Question

[Minh Thư]

Cho A= x^3 -2x+n B=x-2 Tìm n để A chia hết cho B

(x-3)^2-2x+6=0

x^2-5x+6=0

Giúp mình nhanh vs ạ

Answer 1

[Minh Hiếu]

\(A=x^3-2x+n\)

\(B=n-2\)

\(A\text{⋮}B\) ⇒ \(\left(x^3-2x+n\right)\text{⋮}\left(n-2\right)\)

⇒ \(\left[\left(x^3-2x^2\right)+\left(2x^2-4x\right)+\left(2x-4\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)

⇒ \(\left[x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)

⇒ \(\left[\left(x-2\right)\left(x^2+2x+2\right)+\left(n+4\right)\right]\text{⋮}\left(x-2\right)\)

Vì \(\left(x-2\right)\left(x^2+2x+2\right)\text{⋮}\left(n-2\right)\)

Để \(A\text{⋮}B\)

⇒ \(n+4=0\)

⇒ \(n=-4\)