Question

x cube minus 10 x square - 53 x minus 42 solve by the factor theorem

Answer 1

Here uh go!

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${x}^{3} - {10x}^{2} - 53x - 42$

$= {x}^{3} + {x}^{2} - {11x}^{2} - 11x - 42x - 42$

$= {x}^{2} (x + 1) - 11x(x + 1) - 42(x + 1)$

$= (x + 1) \: ( {x}^{2} - 11x - 42)$

$= (x + 1) \: ( {x}^{2} - 14x + 3x - 42)$

$= (x + 1) \: {x(x - 14) + 3(x - 14)}$

$= (x + 1) \: (x + 3) \: (x - 14)$

$HENCE, (x+1), (x+3) \: and \: (x-14) \: are \: the \: factor \: of \: equation</p><p>$

${x}^{3} - {10x}^{2} - 53x - 42$