SOLUTION: If x=1+<2+<3 <=underroot Then what is the value of 2x^4-8x^3-5x^2+26x-28

Algebra ->  Square-cubic-other-roots -> SOLUTION: If x=1+<2+<3 <=underroot Then what is the value of 2x^4-8x^3-5x^2+26x-28      Log On


   

Question 1026742: If x=1+<2+<3
<=underroot
Then what is the value of
2x^4-8x^3-5x^2+26x-28
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
There must be a better trick. Let me know.
Maybe it is easier using synthetic division, but it did not sound so.
Here is my trick.
x=1%2Bc where c=sqrt%282%29%2Bsqrt%283%29
P%28x%29=2x%5E4-8x%5E3-5x%5E2%2B26x-28
P%281%2Bc%29=2%281%2Bc%29x%5E4-8%281%2Bc%29%5E3-5%281%2Bc%29%5E2%2B26%281%2Bc%29-28


That is considerably simpler.


So substituting the expressiond found for c%5E2 and c%5E4 into
P%281%2Bsqrt%282%29%2Bsqrt%283%29%29=P%281%2Bc%29=-13-17c%5E2%2B2c%5E4 , we get


NOTE: Can someone please tell me what is the simpler way to the solution that I overlooked? I will be very, very grateful to whoever tells me.