Question

Bài 8: Cho đa thức: P(x)= \(2x^4+3x^2+4\)

a) Tính P(0), P(1); P(-1),P(2); P(-2); P(\(\dfrac{-2}{3}\))

Answer 1

a: \(P\left(0\right)=2\cdot0^4+3\cdot0^2+4=4\)

\(P\left(1\right)=2\cdot1^4+3\cdot1^2+4=2+3+4=9\)

\(P\left(-1\right)=2\cdot\left(-1\right)^4+3\cdot\left(-1\right)^2+4=9\)

\(P\left(2\right)=2\cdot2^4+3\cdot2^2+4=32+18+4=54\)

\(P\left(-2\right)=2\cdot\left(-2\right)^4+3\cdot\left(-2\right)^2+4=54\)

Answer 2

`P(0)=2.0^4+3.0^2+4=0+0+4=4`

`P(1)=2.1^4+3.1^2+4=2+3+4=9`

`P(-1)=2.(-1)^4+3.(-1)^2+4=2+3+4=9`

`P(2)=2.2^4+3.2^2+4=32+12+4=48`

`P(-2)=2.(-2)^4+3.(-2)^2+4=32+12+4=48`

`P([-2]/3)=2.([-2]/3)^4+3.([-2]/3)^2+4=32/81+4/3+4=464/81`