Question

[Trương Nguyên Đại Thắng]

Cho đa thức bậc 4 P(x) thỏa mãn : P(-1) = 0 và P(x) - P(x-1) = x(x+1)(2x+1). Xác định P(x)

Answer 1

[Nguyễn Việt Lâm]

\(P\left(x\right)-P\left(x-1\right)=x\left(x+1\right)\left(2x+1\right)\)

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

\(P\left(-1\right)-P\left(-1-1\right)=\left(-1\right).\left(-1+1\right)\left(-2+1\right)=0\)

\(\Rightarrow P\left(-1\right)=P\left(-2\right)=0\)

\(\Rightarrow P\left(x\right)\) được viết dưới dạng: \(P\left(x\right)=kx\left(x+1\right)\left(x+2\right)\left(x-a\right)\) (với \(k\ne0\))

\(P\left(x\right)-P\left(x-1\right)=kx\left(x+1\right)\left(x+2\right)\left(x-a\right)-k.\left(x-1\right)x\left(x+1\right)\left(x-a-1\right)\)

\(=k.x\left(x+1\right)\left[\left(x+2\right)\left(x-a\right)-\left(x-1\right)\left(x-a-1\right)\right]\)

\(=x\left(x+1\right)\left(4kx-k\left(3a+1\right)\right)\)

Mà \(P\left(x\right)-P\left(x-1\right)=x\left(x+1\right)\left(2x+1\right)\)

\(\Rightarrow\left\{{}\begin{matrix}4k=2\\-k\left(3a+1\right)=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}k=\dfrac{1}{2}\\a=-1\end{matrix}\right.\)

Vậy \(P\left(x\right)=\dfrac{1}{2}x\left(x+1\right)^2\left(x+2\right)\)