+0  
 
0
46
1
avatar+16 

I got 9x^2-6x+16 +10.333/3x-1 but I'm not sure about the 10.333 part.

 Nov 5, 2020
 #1
avatar+111600 
+1

(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.

https://www.khanacademy.org/math/algebra-home/alg-polynomials/alg-synthetic-division-of-polynomials/v/synthetic-division

 Nov 5, 2020

36 Online Users

avatar
avatar