Phân tích đa thức thành nhân tử:
a)x^4+2x^3+x^2-y^4
b)x^2(x-z)+y^2(y-z)+xyz
c)ac^2-a^2c+b^2c-bc^2+ab(a-b)
d)a^3+2020a^2+2020a+2019
Phân tích đa thức thành nhân tử:
a)x^4+2x^3+x^2-y^4
b)x^2(x-z)+y^2(y-z)+xyz
c)ac^2-a^2c+b^2c-bc^2+ab(a-b)
d)a^3+2020a^2+2020a+2019
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^4} + 2{x^3} + {x^2} – {y^4}\\
= {\left( {{x^2}} \right)^2} + 2.{x^2}.x + {x^2} – {y^4}\\
= {\left( {{x^2} + x} \right)^2} – {\left( {{y^2}} \right)^2}\\
= \left( {{x^2} + x + {y^2}} \right)\left( {{x^2} + x – {y^2}} \right)\\
b,\\
{x^2}.\left( {x – z} \right) + {y^2}\left( {y – z} \right) + xyz\\
= {x^3} – {x^2}z + {y^3} – {y^2}z + xyz\\
= \left( {{x^3} + {y^3}} \right) – \left( {{x^2}z + {y^2}z – xyz} \right)\\
= \left( {x + y} \right)\left( {{x^2} – xy + {y^2}} \right) – z.\left( {{x^2} – xy + {y^2}} \right)\\
= \left( {{x^2} – xy + {y^2}} \right).\left( {x + y – z} \right)\\
c,\\
a{c^2} – {a^2}c + {b^2}c – b{c^2} + ab\left( {a – b} \right)\\
= \left( {a{c^2} – b{c^2}} \right) – \left( {{a^2}c – {b^2}c} \right) + ab\left( {a – b} \right)\\
= {c^2}\left( {a – b} \right) – c\left( {{a^2} – {b^2}} \right) + ab\left( {a – b} \right)\\
= {c^2}\left( {a – b} \right) – c\left( {a – b} \right)\left( {a + b} \right) + ab\left( {a – b} \right)\\
= \left( {a – b} \right).\left( {{c^2} – c.\left( {a + b} \right) + ab} \right)\\
= \left( {a – b} \right).\left( {{c^2} – ca – cb + ab} \right)\\
= \left( {a – b} \right).\left[ {c\left( {c – a} \right) – b.\left( {c – a} \right)} \right]\\
= \left( {a – b} \right)\left( {c – a} \right)\left( {c – b} \right)\\
d,\\
{a^3} + 2020{a^2} + 2020a + 2019\\
= \left( {{a^3} + 2019{a^2}} \right) + \left( {{a^2} + 2019a} \right) + \left( {a + 2019} \right)\\
= {a^2}\left( {a + 2019} \right) + a\left( {a + 2019} \right) + \left( {a + 2019} \right)\\
= \left( {a + 2019} \right).\left( {{a^2} + a + 1} \right)
\end{array}\)