a: \(\left(x^2-2x\right)\left(x^2-2x-1\right)-6\)
\(=\left(x^2-2x\right)^2-\left(x^2-2x\right)-6\)
\(=\left(x^2-2x-3\right)\left(x^2-2x+2\right)\)
\(=\left(x-3\right)\left(x+1\right)\left(x^2-2x+2\right)\)
b: \(\dfrac{f\left(x\right)}{x-2}=\dfrac{4x^3+ax+b}{x-2}\)
\(=\dfrac{4x^3-8x^2+8x^2-16x+\left(a+16\right)x-2a-32+2a+32+b}{x-2}\)
\(=4x^2+8x+a+16+\dfrac{2a+b+32}{x-2}\)
Vì f(x) chia hết cho x-2 nên 2a+b+32=0
=>2a+b=-32(1)
\(\dfrac{f\left(x\right)}{x+1}=\dfrac{4x^3+ax+b}{x+1}\)
\(=\dfrac{4x^3+4x^2-4x^2-4x+\left(a+4\right)x+a+4-a-4+b}{x+1}\)
\(=4x^2-4x+a+4+\dfrac{-a+b-4}{x+1}\)
Vì f(x) chia hết cho x+1 nên -a+b-4=0
=>-a+b=4(2)
Từ (1),(2) ta có hệ phương trình:
\(\left\{{}\begin{matrix}2a+b=-32\\-a+b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+b+a-b=-32-4\\a-b=-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3a=-36\\b=a+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-9\\b=-9+4=-5\end{matrix}\right.\)
