a) \(x^3-1+5x^2-5+3x-3\)
= \(x^3+5x^2+3x-9\)
= \(x^3-x^2+6x^2-6x+9x-9\)
= \(x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2+6x+9\right)\)
= \(\left(x-1\right)\left(x-3\right)^2\)
b) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
= \(\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+1\)
= \(\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\) (1)
Đặt \(x^2+5x+4=a\)
Đa thức (1) \(\Leftrightarrow a\left(a+2\right)+1\)
= \(a^2+2a+1=\left(a+1\right)^2=\left(x^2+5x+4+1\right)^2\)
= \(\left(x^2+5x+6\right)^2\)
c) \(x^8+x^4+1\)
Ta thấy \(\left\{{}\begin{matrix}x^8\ge0\\x^4\ge0\\1>0\end{matrix}\right.\) \(\Rightarrow x^8+x^4+1\ge1\)
\(\Rightarrow\) Không phân tích thành nhân tử đc.
d) \(x^3+x^2+4\)
= \(x^3+2x^2-x^2+4\)
= \(x^2\left(x-2\right)-\left(x^2-4\right)\)
= \(x^2\left(x-2\right)-\left(x-2\right)\left(x+2\right)\)
= \(\left(x-2\right)\left(x^2-x-2\right)\)
