Suppose P is a polynomial with real coefficients and P(a+bi)=0. Prove (a-bi)=0

Solution Summary

Complex conjugates are used to prove that for P (a polynomial with real coefficients and P(a+bi)=0) that P(a-bi)=0. The solution is detailed and well presented.

Polynomial Identities and Proofs Essential Questions: How can polynomial... this activity is to demonstrate the proof of your polynomial identity through ...

... This provides examples of proofs regarding subfields and prime fields, polynomial in a domain, and commutative rings. ...Proof. ... Now consider the polynomial. ...

...Proof: From the condition, we know that is an extension field of that is an extension field of . ... If not, then we can find a nontrivial polynomial , such that . ...

... Show that. Proof: Let and let We want to show that By definition of and for some positive integer n. Since are all polynomial functions in I, we have. ...

... there are ai K ( not all 0 ) such that , ie, there is exist a polynomial such that u is a root of f ( x ) . Therefore u is algebraic over K . Hence the proof. ...

Power Series Proof. ... and multiplication of power series in the same way as for power series with real or complex coeficients,ie extend polynomial addition and ...

... In fact, any polynomial with integer Now, an X n + ... + a1 X + a 0 can be written as X (an X n −1 ... The same proof applies to any ideal (p,X), of course. ...