+0

# ​ Synthetic division

+1
267
4 I know how to set it up but still having a hard time actually solving these types of problems.

Sep 18, 2018

#1
0

RP......Do you want synthetic division  or just regular polynomial division  ???   Sep 18, 2018
edited by CPhill  Sep 18, 2018
#2
+1

Both maybe

RainbowPanda  Sep 18, 2018
#3
+2

OK

Polynomial division....the idea is  to multiply the  "n"  of the divisor by a term that will "match" the first term

For instance.....we want to multiply  "n"  by 9n^4   to match the first term , 9n^5

We just keep carrying this idea through each time

For the second term....we want to multiply "n"  by  -2n^3  to "match"  -2n^4.....etc......

9n^4   - 2n^3      + 10n^2   +  n   + 6

(n + 9)  [  9n^5   + 79n^4    - 8n^3   + 91n^2  + 15n  +  46  ]

-(  9n^5  + 81n^4 )

_________________________________________

-2n^4     - 8n^3

- (-2n^4   - 18n^3)

_____________________________________

10n^3  +  91n^2

-(10n^3 + 90n^2)

____________________________

n^2   +  15n

-(   n^2  +     9n)

______________________

6n  + 46

-( -  6n  + 54)

____________

-8

So....the result  is   9n^4  - 2n^3  + 10n^2  + n  + 6   R [ -8 / (n + 9)  ]

Synthetic is usually  much faster....we  want to divide by  the thing that  makes (n + 9)  = 0....this is  - 9

So we have

-9 [ 9     79     -  8    91   15    46   ]

-81    18    -90   -9    -54

_________________________

9      -2     10       1     6     -8

The residual polynomial is  9n^4 - 2n^3 + 10n^2 + 1n  +  6    R [ -8 / ( n + 9) ]...just as we found before !!!

Here's a good primer on polynomial diivsion :

http://www.mesacc.edu/~scotz47781/mat120/notes/divide_poly/long_division/long_division.html

Here's a good primer on synthetic division : http://mesacc.edu/~scotz47781/mat120/notes/divide_poly/synthetic/synthetic_division.html   Sep 18, 2018
#4
+1

Thank you so much!

RainbowPanda  Sep 18, 2018