+386 I have to factorize a polinomial, then have 2 rationnal roots, say: x=2 and x=-2.
Then I divide the polinomial by (x-2) in reference to rationnal root x=2.
Then I end up with another polinomial and a leftover constant.
What should I do next? Is it possible to divide the new polinomial like by (x+2) in reference to rationnal root x=-2?
How should I handle the leftover throughout this process?
Thanks a lot!
+129840 If you have two rational roots, you shouldn't end up with any remainder......after dividing by (x - 2) and (x + 2), you should have either another polynomial [or 0] as a remainder
Can you give us an example of what you have???
+386 That's what I just realized haha!
Well, I start with:
x^5+0x^4-2x^3+5x^2-8x-20 / x-2
-(x^5-2x^4) x^4
= 2x^4-2x^3+5x^2-8x-20
-(2x^4-4x^3) 2x^3
= 2x^3+5x^2-8x-20
-(2x^3-4x^2) 2x^2
= x^2-8x-20
-(x^2-2x) x
= -6x-20
So from here if I keep going I'll be left with -32... Can't seem to find my error...
+129840 See if this helps, Tony
x^4 + 2x^3 + 2x^2 + 9x + 10
x - 2 [ x^5 + 0x^4 - 2x^ 3 + 5x^2 - 8x - 20 ]
x^5 - 2x^4
----------------------------------------------
2x^4 - 2x^3
2x^4 - 4x^3
-----------------
2x^3 + 5x^2
2x^3 - 4x^2
-----------------
9x^2 - 8x
9x^2 -18x
--------------
10x - 20
10x - 20
------------
0
x^3 + 2x + 5
x + 2 [ x^4 + 2x^3 + 2x^2 + 9x + 10 ]
x^4 + 2x^3
--------------------------------------
2x^2 + 9x
2x^2 + 4x
---------------
5x + 10
5x + 10
________
0
The remaining polynomial, x^3 + 2x + 5 doesn't have any rational roots........just one "real" root and 2 complex roots
+386 My error was that I didn't change -4x^2 sign. Sorry, that was a no brainer... meh
Huge thanks!
