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>>Hey! So I couldn't really understand this question and how to solve it... could anyone help me here?

 

P(x) is a polynomial with real coefficients such that P(2+i)=4-3i

There is a linear function Q(x)= ax+b with real coefficients such that Q(2+i)=P(2+i).

Find Q(x). Make sure to express your answer in the form ax+b. 

 

Thanks! I would appreciate an explanation too, if you could! :)

 Dec 28, 2018
 #1
avatar+98005 
+1

P(2 + i) = 4 - 3i

and

Q(2 + i) = P(2 + i)

 

Which implies that

 

Q(2 + i) = 4 - 3i

 

And Q (x ) = ax+ b

 

So

 

Q ( 2 + i)  =  a (2 + i) + b  =   2a +ai + b  

 

And

 

2a + ai + b  =   4 - 3i

 

So.....this implies that

 

(2a + b) + ai   =   4 - 3i         equate co-efficients

 

2a + b = 4

a = - 3

 

So

 

2(-3) + b = 4

 

-6 + b = 4

 

b = 10

 

So

 

Q(x)  =  -3x + 10

 

Check

 

Q(2 + i)  =  -3(2 + i) + 10   =  -6 - 3i + 10  =  4 - 3i

 

 

cool cool cool

 Dec 28, 2018
edited by CPhill  Dec 28, 2018
edited by CPhill  Dec 28, 2018
 #2
avatar+142 
+1

Wow! Thanks so much! I finally get it now. This was a big help for me. Thanks especially for doing it so quickly! :)

noobieatmath  Dec 28, 2018
 #3
avatar+98005 
+1

No prob....every now and then I manage to get one correct....LOL!!!

 

BTW....welcome aboard  !!!

 

 

cool cool cool

CPhill  Dec 28, 2018
 #4
avatar+142 
+1

Thanks, I appreciate it! Though I'm so glad that I was able to find this website!! So I should be the one who's thanking you!!

noobieatmath  Dec 28, 2018

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