Giải thích các bước giải:
$\begin{array}{l} 90)\\ a){x^4} - 2{x^3} + 4{x^2} - 8x\\ = {x^4} + 4{x^2} - 2{x^3} - 8x\\ = {x^2}\left( {{x^2} + 4} \right) - 2x\left( {{x^2} + 4} \right)\\ = \left( {{x^2} + 4} \right)\left( {{x^2} - 2x} \right)\\ \Rightarrow \left( {{x^4} - 2{x^3} + 4{x^2} - 8x} \right):\left( {{x^2} + 4} \right)\\ = \left( {{x^2} + 4} \right)\left( {{x^2} - 2x} \right):\left( {{x^2} + 4} \right)\\ = {x^2} - 2x\\ b){x^4} - 4{x^3} + 16x - 16\\ = {x^4} - 4{x^2} - 4{x^3} + 16x + 4{x^2} - 16\\ = {x^2}\left( {{x^2} - 4} \right) - 4x\left( {{x^2} - 4} \right) + 4\left( {{x^2} - 4} \right)\\ = \left( {{x^2} - 4} \right)\left( {{x^2} - 4x + 4} \right)\\ \Rightarrow \left( {{x^4} - 4{x^3} + 16x - 16} \right):\left( {{x^2} - 4} \right)\\ = {x^2} - 4x + 4\\ c){x^4} - {x^3} - 3{x^2} + x + 2\\ = {x^4} - {x^2} - {x^3} + x - 2{x^2} + 2\\ = {x^2}\left( {{x^2} - 1} \right) - x\left( {{x^2} - 1} \right) - 2\left( {{x^2} - 1} \right)\\ = \left( {{x^2} - 1} \right)\left( {{x^2} - x - 2} \right)\\ \Rightarrow \left( {{x^4} - {x^3} - 3{x^2} + x + 2} \right):\left( {{x^2} - 1} \right)\\ = {x^2} - x - 2\\ d)2{x^4} - 10{x^3} - 5{x^2} + 15x - 3\\ = 2{x^4} - 3{x^2} - 10{x^3} + 15x - 2{x^2} + 3 - 6\\ = {x^2}\left( {2{x^2} - 3} \right) - 5x\left( {2{x^2} - 3} \right) - \left( {2{x^2} - 3} \right) - 6\\ = \left( {2{x^2} - 3} \right)\left( {{x^2} - 5x - 1} \right) - 6\\ \Rightarrow \left( {2{x^4} - 10{x^3} - 5{x^2} + 15x - 3} \right):\left( {2{x^2} - 3} \right)\\ = {x^2} - 5x - 1\,\text{dư}\, - 6\\ 91)A:\left( {{x^2} + x - 2} \right) = 2{x^2} - 1\,\text{dư}\,3x - 2\\ \Rightarrow A = \left( {{x^2} + x - 2} \right)\left( {2{x^2} - 1} \right) + 3x - 2\\ = 2{x^4} - {x^2} + 2{x^3} - x - 4{x^2} + 4 + 3x - 2\\ = 2{x^4} + 2{x^3} - 5{x^2} + 2x + 2 \end{array}$
