Question

[Leo Messai]

tìm đa thức bậc 3 P(x) biết P(x) chia cho (x-1),(x+2),(x-4) đều dư 6 và P(-1)=16

Answer 1

[Nguyễn Hoàng Minh]

Gọi \(P\left(x\right)=ax^3+bx^2+cx+d\)

Ta có \(P\left(x\right):\left(x-1\right)R6\Leftrightarrow P\left(1\right)=a+b+c+d=6\left(1\right)\)

\(P\left(x\right):\left(x+2\right)R6\Leftrightarrow P\left(-2\right)=-8a+4b-2c+d=-2\left(2\right)\)

\(P\left(x\right):\left(x-4\right)R6\Leftrightarrow P\left(4\right)=64a+16b+4c+d=6\left(3\right)\)

\(P\left(-1\right)=16\Leftrightarrow-a+b-c+d=16\left(4\right)\)

Từ \(\left(1\right)\left(2\right)\left(3\right)\left(4\right)\Leftrightarrow\left\{{}\begin{matrix}a+b+c+d=6\\-8a+4b-2c+d=6\\64a+16b+4c+d=6\\-a+b-c+d=16\end{matrix}\right.\)

\(\Leftrightarrow a=1;b=-3;c=-6;d=14\)

Vậy \(P\left(x\right)=x^3-3x^2-6x+14\)