Phân tích các đa thức sau thành nhân tử(thêm bớt cùng một hạng tử) a, x^4+4 b, x^4+64 c, x^8+x^7+1 d, x^8+x^4+1 e, x^5+x+1 f, x^3+x^2+4 g, x^4+2x^2-24 h, x^3-2x-4 i, a^4+4b^4

$$\eqalign{ & a)\,\,{x^4} + 4 \cr & = {x^4} + 4{x^2} + 4 - 4{x^2} \cr & = {\left( {{x^2} + 2} \right)^2} - 4{x^2} \cr & = \left( {{x^2} + 2 - 2x} \right)\left( {{x^2} + 2 + 2x} \right) \cr & b)\,\,{x^4} + 64 \cr & = {x^4} + 16{x^2} + 64 - 16{x^2} \cr & = {\left( {{x^2} + 8} \right)^2} - {\left( {4x} \right)^2} \cr & = \left( {{x^2} + 8 + 4x} \right)\left( {{x^2} + 8 - 4x} \right) \cr & c)\,\,{x^8} + {x^7} + 1 \cr & = {x^8} + {x^7} + {x^6} - {x^6} + 1 \cr & = {x^6}\left( {{x^2} + x + 1} \right) - \left( {{x^3} - 1} \right)\left( {{x^3} + 1} \right) \cr & = {x^6}\left( {{x^2} + x + 1} \right) - \left( {x - 1} \right)\left( {{x^3} + 1} \right)\left( {{x^2} + x + 1} \right) \cr & = \left( {{x^2} + x + 1} \right)\left( {{x^6} - \left( {{x^4} + x - {x^3} - 1} \right)} \right) \cr & = \left( {{x^2} + x + 1} \right)\left( {{x^6} - {x^4} + {x^3} - x + 1} \right) \cr & d)\,\,{x^8} + {x^4} + 1 \cr & = {x^8} + 2{x^4} + 1 - {x^4} \cr & = {\left( {{x^4} + 1} \right)^2} - {\left( {{x^2}} \right)^2} \cr & = \left( {{x^4} + 1 - {x^2}} \right)\left( {{x^4} + 1 + {x^2}} \right) \cr & e)\,\,{x^5} + x + 1 \cr & = {x^5} - {x^2} + {x^2} + x + 1 \cr & = {x^2}\left( {{x^3} - 1} \right) + \left( {{x^2} + x + 1} \right) \cr & = {x^2}\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) + \left( {{x^2} + x + 1} \right) \cr & = \left( {{x^2} + x + 1} \right)\left( {{x^3} - {x^2} + 1} \right) \cr & f)\,{x^3} + {x^2} + 4 \cr & = {x^3} + 2{x^2} - {x^2} + 4 \cr & = {x^2}\left( {x + 2} \right) - \left( {x - 2} \right)\left( {x + 2} \right) \cr & = \left( {x + 2} \right)\left( {{x^2} - x + 2} \right) \cr & g)\,\,{x^4} + 2{x^2} - 24 \cr & = {x^4} + 6{x^2} - 4{x^2} - 24 \cr & = {x^2}\left( {{x^2} + 6} \right) - 4\left( {{x^2} + 6} \right) \cr & = \left( {{x^2} + 6} \right)\left( {{x^2} - 4} \right) \cr & = \left( {{x^2} + 6} \right)\left( {x - 2} \right)\left( {x + 2} \right) \cr & h)\,\,{x^3} - 2x - 4 \cr & = {x^3} - 8 - 2x + 4 \cr & = \left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) - 2\left( {x - 2} \right) \cr & = \left( {x - 2} \right)\left( {{x^2} + 2x + 2} \right) \cr & i)\,\,{a^4} + 4{b^4} \cr & = {a^4} + 4{a^2}{b^2} + 4{b^4} - 4{a^2}{b^2} \cr & = {\left( {{a^2} + 2{b^2}} \right)^2} - {\left( {2ab} \right)^2} \cr & = \left( {{a^2} + 2{b^2} - 2ab} \right)\left( {{a^2} + 2{b^2} + 2ab} \right) \cr} $$