deo biet

a: \(x^4-9x^3+21x^2+ax+b\) ⋮\(x^2-x-2\)

=>\(x^4-x^3-2x^2-8x^3+8x^2+16x+15x^2-15x-30\) +(a-1)x+b+30⋮\(x^2-x-2\)

=>a-1=0 và b+30=0

=>a=1 và b=-30

b: \(x^4-7x^3+10x^2+\left(a-1\right)x+b-a\) ⋮\(x^2-6x+5\)

=>\(x^4-6x^3+5x^2-x^3+6x^2-5x-x^2+6x-5\) +(a-2)x+b-a+5⋮\(x^2-6x+5\)

=>a-2=0 và b-a+5=0

=>a=2 và b=a-5=2-5=-3

a: \(\Leftrightarrow2x^2+8x+\left(a-8\right)x+4\left(a-8\right)-4a+28⋮x+4\)

hay a=7

\(a,4x^3+ax+b⋮x-2\\ \Leftrightarrow4x^3+ax+b=\left(x-2\right)\cdot a\left(x\right)\)

Thay \(x=2\Leftrightarrow32+2a+b=0\Leftrightarrow2a+b=-32\left(1\right)\)

\(4x^3+ax+b⋮x+1\\ \Leftrightarrow4x^3+ax+b=\left(x+1\right)\cdot b\left(x\right)\)

Thay \(x=-1\Leftrightarrow-4-a+b=0\Leftrightarrow a-b=-4\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\) ta có hệ \(\left\{{}\begin{matrix}2a+b=-32\\a-b=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a=-36\\b=a+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-12\\b=-8\end{matrix}\right.\)

Hay  a − 1 = 0 b + 30 = 0 ⇒ a = 1 b = − 30 .

thank bạn