Bài `3`:
`a) C(x) = x^3 - x^2 - 5x + 10`
`D(x) = -2x^3 - x^2 + 7x + 10`
`=> E(x) = C(x) + D(x) = (x^3 - x^2 - 5x + 10) + (-2x^3 - x^2 + 7x + 10)`
`= (x^3 - 2x^3) - (x^2 + x^2) + (-5x + 7x) + (10 + 10)`
`= -x^3 - 2x^2 + 2x + 20`
Vậy `E(x) = -x^3 - 2x^2 + 2x + 20`
`b) F(x) = C(x) - D(x) = (x^3 - x^2 - 5x + 10) - (-2x^3 - x^2 + 7x + 10)`
`= x^3 - x^2 - 5x + 10 + 2x^3 + x^2 - 7x - 10`
`= (x^3 + 2x^3) + (-x^2 + x^2) - (5x + 7x) + (10 - 10)`
`= 3x^3 - 12x`
`= 3x(x^2 - 4)
Đặt `F(x) = 0 => 3x(x^2 - 4) = 0`
`=>` \(\left[ \begin{array}{l}3x=0\\x^2 - 4 = 0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=0\\x^2 = 4\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=0\\x=2\\x=-2\end{array} \right.\)
Vậy `x = 0, x = 2, x = - 2` là nghiệm của đa thức `F(x)`
