Answer to Question #104443 in Algebra for maria

Question

Find the value of a for which the polynomial x^5-ax²-ax+1has −1 as a root with multiplicity at least 2.

Answer 1

Solution

If -1 is a root with a multiplicity of at least 2, then this polynomial must be divisible by $(x+1)^2$

$(x+1)^2=x^2+2x+1$,

$(x^5-ax-ax+1)/(x^2+2x+1)=\\=x^3-2x^2+3x+\\(-(a+4)x^2-(3+a)+1)/(x^2+2x+1)$

In order for a polynomial to be divisible by $(x^2+2x+1)$ , it is necessary that $(-(a+4)x^2-(3+a)+1)$ to be divisible by $(x^2+2x+1)$ .

Then $-(a+4)=1$ and $-(3+a)=2$ , hence $a=-5.$

Answer

$a=-5.$