Question

Answer 1

\(P\left(-1\right)=a+b+c\)

\(P\left(3\right)=9a-3b+c\\ =\left(5a-b+c\right)+\left(4a-2b\right)\\ =4a-2b\)

\(5a-b+c=0\\ \Rightarrow b=5a+c\)

\(P\left(-1\right)=a+b+c\\ =a+5a+c+c\\ =6a+2c\)

\(P\left(3\right)=4a-2b=4a-2\left(5a+c\right)\\ =4a-10a-2c\\ =-6a-2c\)

\(\Rightarrow P\left(-1\right)+P\left(3\right)=0\\ \Rightarrow P\left(-1\right)=-P\left(3\right)\)

\(P\left(-1\right).P\left(-3\right)=-P\left(-3\right).P\left(-3\right)=-\left(P\left(-3\right)\right)^2\le0\)