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
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} $$