Phân tích các đa thức sau thành nhân tử bằng cách phối hợp nhiều phương pháp :
d) \(k\left(x\right)=27x^4-9x^3+14x^2-4\)
e) \(l\left(x\right)=\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
f) \(m\left(x\right)=x^6+27\)
g) \(n\left(x\right)=x^4+3x^2+4\)
h) \(p\left(x\right)=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
h)Ta có : \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt\(x^2+7x+11=y\)
\(=>p\left(x\right)=\left(y-1\right)\left(y+1\right)-24=y^2-1-24=y^2-25=\left(y-5\right)\left(y+5\right)\)
Thay \(y=x^2+7x+11\) vào ta có : \(p\left(x\right)=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(f)m\left(x\right)=x^6+27=\left(x^2+3\right)\left(x^4-3x^2+9\right)\)
e)\(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=\left(x^2+x\right)^2-2\left(x^2+x\right)+6\left(x^2+x\right)-12=\left(x^2+x\right)\left(x^2+x-2\right)+6\left(x^2+x-12\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)=\left(x^2+x+6\right)\left(x^2-x+2x-2\right)=\left(x^2+x+6\right)\left[x\left(x-1\right)+2\left(x-1\right)\right]=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
