+16 I got 9x^2-6x+16 +10.333/3x-1 but I'm not sure about the 10.333 part.
+118609 (9x^3-9x^2+18x+5)/(3x-1)
It beats me why any teacher would bother teaching synthetic division.
Normal algebraic division is more logical and more versatile.
First the divisor has to have a coefficient of 1
\(\frac{9x^3-9x^2+18x+5}{3x-1}\\ =\frac{3(3x^3-3x^2+6x+\frac{5}{3}}{3(x-\frac{1}{3})}\\ =\frac{3x^3-3x^2+6x+\frac{5}{3}}{x-\frac{1}{3}}\)
1/3 | 3 -3 +6 5/3
|________1____ -2/3_____16/9____
3 -2 16/3 31/9
So the answer is
\(3x^2-2x+\frac{16}{3}+ \frac{31}{9(x-\frac{1}{3})}\\ =3x^2-2x+\frac{16}{3}+ \frac{31}{3(3x-1)}\\\)
Here is a video people can watch which will explain it.
