Bài 1. Phân tích các đa thức sau thành nhân tử:
a) \({a^3} + {a^2}b – {a^2}c – abc\)
b) \({a^4} + {a^3} – {a^2} – a\)
c) \({b^4} – 4{b^3} – b + 4.\ )
Bài 2. Tìm x, biết: \(2\left( {x + 3} \right) – {x^2} – 3x = 0.\)
Bài 1. a) \({a^3} + {a^2}b – {a^2}c – abc \)
\(= a\left( {{a^2} + ab – ac – bc} \right) \)
\(= a\left[ {a\left( {a + b} \right) – c\left( {a + b} \right)} \right]\)
\( = a\left( {a + b} \right)\left( {a – c} \right).\)
Advertisements (Quảng cáo)
b) \({a^4} + {a^3} – {a^2} – a\)
\(= {a^3}\left( {a + 1} \right) – a\left( {a + 1} \right) \)
\(= \left( {a + 1} \right)\left( {{a^3} – a} \right)\)
\( = a\left( {a + 1} \right)\left( {{a^2} – 1} \right) \)
\(= a{\left( {a + 1} \right)^2}\left( {a – 1} \right).\)
c) \({b^4} – 4{b^3} – b + 4 \)
Advertisements (Quảng cáo)
\(= {b^3}\left( {b – 4} \right) – \left( {b – 4} \right) \)
\(= \left( {b – 4} \right)\left( {{b^3} – 1} \right)\)
\( = \left( {b – 4} \right)\left( {b – 1} \right)\left( {{b^2} + b + 1} \right).\)
Bài 2. Ta có:
\(2\left( {x + 3} \right) – {x^2} – 3x \)
\(= 2\left( {x + 3} \right) – x\left( {x + 3} \right) \)
\(= \left( {x + 3} \right)\left( {2 – x} \right)\)
Vậy \(\left( {x + 3} \right)\left( {2 – x} \right) = 0\)
\(\Rightarrow x + 3 = 0\) hoặc \(2 – x = 0\)
\(\Rightarrow x = – 3\) hoặc \(x = 2.\)
