Phân tích đa thức thành nhân tử: a, -6x+8y , b,1- x ³ c, x ² -25 c, x ² (x-5) -x +5 d, (x+y) ² – 4x ² e, x ³ (x+2) – x – 2 f, x ³ + 3x ² +3x + 1 g, 7.

Đáp án:

k) \(\left( {x – 2} \right)\left( {2x + 1} \right)\)

Giải thích các bước giải:

\(\begin{array}{l}
a)2\left( { – 3x + 4y} \right)\\
b)1 – {x^3} = \left( {1 – x} \right)\left( {1 + x + {x^2}} \right)\\
c){x^2} – 25 = \left( {x – 5} \right)\left( {x + 5} \right)\\
c){x^2}\left( {x – 5} \right) – \left( {x – 5} \right) = \left( {x – 5} \right)\left( {{x^2} – 1} \right)\\
 = \left( {x – 5} \right)\left( {x – 1} \right)\left( {x + 1} \right)\\
d){\left( {x + y} \right)^2} – 4{x^2}\\
 = \left( {x + y – 2x} \right)\left( {x + y + 2x} \right)\\
 = \left( {y – x} \right)\left( {3x + y} \right)\\
e){x^3}\left( {x + 2} \right) – \left( {x + 2} \right)\\
 = \left( {x + 2} \right)\left( {{x^3} – 1} \right)\\
 = \left( {x + 2} \right)\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)\\
f){x^3} + 3{x^2} + 3x + 1 = {\left( {x + 1} \right)^3}\\
g)7\left( {2x – 3} \right) + x\left( {2x – 3} \right)\\
 = \left( {2x – 3} \right)\left( {7 + x} \right)\\
h){x^2} – x + 5x – 5\\
 = x\left( {x – 1} \right) + 5\left( {x – 1} \right)\\
 = \left( {x – 1} \right)\left( {x + 5} \right)\\
i){x^2} – x + 3x – 3\\
 = x\left( {x – 1} \right) + 3\left( {x – 1} \right)\\
 = \left( {x – 1} \right)\left( {x + 3} \right)\\
j)3{x^2} – 7x – 6\\
 = 3{x^2} – 9x + 2x – 6\\
 = 3x\left( {x – 3} \right) + 2\left( {x – 3} \right)\\
 = \left( {x – 3} \right)\left( {3x + 2} \right)\\
k)2{x^2} – 4x + x – 2\\
 = 2x\left( {x – 2} \right) + \left( {x – 2} \right)\\
 = \left( {x – 2} \right)\left( {2x + 1} \right)
\end{array}\)