Ta có: \(P\left(1\right)=1^2+a\cdot1+b=3=a+b+1=3\Leftrightarrow a+b=2\)(1)
\(P\left(-2\right)=\left(-2\right)^2+\left(-2\right)a+b=4-2a+b=4\Rightarrow b-2a=0\)
\(\Rightarrow\left(a+b\right)-\left(b-2a\right)=2-0=2\)
\(\Rightarrow a+b-b+2a=2\)
\(\Rightarrow3a=2\Rightarrow a=\dfrac{2}{3}\)
Thay \(a=\dfrac{2}{3}\) vào (1) ta được:
\(\dfrac{2}{3}+b=2\Rightarrow b=2-\dfrac{2}{3}=\dfrac{6}{3}-\dfrac{2}{3}=\dfrac{4}{3}\)
Vậy \(a=\dfrac{2}{3};b=\dfrac{4}{3}\)
p(1) = 1 => 1+ a + b = 1 => a + b = 0
p(-1) = 3 => 1 - a + b = 3 => b - a = 4
=> a = (0 - 4 ) : 2 = -2
b = (0 + 4) : 2 = 2
