Respuesta :
[tex]\frac{3x^3+9x^2-6x}{-3x}[/tex]
Since the denominator is a monomial, we divide each term of the trinomial in the numerator by the denominator.
[tex]\frac{3x^3}{-3x} = -1x^2[/tex], because 3/-3 = 1 and x³/x = x².
[tex]\frac{9x^2}{-3x}=-3x[/tex]; 9/-3 = -3 and x²/x = x.
[tex]\frac{-6x}{-3x} = 2[/tex]; -6/-3 = 2 and x/x = 1, so we have 2*1 = 2.
Together this gives us [tex]-x^2-3x+2[/tex]
Since the denominator is a monomial, we divide each term of the trinomial in the numerator by the denominator.
[tex]\frac{3x^3}{-3x} = -1x^2[/tex], because 3/-3 = 1 and x³/x = x².
[tex]\frac{9x^2}{-3x}=-3x[/tex]; 9/-3 = -3 and x²/x = x.
[tex]\frac{-6x}{-3x} = 2[/tex]; -6/-3 = 2 and x/x = 1, so we have 2*1 = 2.
Together this gives us [tex]-x^2-3x+2[/tex]