Question

Tìm a, b sao cho đa thức

a) f(x)=x⁴-x³-3x²+a x+b chia cho đa thức x²-x-2 dư 2x-3

b) g(x)=x⁴+a x +b chia cho đa thức x²-4

Answer 1

a) Ta có: \(x^2-x-2=0\)

\(\Leftrightarrow x^2+x-2x-2=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

Ap dung Be du ta co:

\(\left\{{}\begin{matrix}2^4-2^3-3.2^2+2a+b=2.2-3\\\left(-1\right)^4-\left(-1\right)^3-3.\left(-1\right)^2-a+b=2.\left(-1\right)-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2a+b=5\\-a+b=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=-1\end{matrix}\right.\)

Câu b tương tự rồi nhé