the complete question in the attached figure
we have that
(3x³ – 2x² + 4x – 3) is divided by (x²
+ 3x + 3)
(3x³ – 2x² + 4x – 3) is equals to------> 3x³+9x²+9x−11x²−33x−33+28x+
30
=3x(x²+3x+3)−11(x²+3x+3)+28x+
30
=(x²+3x+3)(3x−11)+28x+30
Hence
[3x³−2x²+4x−3]/[x²+3x+3]=[(x²+3x+3)(3x−11)+28x+30]/[x²+3x+3]
=[(3x−11)]+{[28x+30]/[x²+3x+3]}
So the remainder of division is
28x+30
the answer is the option D) 28x+30
another way to make it more direct
(3x³ – 2x² + 4x – 3) is divided by║ (x² + 3x + 3)
-------------------------------------------- ║(3x-11)
-3x³-9x²-9x
--------------------------------------------
-11x²-5x-3
----------------------------------------------
+11x²+33x+33
----------------------------------------------
28x+30----------> this is the remainder
