Phân tích đa thức thành nhân tử a. a^3 -7a – 6 b. a^3 + 4a^2 – 7a – 10 c. a( a +b )^2 + b( c + a )^2 + c( a + b )^2 – 4ab d. ( a^2 + a )^2 + 4( a^2 +

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
{a^3} – 7a – 6\\
 = \left( {{a^3} + {a^2}} \right) – \left( {{a^2} + a} \right) – \left( {6a + 6} \right)\\
 = {a^2}\left( {a + 1} \right) – a\left( {a + 1} \right) – 6\left( {a + 1} \right)\\
 = \left( {a + 1} \right)\left( {{a^2} – a – 6} \right)\\
 = \left( {a + 1} \right).\left[ {\left( {{a^2} – 3a} \right) + \left( {2a – 6} \right)} \right]\\
 = \left( {a + 1} \right).\left[ {a\left( {a – 3} \right) + 2\left( {a – 3} \right)} \right]\\
 = \left( {a + 1} \right)\left( {a – 3} \right)\left( {a + 2} \right)\\
b,\\
{a^3} + 4{a^2} – 7a – 10\\
 = \left( {{a^3} + {a^2}} \right) + \left( {3{a^2} + 3a} \right) – \left( {10a + 10} \right)\\
 = {a^2}\left( {a + 1} \right) + 3a\left( {a + 1} \right) – 10\left( {a + 1} \right)\\
 = \left( {a + 1} \right)\left( {{a^2} + 3a – 10} \right)\\
 = \left( {a + 1} \right)\left[ {\left( {{a^2} + 5a} \right) – \left( {2a + 10} \right)} \right]\\
 = \left( {a + 1} \right)\left[ {a\left( {a + 5} \right) – 2\left( {a + 5} \right)} \right]\\
 = \left( {a + 1} \right)\left( {a + 5} \right)\left( {a – 2} \right)\\
d,\\
{\left( {{a^2} + a} \right)^2} + 4.\left( {{a^2} + a} \right) – 12\\
 = \left[ {{{\left( {{a^2} + a} \right)}^2} + 6\left( {{a^2} + a} \right)} \right] – \left[ {2\left( {{a^2} + a} \right) + 12} \right]\\
 = \left( {{a^2} + a} \right).\left[ {\left( {{a^2} + a} \right) + 6} \right] – 2.\left[ {\left( {{a^2} + a} \right) + 6} \right]\\
 = \left( {{a^2} + a + 6} \right)\left[ {\left( {{a^2} + a} \right) – 2} \right]\\
 = \left( {{a^2} + a + 6} \right).\left[ {\left( {{a^2} – a} \right) + \left( {2a – 2} \right)} \right]\\
 = \left( {{a^2} + a + 6} \right).\left[ {a\left( {a – 1} \right) + 2\left( {a – 1} \right)} \right]\\
 = \left( {{a^2} + a + 6} \right).\left( {a – 1} \right)\left( {a + 2} \right)\\
e,\\
\left( {{x^2} + x + 1} \right)\left( {{x^2} + x + 2} \right) – 12\\
 = \left( {{x^2} + x + 1} \right).\left[ {\left( {{x^2} + x + 1} \right) + 1} \right] – 12\\
 = {\left( {{x^2} + x + 1} \right)^2} + \left( {{x^2} + x + 1} \right) – 12\\
 = \left[ {{{\left( {{x^2} + x + 1} \right)}^2} + 4.\left( {{x^2} + x + 1} \right)} \right] – \left[ {3.\left( {{x^2} + x + 1} \right) + 12} \right]\\
 = \left( {{x^2} + x + 1} \right).\left[ {\left( {{x^2} + x + 1} \right) + 4} \right] – 3.\left[ {\left( {{x^2} + x + 1} \right) + 4} \right]\\
 = \left( {{x^2} + x + 5} \right).\left[ {\left( {{x^2} + x + 1} \right) – 3} \right]\\
 = \left( {{x^2} + x + 5} \right).\left( {{x^2} + x – 2} \right)\\
 = \left( {{x^2} + x + 5} \right).\left[ {\left( {{x^2} – x} \right) + \left( {2x – 2} \right)} \right]\\
 = \left( {{x^2} + x + 5} \right).\left[ {x\left( {x – 1} \right) + 2\left( {x – 1} \right)} \right]\\
 = \left( {{x^2} + x + 5} \right).\left( {x – 1} \right)\left( {x + 2} \right)
\end{array}\) 

Em xem lại đề câu c nhé!