Question

Thực hiện phép chia bằng cách phân tích đa thức thành nhân tử

a) \(\left(x^5+x^3+x^2+1\right):\left(x^3+1\right)\)

b) \(\left(x^2+5x+6\right):\left(x+3\right)\)

c) \(\left(x^3+x^2-12\right):\left(x-2\right)\)

Answer 1

\(a,\left(x^5+x^3+x^2+1\right):\left(x^3+1\right)\)

\(=\left[x^3.\left(x^2+1\right)+\left(x^2+1\right)\right]:\left(x^3+1\right)\)

\(=\left(x^2+1\right)\left(x^3+1\right):\left(x^3+1\right)\)

\(=x^2+1\)

\(b,\left(x^2+5x+6\right):\left(x+3\right)\)

\(=\left(x^2+2x+3x+6\right):\left(x+3\right)\)

\(=\left[x\left(x+2\right)+3\left(x+2\right)\right]:\left(x+3\right)\)

\(=\left(x+2\right)\left(x+3\right):\left(x+3\right)\)

\(=x+2\)

Answer 2

c. \(\left(x^3+x^2-12\right):\left(x-2\right)=\left(x^3-2x^2+3x^2-6x+6x-12\right):\left(x-2\right)=\left[x^2\left(x-2\right)+3x\left(x-2\right)+6\left(x-2\right)\right]:\left(x-2\right)=\left(x-2\right)\left(x^2+3x+6\right):\left(x-2\right)=x^2+3x+6\)