Question

factorization of x^3-7x^2+15x-9

Answer 1

BY inception we found that (x-1) is the factor of
$f(x) = {x}^{3} - 7 {x}^{2} + 15x - 9$
$f(1) = {1}^{3} - 7 + 15 - 9 = 0 \\ hence \: (x - 1) \: is \: the \: factor \: of \: f(x) \\ now \: by \: factor \: theorem \\ f(x) \: can \: be \: splitted \: as \\ {x}^{2} (x - 1) - 6x(x - 1) + 9(x - 1) \\ taking \: (x - 1) \: common \: we \: get \\ (x - 1)( {x}^{2} - 6x + 9) \\ (x - 1)( {x}^{2} - 3x - 3x + 9) \\ (x - 1) (x(x - 3) - 3(x - 3)) \\ (x - 1)(x - 3)(x - 3) \\ hence \: (x - 1)(x - 3)(x - 3) \: are \: factors$

Answer 2

Answer:

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