Question

(x-1)(x-2)(x-3)(x-4)=12
solve correctly​

Answer 1

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$\diamondsuit\huge\underline{\displaystyle\sf\color{green} \: Answer}$

$\: \\ \implies \displaystyle \tt{(x - 1)(x - 2)(x - 3)(x - 4) = 12 \: \: ...given \: equation}$

$\: \\ \implies \displaystyle \tt{x(x - 2) - 1(x - 2)(x - 3)(x - 4) = 12}$

$\: \\ \implies \displaystyle \tt{( {x}^{2} - 2x - x + 2)(x - 3)(x - 4) = 12}$

$\: \\ \implies \displaystyle \tt{( {x}^{2} - 3x - 2)(x - 3)(x - 4) = 12}$

$\: \\ \implies \displaystyle \tt{x( {x}^{2} - 3x + 2) - 3( {x}^{2} - 3x + 2)(x - 4) = 12}$

$\: \\ \implies \displaystyle \tt{ {x}^{3} - {3x}^{2} + 2x - {3x}^{2} + 9x - 6)(x - 4) = 12}$

$\: \\ \implies \displaystyle \tt{ ({x}^{3} - 6 {x}^{2} + 11x - 6)(x - 4) = 12}$

$\: \\ \implies \displaystyle \tt{ {x}^{3} (x - 4) - 6 {x}^{2} (x - 4) + 11x(x - 4) - 6(x - 4) = 12}$

$\: \: \\ \implies \displaystyle \tt{ {x}^{4} - {4x}^{3} - {6x}^{3} + 24 {x}^{2} + 11 {x}^{2} - 44x - 6x + 24 = 12}$

$\: \\ \implies \displaystyle \tt{ {x}^{4} - 10 {x}^{3} + 35 {x}^{2} - 50x + 24 = 12}$

$\: \\ \implies \displaystyle \tt{ {x}^{4} - {10x}^{3} + 35 {x}^{2} - 50x + 24 - 12 = 0}$

$\: \\ \implies \displaystyle \tt{ {x}^{4} - {10x}^{3} + 35 {x}^{2} - 50x + 12 = 0}$

$\: \divideontimes \boxed {\displaystyle \bf{ {x}^{4} - 10 {x}^{3} + 35 {x}^{2} - 50x + 12 = 0 \: is \: the \: answer}}$