1 [2 (2 x) - 9 x] = 0$

$\Rightarrow x [- 3 x^{2} - 6 - 2 x^{2} + 6] + 6 [2 x + 4 + 3 x - 9] - [4 x - 9 x] = 0$

$\Rightarrow x (- 5 x^{2}) + 6 (5 x - 5) + 5 x = 0$

$\Rightarrow - 5 x^{3} + 30 (x - 1) + 5 x = 0$

$\Rightarrow - x^{3} + 6 x - 6 + x = 0$

$\Rightarrow - x^{3} + 7 x - 6 = 0$

$\Rightarrow x^{3} - 7 x + 6 = 0$

Let $f (x) = x^{3} - 7 x + 6$

$f (2) = 2^{3} - 7.2 + 6$

$= 8 - 14 + 6$

$= 14 - 14$

$= 0$

Hence $x = 2$ is a root of the equation $x^{3} - 7 x + 6 = 0$

[see image]

$(x - 2) [x (x + 3) - 1 (x + 3)] = 0$
$(x - 2) [x^{2} + 3 x - x - 3] = 0$

$(x - 2) [x (x + 3) - 1 (x + 3)] = 0$
$(x - 2) (x - 1) (x + 3) = 0$

$∴ x = 2, 1, - 3$

Answer: Roots of the equation are $x = 2, 1, - 3$

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