Question

solve this problem please

Answer 1

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i)

$= > \bf \: g(x) = (x - 3) \\ \\ = > \bf \: p(x) = {x}^{3} - 4 {x}^{2} + x + 6 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {x}^{3} - 3 {x}^{2} - {x}^{2} + 3x - 2x + 6 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = {x}^{2} (x - 3) - x(x - 3) - 2(x - 3) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = (x - 3)( {x}^{2} - x - 2) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = g(x) \times ( {x}^{2} -x - 2)$
Thus..

g(x) is a factor of p(x)

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ii)

$= > \bf \: g(x) = (x + 2) \\ \\ = > \bf \: p(x) = {x}^{3} + 3 {x}^{2} + 3x + 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {(x)}^{3} + 3 {(x)}^{2} + 3x(1) ^{ 3 } + {(1)}^{3} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = (x + 1) ^{3}$
Thus g(x) is not a factor of p(x)

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