Question

Bài 1: phân tích đa thức sau thành nhân tử( thêm bớt cùng 1 hạng tử)

x^3+x^2+4

Answer 1

\(x^3+x^2+4\)

\(=x^3+2x^2-x^2+2x-2x+4\)

\(=\left(x^3+2x^2\right)-\left(x^2+2x\right)+\left(2x+4\right)\)

\(=x^2\left(x+2\right)-x\left(x+2\right)+2\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-x+2\right)\)

Answer 2

\(x^3+x^2+4\)

\(=x^3+x^2+4+4x-4x\)

\(=x^3-4x+x^2+4+4x\)

\(=x\left(x^2-4\right)+\left(x+2\right)^2\)

\(=x\left(x-2\right)\left(x+2\right)+\left(x+2\right)^2\)

\(=\left(x+2\right)\left[x\left(x-2\right)+\left(x+2\right)\right]\)

\(=\left(x+2\right)\left[x^2-2x+x+2\right]\)

\(=\left(x+2\right)\left[x^2-x+2\right]\)

\(=\left(x+2\right)\left[\left(x^2-x+0,25\right)+0,75\right]\)

\(=\left(x+2\right)\left[\left(x-0,5\right)^2+0,75\right]\)

Answer 3

\(x^3+x^2+4\\ =x^3+2x^2-x^2-2x+2x+4\\ =x^2\left(x+2\right)-x\left(x+2\right)+2\left(x+2\right)\\ =\left(x+2\right)\left(x^2-x+2\right)\)