a) Thay x=1, ta có:
f(1) = a.1 + b = 1 => a + b =1 (1)
Thay x = -1, ta có:
f(-1) = a.(-1) + b = -5 => -a + b = -5 (2)
(1)(2) => \(\left\{{}\begin{matrix}a=3\\b=-2\end{matrix}\right.\)
b) Thay x = -1, ta có:
g(-1) = \(3\left(-1\right)^3-5\left(-1\right)^2+a\left(-1\right)+b\) = 8
=> \(-a+b=16\) (1)
Thay x = 2, ta có:
g(2) = \(3.2^3-5.2^2+a.2+b=3\)
=> \(2a+b=-1\) (2)
(1)(2) => \(\left\{{}\begin{matrix}a=\frac{-17}{3}\\b=\frac{31}{3}\end{matrix}\right.\)