Appendix C - Solving Polynomial Equations - Exercise Set C - Page A33: 14

\[p\left( x \right) = \left( {x - 1} \right)\left( {2x + 3} \right)\left( {{x^2} + 3} \right)\]

$$\eqalign{ & p\left( x \right) = 2{x^4} + {x^3} + 3{x^2} + 3x - 9 \cr & {\text{The constant term is 9 }}\left( {{\text{which has divisors }} \pm 1, \pm 3, \pm 9} \right) \cr & {\text{Evaluating the divisors}} \cr & p\left( { - 1} \right) = 2{\left( { - 1} \right)^4} + {\left( { - 1} \right)^3} + 3{\left( { - 1} \right)^2} + 3\left( { - 1} \right) - 9 \cr & p\left( { - 1} \right) = - 8 \cr & p\left( 1 \right) = 2{\left( 1 \right)^4} + {\left( 1 \right)^3} + 3{\left( 1 \right)^2} + 3\left( 1 \right) - 9 \cr & p\left( 1 \right) = 0 \cr & \cr & p\left( x \right) = 0{\text{ for }}x = 1,{\text{ then }}\left( {x - 1} \right){\text{ is a factor of }}p\left( x \right) \cr & p\left( x \right) = 2{x^4} + {x^3} + 3{x^2} + 3x - 9 = \left( {x - 1} \right)q\left( x \right) \cr & {\text{By the long division}}\,\,{\text{ }}x - 1\left){\vphantom{1{2{x^4} + {x^3} + 3{x^2} + 3x - 9}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{2{x^4} + {x^3} + 3{x^2} + 3x - 9}}}{\text{ we obtain}} \cr & 2{x^4} + {x^3} + 3{x^2} + 3x - 9 = \left( {x - 1} \right)\left( {2{x^3} + 3{x^2} + 6x + 9} \right) \cr & p\left( x \right) = \left( {x - 1} \right)\left( {2{x^3} + 3{x^2} + 6x + 9} \right) \cr & {\text{with }}q\left( x \right) = 2{x^3} + 3{x^2} + 6x + 9 \cr & {\text{Factoring }}q\left( x \right){\text{ by grouping terms}} \cr & 2{x^3} + 3{x^2} + 6x + 9 = {x^2}\left( {2x + 3} \right) + 3\left( {2x + 3} \right) \cr & 2{x^3} + 3{x^2} + 6x + 9 = \left( {2x + 3} \right)\left( {{x^2} + 3} \right) \cr & \cr & {\text{Thus}}{\text{,}} \cr & p\left( x \right) = \left( {x - 1} \right)\left( {2x + 3} \right)\left( {{x^2} + 3} \right) \cr} $$