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.

Expert's answer

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,

(x5−ax−ax+1)/(x2+2x+1)==x3−2x2+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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