Thu gọn các đa thức trên.
\(f\left(x\right)=x^3-2x^2+2x-5\)
\(g\left(x\right)=-x^3+3x^2-2x+4\)
\(h\left(x\right)=f\left(x\right)+g\left(x\right)\)
\(=\left(x^3-2x^2+2x-5\right)+\left(-x^3+3x^2-2x+4\right)\)
\(=x^2-1\)
Ta có:
\(2g\left(x\right)=2\left(-x^3+3x^2-2x+4\right)\)
\(=-2x^3+6x^2-4x+8\)
\(\Rightarrow f\left(x\right)-2g\left(x\right)=\left(x^3-2x^2+2x-5\right)-\left(-2x^3+6x^2-4x+8\right)\)
\(=\left(x^3+2x^3\right)-\left(2x^2+6x^2\right)+\left(2x+4x\right)-\left(5+8\right)\)
\(=3x^3-8x^2+6x-13\)