Prove that `x=a` is a root of a multiplicity `k ` of the polynomial `P(x)` , `deg[P(x)]=n,k<n` if and only if the sequence of the derivatives `P^((m))(x),m<=k<n` satisfies the condition: `P'(x)=0, P''(x)=0, P'''(x)=0,..., P^((k-1))(x)=0, P^((k))(x)!=0 ` .

Proved using mathematical induction and factor theorem.

Expert Answers

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By definition, a polynomial `P ( x ) ` has a root of multiplicity `k ` at `x = a ` if and only if `P ( x ) = ( x - a )^k Q ( x ) ` and `Q ( a ) != 0 .`

Prove by induction that the definition implies the property. The base case is `k = 1 ` and if `P ( x ) = ( x - a ) Q ( x ) , ` then `P ( a ) = 0` and `P' ( x ) = Q ( x ) + ( x - a ) Q' ( x ) , ` so `P' ( a ) = Q ( a ) != 0 .`

Induction step: let `P ( x ) = ( x - a )^( k + 1 ) Q ( x ) , ` then `P' ( x ) = (k+1)( x - a )^k Q ( x ) + ( x - a )^( k + 1 ) Q' ( x ) = ( x - a )^k ((k+1) Q ( x ) + ( x - a ) Q' ( x ))`has a root of multiplicity `k ` and by the induction assumption has zero derivatives at `x = a ` up to `(k-1)` -th. This way, `P ` also has all required zero derivatives.

To prove that the property implies the definition, we need an additional statement that if `P(x)=(x-a)Q(x) , ` then

`P^((m))(x) = m Q^((m-1))(x) + (x-a)Q^((m))(x), ` so `P^((m))(a) = m Q^((m-1))(a).`

It is proved by induction easily because `P^((m+1)) = (P^((m)))'.` Try to prove it yourself.

Now prove that the property implies the definition. The base case is again `k = 1 ` and if `P(a)=0 , ` then by the factor theorem `P(x)=(x-a)Q(x). ` And because `P'(x)=Q(x)+(x-a)Q'(x), ` we see `Q(a) = P'(a)!=0.`

Induction step: let `P(a)=P'(a)=...=P^((k))(a)=0, ` `P^((k+1))(a) != 0. ` Then `P(x)=(x-a)Q(x). ` By the statement we use, `Q^((m))(a)=1/(m+1) P^((m+1))(a), ` so `Q^((m))(a)=0 ` for `0lt=mlt=k-1 ` and `Q^((k))(a)!=0.`

This means that `Q ` satisfies the induction assumption for `k ` and `Q(x)=(x-a)^(k-1)R(x), ` so `P(x)=(x-a)^(k+1)R(x), ` which is what we need.

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