Given:
Dividend = [tex]3x^3-3x^2-4x+3[/tex]
Divisor = [tex]x+3[/tex]
To find:
The quotient and remainder.
Solution:
On dividing [tex]3x^3-3x^2-4x+3[/tex] by [tex]x+3[/tex], we get
[tex]x+3\ |\overline{3x^3-3x^2-4x+3} |\ 3x^2-12x+32[/tex]
[tex]3x^3+9x^2[/tex]
[tex]\underline{(-)\quad (-)\quad \quad \quad \quad \quad }[/tex]
[tex]-12x^2-4x+3[/tex]
[tex]-12x^2-36x[/tex]
[tex]\underline{(+)\quad (+)\quad \quad \quad }[/tex]
[tex]32x+3[/tex]
[tex]32x+96[/tex]
[tex]\underline{(-)\quad (-)\quad }[/tex]
[tex]\underline{\quad -93\quad }[/tex]
Here,
Quotient = [tex]3x^2-12x+32[/tex]
Remainder = [tex]-93[/tex]
Therefore, the correct option is D.
