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Write the polynomial functions with least degree and a leading cofficent of one that contains these zeros.

I have two problems

1. 1,-1(multiplicity of 3), and 3i

2. 3(multiplicity of 2) and 3i

 

my problem is that I dont know how to solve a problem like this that contains a multiplicity.

 Oct 18, 2015

Best Answer 

 #2
avatar+6251 
+5

\(\mbox{For (2) we have, for the same reasons as before }\\ p(x)=(x-3)^2(x-\imath 3)(x+\imath 3)\)

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 Oct 18, 2015
 #1
avatar+6251 
+5

\(\mbox{This problem leaves out a vital piece of information.}\\ \mbox{Are the coefficients of your polynomial restricted to the real numbers or can they be complex numbers?}\)

\(\mbox{Assuming they are restricted to the reals.}\\ \mbox{From the description you have polynomial factors }\\ (x-1), (x+1)^3, (x-\imath 3)\\ \mbox{Note the power of 3 for the }(x+1) \mbox{ factor comes from the meaning of multiplicity 3.}\\ \mbox{Now given the restriction on real coefficients all complex roots must come in conjugate pairs.} \\ \mbox{So you must have a further root of }(x+\imath 3).\\ \mbox{Thus }p(x)=(x-1)(x+1)^3(x-\imath 3)(x+\imath 3)\\ \mbox{I leave it to you to multiply all that out.}\)

.
 Oct 18, 2015
 #2
avatar+6251 
+5
Best Answer

\(\mbox{For (2) we have, for the same reasons as before }\\ p(x)=(x-3)^2(x-\imath 3)(x+\imath 3)\)

Rom Oct 18, 2015

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