Question

if you divide 9x^3+3x^2-6x-3 to 3x-1 using synthetic division, how it is going to solve?

Answer 1

\[\frac{9x^3+3x^2-6x-3}{3x-1}\]
You can't use synthetic division right away, because the divisor (the bottom part) has to be in the format of \(x-k\), without a number out in front of it. First factor out a 3 from the denomenator: \[\frac{9x^3+3x^2-6x-3}{3(x-1)} = \frac{1}{3}\times\frac{9x^3+3x^2-6x-3}{x-\dfrac{1}{3}}\]
Now try synthetic division.

Answer 2

9 6 \[\frac{ -4 }{ 3 }\] i got this answer and i have no idea what to do next because i encouter a fraction. HOw is it going to be?

Answer 3

Well you can first cancel the \(\large^1/_3\) and the numerator to \(\large3x^3+x^2-2x-1\).
Then divide:\[

\large{\left.\underline{^1/_3}\right\vert~~~~~~3~~~~1~~ -2 ~ ~~-1 \\
~~~~~~~~~~~~~~~~~~~1~~^{-2}/_3~~~^{-8}/_9 \\
~~~~~~~~~~~~~\overline{3~~~~2~~^{-8}/_3~~\left\vert^{-11}/_3\right.}}\]
So\[3x^2 + 2x - \frac{8}{3} - \frac{\dfrac{11}{3} }{x - \dfrac{1}{3} }\]

Answer 4

so
\[3x ^{2}+2x-\frac{4}{3}-\frac{\frac{13}{9}}{x-\frac{1}{3}}\]

Answer 5

Oh yeah, true 2/3 = \(\dfrac{2}{3}\), not \(-\dfrac{2}{3}\)