+0

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

0
786
5

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

Aug 17, 2015

#2
+101149
+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
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
+101149
+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]

CPhill Aug 18, 2015
#3
+101741
0

Very nice presentation Chris :)

Aug 18, 2015
#4
+22290
+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

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

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

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

Aug 18, 2015
#5
+101741
+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