Question

The equation whose roots are 1,2 and 3 is

Answer 1

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Answer 2

$\Huge{\textbf{\textsf{{\purple{Ans}}{\pink{wer}}{\color{pink}{:}}}}}$

Method -1:

Most general formula :

If a,b,c, ... p,q are the roots of any equation then the equation is

(x-a)(x-b)(x-c)...(x-p)(x-q) = 0

Now,

If 1,2,3 are the roots of any equation then the equation is

$(x-1)(x-2)(x-3) = 0 \\ \implies \: (x - 1)</p><p>[ \: {x}^{2} - 3x - 2x + 6 ] = 0 \\ \implies \: (x - 1)( {x}^{2} - 5x + 6) = 0 \\ \implies \: ( {x}^{3} - 5 {x}^{2} + 6x - {x}^{2} + 5x - 6) = 0 \\ \implies \: {x}^{3} - 6 {x}^{2} + 11x - 6 = 0$

Note :

You can use this rule anywhere.

Method -2:

$if \: \alpha, \: \: \beta \: and \: \: \gamma\:\: are \: the \: roots \: \\ of \: any \: equation \: the \: eqution \: is \\ \\ {x}^{3} - ( \alpha + \beta + \gamma ) {x}^{2} + ( \alpha \beta + \beta \gamma + \gamma \alpha )x - \alpha \beta \gamma = 0$