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Last 2

Edit: I solved the first one, it's x^2-9x-9 R=0/7x+8

RainbowPanda  Sep 19, 2018
edited by RainbowPanda  Sep 19, 2018
 #1
avatar+2291 
0

Is the last one -4/9| 1   49/9  -7/9   -34/3   -41/9

                                    -4/9  -20/9   4/3      40/9

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

                               1    5       -3      10       -1/9

n^3+5n^2-3n-10+ -1/9n+4

RainbowPanda  Sep 19, 2018
 #2
avatar+92674 
+2

Correct,RP!!!...good job  !!!

 

[Just a note...remember to write the remainder as   -1 /[9n + 4]....what you have is  -1/9n + 4...which translates to :

 

(-1/9) n   +  4 .... ]

 

 

Otherwise....very nice!!!

 

 

cool cool cool

CPhill  Sep 19, 2018
 #3
avatar+3446 
+1

Divide 9n^4+49n^3-7n^2-102n-41 by -4/9, 9  49  -7 -102 -41 divided by -4/9,

Solving synthetic, 9x^3+45x^2-27x-90-1/x-4/9?

Is it wrong?

tertre  Sep 19, 2018
edited by tertre  Sep 19, 2018
 #4
avatar+92674 
+2

Almost, terte

 

9n + 4   = 0   ⇒   n  = -4/9    and this is what we need to divide by

 

-4/9   [  9    49    - 7   - 102    - 41  ]

                   -4    -20      12       40

         ________________________

           9    45     -27     -90      - 1

 

Your remainder of  -1   is correct

 

The apparent residual plynomial is    9n^3  + 45n^2  - 27n  - 90

 

Note....to find the correct residual polynomial,  divide  the  apparent  residual polynomial by the  "9"  in the linear divisor of  9x - 4

 

So...we actually have

 

n^3  + 5n^2  - 3n  - 10  R ( -1/ [ 9n  - 4 ] )

 

To see why this is true...note the polynomial division  ;

 

                  n^3

(9n - 4)  [ 9n^4.........            ]

               -(9n^4)

               ___________

 

etc. ........

 

Note that the first term is  n^3 .....not 9n^3...

 

 

cool cool cool

CPhill  Sep 19, 2018
 #5
avatar+3446 
+1

Oh, thank you, CPhill! My small error, oops!

tertre  Sep 19, 2018

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