`a) 3x^4-8x^3-10x^2+ax-b`
`=3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5+8x+ax+5-b`
`=x^2(3x^2-2x+1)-2x(3x^2-2x+1)-5(3x^2-2x+1)+(a+8)x+5-b`
Để `(3x^4-8x^3-10x^2+ax-b)`chia hết
`(3x^2-2x+1)` thì:
`(a+8)x+5-b=0` với mọi $x$
`<=>`$\begin{cases}a+8=0\\5-b=0\end{cases}$
`<=>`$\begin{cases}a=-8\\b=5\end{cases}$
$\\$
`b) x^4+2x^3-3x^2+ax+b`
`=x^4-x^3-2x^2+3x^3-3x^2-6x+2x^2-2x-4+8x+4+ax+b`
`=x^2(x^2-x-2)+3x(x^2-x-2)+2(x^2-x-2)+(a+8)x+b+4`
`=(x^2-x-2)(x^2+3x+2)+(a+8)x+b+4`
Để `(x^4+2x^3-3x^2+ax+b)` chia `(x^2-x-2)` dư `-4x-1`:
`<=>`$\begin{cases}a+8=-4\\b+4=-1\end{cases}$
`<=>`$\begin{cases}a=-12\\b=-5\end{cases}$
Vậy `a=-12;b=-5`
