According to remainder theorem, when a polynomial f (x) is divided by x – a, then the remainder is f(a).
Here, f(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{5x}\text{ }\text{ }\mathbf{6}\]
(i) \[f\text{ }\left( -1 \right)\text{ }=\text{ }2{{\left( -1 \right)}^{3}}~+\text{ }3{{\left( -1 \right)}^{2}}~\text{ }5\left( -1 \right)\text{ }\text{ }6\text{ }=\text{ }-2\text{ }+\text{ }3\text{ }+\text{ }5\text{ }\text{ }6\text{ }=\text{ }0\]
⇒ Remainder is zero for \[x\text{ }=\text{ }-1\]
Therefore, \[\left( x\text{ }+\text{ }1 \right)\] is a factor of the polynomial f(x).
(ii) \[f\left( 1/2 \right)\text{ }=\text{ }2{{\left( 1/2 \right)}^{3}}~+\text{ }3{{\left( 1/2 \right)}^{2}}~\text{ }5\left( 1/2 \right)\text{ }\text{ }6\]
= \[\text{1/4 }+\text{ 3/4 }\text{ }5/2\text{ }\text{ }6\]
= \[-5/2\text{ }\text{ }5\text{ }=\text{ }-15/2\]
⇒ Remainder is not equals to zero for \[x\text{ }=\text{ }1/2\]
Therefore, \[\left( 2x\text{ }\text{ }1 \right)\]is not a factor of the polynomial f(x).
(iii) \[f\text{ }\left( -2 \right)\text{ }=\text{ }2{{\left( -2 \right)}^{3}}~+\text{ }3{{\left( -2 \right)}^{2}}~\text{ }5\left( -2 \right)\text{ }\text{ }6\text{ }=\text{ }-16\text{ }+\text{ }12\text{ }+\text{ }10\text{ }\text{ }6\text{ }=\text{ }0\]
⇒ Remainder is zero for \[x\text{ }=\text{ }-2\]
Therefore, \[\left( x\text{ }+\text{ }2 \right)\] is a factor of the polynomial f(x).