+2448 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
+129852 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!!!
+4622 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?
+129852 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...
