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[tex]\large\underline{\sf{Solution-}}[/tex]

Given polynomial is

[tex]\rm \: p(x) = {x}^{3} - {6x}^{2} + 2x - 3 \\ [/tex]

and divisor is

[tex]\rm \: g(x) = 3x - 1 \\ [/tex]

We know,

Remainder Theorem :- Remainder Theorem states that if a polynomial p(x) of degree greater than or equals to 1 is divided by linear polynomial x - a, then remainder is p(a).

So, Using Remainder Theorem, remainder when p(x) is divided by g(x) is

[tex]\rm \: p\bigg(\dfrac{1}{3} \bigg) \\ [/tex]

[tex]\rm \: = {\bigg(\dfrac{1}{3} \bigg)}^{3} - {6\bigg(\dfrac{1}{3} \bigg)}^{2} + 2\bigg(\dfrac{1}{3} \bigg) - 3 \\ [/tex]

[tex]\rm \: = \: \dfrac{1}{27} - \dfrac{6}{9} + \dfrac{2}{3} - 3 \\ [/tex]

[tex]\rm \: = \: \dfrac{1}{27} - \dfrac{2}{3} + \dfrac{2}{3} - 3 \\ [/tex]

[tex]\rm \: = \: \dfrac{1 - 81}{27} \\ [/tex]

[tex]\rm \: = \: - \: \dfrac{80}{27} \\ \\ [/tex]

Hence,

[tex]\bf\implies \:Remainder= \: - \: \dfrac{80}{27} \\ \\ [/tex]

[tex]\rule{190pt}{2pt} \\ [/tex]

[tex] { \red{ \mathfrak{Additional\:Information}}}[/tex]

Factor Theorem :-

This theorem states that if a polynomial p(x) is a polynomial of degree greater than or equals to 1 is divided by linear polynomial x - a, then remainder is 0.

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]