Phân tích đa thức thành nhân tử a, (x ²+x) ²+4x ²+4x-12 b, x(x+1)(x+2)(x+3)+1 c, (x+2)(x+3)(x+4)(x+5)-24 Giải giúp mk với ạ

Đáp án:

$\begin{array}{l}
a)A = {\left( {{x^2} + x} \right)^2} + 4{x^2} + 4x – 12\\
 = {\left( {{x^2} + x} \right)^2} + 4\left( {{x^2} + x} \right) – 12\\
Dat:\left( {{x^2} + x} \right) = a\\
 \Rightarrow A = {a^2} + 4a – 12\\
 = {a^2} + 6a – 2a – 12\\
 = \left( {a + 6} \right)\left( {a – 2} \right)\\
 = \left( {{x^2} + x + 6} \right)\left( {{x^2} + x – 2} \right)\\
b)B = x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) + 1\\
 = x\left( {x + 3} \right)\left( {x + 1} \right)\left( {x + 2} \right) + 1\\
 = \left( {{x^2} + 3x} \right)\left( {{x^2} + 3x + 2} \right) + 1\\
Dat:{x^2} + 3x = b\\
 \Rightarrow B = b\left( {b + 2} \right) + 1\\
 = {b^2} + 2b + 1\\
 = {\left( {b + 1} \right)^2}\\
 = {\left( {{x^2} + 3x + 1} \right)^2}\\
c)C = \left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 4} \right)\left( {x + 5} \right) – 24\\
 = \left( {x + 2} \right)\left( {x + 5} \right)\left( {x + 3} \right)\left( {x + 4} \right) – 24\\
 = \left( {{x^2} + 7x + 10} \right)\left( {{x^2} + 7x + 12} \right) – 24\\
Dat:{x^2} + 7x + 10 = a\\
 \Rightarrow C = a\left( {a + 2} \right) – 24\\
 = {a^2} + 2a – 24\\
 = {a^2} + 6a – 4a – 24\\
 = \left( {a + 6} \right)\left( {a – 4} \right)\\
 = \left( {{x^2} + 7x + 10 + 6} \right)\left( {{x^2} + 7x + 10 – 4} \right)\\
 = \left( {{x^2} + 7x + 16} \right)\left( {{x^2} + 7x + 6} \right)
\end{array}$