Find the values for a and b to force the polynomial x¹⁰ + ax⁵ + b to have x²+2x+3 as a factor.

(In reply to Another approach by Steve Herman)

I used Steve Herman's other approach doing a very tedious long division.

The final remainder was:

(-11a + 22)x + (-12a - 219 + b)
Which leads to:
a=2
b=243
which agrees with the first solution, by Daniel.

Also, in the quotient obtained by long division, the coefficients for the powers higher than x^3 came out just as in Daniel's solution.  After that:
(a+10)x^3 + (13-2a)x^2 + (a-56)x + (4a+73) i.e. with same answer, only in terms of "a"

Posted by Larry on 2013-08-04 20:05:43