Question

tìm a và b

2x²+ax+b chia hết chp (x-1).(x-2) dư 6

Answer 1

\(\left(2x^2+ax+b\right):\left(x-1\right)\left(x-2\right).dư.6\\ \Leftrightarrow2x^2+ax+b=\left(x-1\right)\left(x-2\right)+6\)

Thay \(x=1\)

\(\Leftrightarrow2\cdot1^2+a+b=6\\ \Leftrightarrow a+b=4\left(1\right)\)

Thay \(x=2\)

\(2\cdot2^2+2a+b=6\\ \Leftrightarrow2a+b=-2\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\) ta có hpt: \(\left\{{}\begin{matrix}a+b=4\\2a+b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-6\\2a+b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-6\\b=10\end{matrix}\right.\)