diketahui suku banyak f(x)=4x³-x²+2x+5dan 9(x)=2x²+x-4 tent nilai f(2)dan 9(-3)

f(x) = 4x³ - x² + 2x + 5
f(2) = 4(2³) - 2² + 2(2) + 5
f(2) = 4(8) - 4 + 4 + 5
f(2) = 32 + 5
f(2) = 37

g(x) = 2x² + x - 4
g(-3) = 2(-3)² + (-3) - 4
g(-3) = 2(9) - 3 - 4
g(-3) = 18 - 3 - 4
g(-3) = 11

f(x)=4x³-x²+2x+5
f(2) = 4(2)³-(2)²+2(2)+5
= 32-4+4+5
= 37


g(x)=2x²+x-4
g(-3) = 2(-3)²+(-3)-4
= 11