value of k

[man111]

find value of [tex]k[/tex] if [tex]x^6-15x^3-8x^2+2[/tex] is Divisable by [tex](x^2+kx+1)[/tex]

[martin123456]

do something like that

we divide the two polynomials using the division algorithm
[tex]x^6-15x^3-8x^2+2 : x^2+kx+1 = x^4+...[/tex]
now we extract [tex]x^4(x^2+kx+1)=x^6+kx^5+x^4[/tex]
we get [tex]x^6-15x^3-8x^2+2-x^6-kx^5-x^4 = -kx^5-x^4-15x^3-8x^2+2[/tex]
this polynomial is divisible by [tex]x^2+kx+1[/tex] so we divide it by it
first, lets check if [tex]k=0[/tex] is ok. if [tex]x^6-15x^3-8x^2+2[/tex] was divisible by [tex]x^2+1[/tex] that would meant that the roots of [tex]x^2+1[/tex] are roots of the 1st polynomial, e.g. [tex]x_{1,2}=\pm {i}[/tex] are roots of the `st. We substitude [tex]x_1=i[/tex] and get [tex]1+15i+8+2=0[/tex] but this is not true. so [tex]k \ne 0[/tex].
[tex]-kx^5-x^4-15x^3-8x^2+2 : x^2+kx+1 = -kx^3+...[/tex]
we extract [tex]-kx^3(x^2+kx+1)=-kx^5-k^2x^4-kx^3[/tex]
we get [tex]-kx^5-x^4-15x^3-8x^2+2+kx^5+k^2x^4+kx^3=(k^2-1)x^4 + (k-15)x^3-8x^2+2[/tex]
now we have to check if [tex]k=\pm 1[/tex] works for us...do it by urself
we have that [tex](k^2-1)x^4 + (k-15)x^3-8x^2+2[/tex] is divisible by [tex]x^2+kx+1[/tex]
we divide them and we get that [tex](k^2-1)x^4 + (k-15)x^3-8x^2+2 : x^2+kx+1 = (k^2-1)x^2+...[/tex]
we extract [tex](k^2-1)x^2(x^2+kx+1)=(k^2-1)x^4+(k^2-1)x^3+(k^2-1)x^2[/tex]
we get [tex](k-14-k^2)x^3-(k^2+7)x^2+2[/tex]
now check by urself if the roots of [tex]k^2-k+14[/tex] work for us
this is divisible by[tex](k-14-k^2)x^3-(k^2+7)x^2+2 : x^2+kx+1 = (k-14-k^2)x[/tex]
we extract [tex](k-14-k^2)x(x^2+kx+1)=(k-14-k^2)x^3+k(k-14-k^2)x^2+(k-14-k^2)x[/tex]
we get [tex](-2k^2-7+14k+k^3)x^2-(k-14-k^2)x+2[/tex]
now we have [tex](-2k^2-7+14k+k^3)x^2-(k-14-k^2)x+2[/tex] is divisible by [tex]x^2+kx+1[/tex]
these polynomials have equal degrees so they are scalar multiples. We must have [tex]-2k^2-7+14k+k^3=2[/tex] and [tex]-(k-14-k^2)=2k[/tex]