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(3x^3-5x^2+10x-3)/(3x-4)

 Aug 17, 2015

Best Answer 

 #2
avatar+97967 
+10

Using polynomial division, we have

 

                           x^2   - (1/3)x  + (26/9)

3x - 4     [ 3x^3  - 5x^2  + 10x    -  3]

                3x^3  - 4x^2

                ------------------

                          -1x^2  + 10x

                          -1x^2  +(4/3)x

                          -------------------

                                        (26/3)x  - 3

                                        (26/3)x  - (104/9)

                                        ---------------------

                                                        (77/9)

 

So.....the answer is..........  [ x^2   - (1/3)x  + (26/9) ] +  [ 77/9]/[3x - 4]

 

 

 

 Aug 18, 2015
 #1
avatar
0

-3/(-4 + 3 x) + (10 x)/(-4 + 3 x) + (-5 x^2)/(-4 + 3 x) + (3 x^3)/(-4 + 3 x)

(-3 + x (10 + x (-5 + 3 x)))/(-4 + 3 x)

(3 - 10 x + 5 x^2 - 3 x^3)/(4 - 3 x)

 Aug 18, 2015
 #2
avatar+97967 
+10
Best Answer

Using polynomial division, we have

 

                           x^2   - (1/3)x  + (26/9)

3x - 4     [ 3x^3  - 5x^2  + 10x    -  3]

                3x^3  - 4x^2

                ------------------

                          -1x^2  + 10x

                          -1x^2  +(4/3)x

                          -------------------

                                        (26/3)x  - 3

                                        (26/3)x  - (104/9)

                                        ---------------------

                                                        (77/9)

 

So.....the answer is..........  [ x^2   - (1/3)x  + (26/9) ] +  [ 77/9]/[3x - 4]

 

 

 

CPhill Aug 18, 2015
 #3
avatar+99109 
0

Very nice presentation Chris :)

 Aug 18, 2015
 #4
avatar+21816 
+5

(3x^3-5x^2+10x-3)/(3x-4)

 

In mathematics, Horner's method (also known as Horner scheme in the UK or Horner's rule in the U.S.) an algorithm for calculating polynomials

see:  https://en.wikipedia.org/wiki/Horner%27s_method#cite_note-HornerRule-2

$$\small{
\begin{array}{lrrl}
& f_1{(x)} &=& 3x^3-5x^2+10x-3\\
& f_2{(x)} &=& 3x-4\\
&\text{Divide } f_1{(x)} \text{ by } f_2{(x)} \text{ using Honer's method }
\end{array}
}$$

$$\small{\text{$
\begin{array}{r|crcrcrc|r}
3 &\quad& 3 &\quad& -5 &\quad& 10 &\quad& -3 \\
\hline
&&&&&&&\\
4 &\quad& &\quad& 4 &\quad& -\dfrac{4}{3} &\quad& 4\cdot \dfrac{26}{9}\\
\hline
&&&&&&&\\
&\quad& 1 &\quad& -1\cdot \dfrac{1}{3} &\quad&\dfrac{26}{3}\cdot \dfrac{1}{3} &\quad & \dfrac{77}{9}
\end{array}
$}}$$

 

$$1\cdot x^2 - \dfrac{1}{3}\cdot x + \dfrac{26}{9}
+ \dfrac{77}{9(3x-4)}$$

 

 

 

 Aug 18, 2015
 #5
avatar+99109 
+3

Thanks Heureka.   I have not seen Horner's Method before      

 

I have included this thread in our Reference Material Sticky Topic.  

 Aug 18, 2015

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