Using factor theorem, show that \[g\left( x \right)\] is a factor of \[p\left( x \right)\] , when \[p\left( x \right) = 2{x^4} + {x^3} - 8{x^2} - x + 6\] and \[g\left( x \right) = 2x - 3\]

Hint: We know that according to Factor Theorem, a polynomial \[f\left( x \right)\] has a factor \[\left( {x - m} \right)\] if and only if \[f\left( m \right) = 0\] . In order to solve this question, first of all we will equate \[g\left( x \right)\] to zero and then we will substitute that value of \[x\] in \[p\left( x \right)\] and solve the polynomial. Then if the result comes out to be zero, we will get our required result.

Complete step by step answer:
We have given:
\[p\left( x \right) = 2{x^4} + {x^3} - 8{x^2} - x + 6{\text{ }} - - - \left( i \right)\]
And \[g\left( x \right) = 2x - 3{\text{ }} - - - \left( {ii} \right)\]
Now first of all we will put the equation \[\left( {ii} \right)\] equal to zero
i.e., \[g\left( x \right) = 0\]
\[ \Rightarrow 2x - 3 = 0\]
On adding \[3\] both sides, we get
\[ \Rightarrow 2x = 3\]
On dividing by \[2\] both sides, we get
\[ \Rightarrow x = \dfrac{3}{2}\]
Thus, we obtain the value of \[x\] as \[\dfrac{3}{2}\]
Now we know that for \[g\left( x \right)\] to be a factor of \[p\left( x \right)\] , the value of \[x\] should satisfy the equation below:
\[p\left( {\dfrac{3}{2}} \right) = 0\]
So, substituting \[x = \dfrac{3}{2}\] in equation \[\left( i \right)\] we get
\[ \Rightarrow p\left( {\dfrac{3}{2}} \right) = 2{\left( {\dfrac{3}{2}} \right)^4} + {\left( {\dfrac{3}{2}} \right)^3} - 8{\left( {\dfrac{3}{2}} \right)^2} - \left( {\dfrac{3}{2}} \right) + 6\]
On simplifying, we get
\[ = 2\left( {\dfrac{{81}}{{16}}} \right) + \left( {\dfrac{{27}}{8}} \right) - 8\left( {\dfrac{9}{4}} \right) - \left( {\dfrac{3}{2}} \right) + 6\]
\[ = \left( {\dfrac{{81}}{8}} \right) + \left( {\dfrac{{27}}{8}} \right) - 2\left( {\dfrac{9}{1}} \right) - \left( {\dfrac{3}{2}} \right) + 6\]
On multiplying, we get
\[ = \left( {\dfrac{{81}}{8}} \right) + \left( {\dfrac{{27}}{8}} \right) - \left( {\dfrac{{18}}{1}} \right) - \left( {\dfrac{3}{2}} \right) + 6\]
On taking L.C.M we get
\[ = \dfrac{{81 + 27 - 144 - 12 + 48}}{8}\]
On simplifying, we get
\[ = \dfrac{0}{8}\]
\[ = 0\]
So, we get the value of \[p\left( {\dfrac{3}{2}} \right)\] equals to zero.
Hence, \[g\left( x \right)\] is a factor of \[p\left( x \right)\]
Hence proved.

Note: Factor Theorem is a special case of a Polynomial Remainder Theorem. To apply the factor theorem, we must always have a note that for a polynomial \[f\left( x \right)\] the degree of the polynomial should be greater than or equal to one. The degree of the polynomial is nothing but the highest power of the term in the expression. Also remember if the resultant value of \[p\left( x \right)\] is zero, then only \[g\left( x \right)\] is a factor of \[p\left( x \right)\] otherwise it is not. The value which solves the equation is known as a polynomial value.