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

if x^2+2x+3 is a factor of x^10+ax^5+b then the zeros of x^2+2x+3 are also the zeros of x^2+2x+3.
x^2+2x+3=0
x= (-2+-sqr(4-4*3))/2
x=(-2+-sqr(-8))/2
x=(-2+-2i*sqr(2))/2
x=-1+-i*sqr(2)
x1=-1+i*sqrt(2) and x2=-1-i*sqr(2)

plugging x1 int x^10+ax^5+b=0 gives
-241+22*i*sqr(2)-a-11*i*sqr(2)*a+b=0

plugging x2 into x^10+ax^5+b=0 gives
-241-22*i*sqr(2)-a+11*i*sqr(2)*a+b=0

adding these two equations gives
-482-2a+2b=0
b-a=241
b=a+241
plugging this into the first equation gives
22*i*sqr(2)-11*i*sqr(2)*a=0
11*i*sqr(2)*a=22*i*sqr(2)
a=2
b=243
thus we have
x^10+2x^5+243=(x^2+2x+3)(x^8-2x^7+x^6+4x^5-11x^4+12x^3+9x^2-54x+81)

Posted by Daniel on 2013-08-03 22:45:59