Question #104443
Find the value of a for which the polynomial x^5-ax²-ax+1has −1 as a root with multiplicity at least 2.
1
Expert's answer
2020-03-04T19:04:34-0500

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+1)2=x2+2x+1(x+1)^2=x^2+2x+1,

(x5axax+1)/(x2+2x+1)==x32x2+3x+((a+4)x2(3+a)+1)/(x2+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 (x2+2x+1)(x^2+2x+1) , it is necessary that ((a+4)x2(3+a)+1)(-(a+4)x^2-(3+a)+1) to be divisible by (x2+2x+1)(x^2+2x+1) .

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

Answer

a=5.a=-5.


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