Search 75,984 tutors
FIND TUTORS
Ask a question
0 0

Find the polynomial having the following roots with multiplicity and find the missing roots x=-3 multiplicity 3, x=3i , x=3 multiplicity of 2

Tutors, please sign in to answer this question.

1 Answer

Hi, if x=3 is a root then (x-3) is a factor and since the multiplicity is 2, (x-3)^2 is a factor.

 

Next, since -3 is a root, then (x- (-3)) is a factor, that is, (x+3) is a factor, and in fact, (x+3)^3 is a factor since the multiplicity is 3.

 

Now, what about x=3i being a root? It means (x-3i) is a factor. But for our polynomial to have real coefficients, the "complex conjugate" of 3i must be a root as well, The complex conjugate of a+bi is a-bi. So the complex conjugate of 3i is just -3k (since a=0 here)

 

So both (x-3i) and (x+3i) are factors; let's multiply them and see what we get: we get,

 

(x-3i)(x+3i) = x^2 -3ix + 3ix -9i^2 = x^2 +9

 

So our polynomial has (x^2 + 9) as a factor and the smallest (lower degree) real polynomial having all the desired roots and multiplicities is:

 

** p(x) = (x+3)^3 *  (x-3)^2 * (x^2 + 9)

 

where the * symbol just stands for "times" to clarify we are multiplying these (not needed mathematically but when reading computer format it helps to clarify for the reading eyes!) I suspect they want yo to leave the polynomial factored like this. Otherwise, you'll need to multiply all these together. Notice that (x+3)(x-3) = (x^2 - 9) so

 

(x+3)^3 * (x-3)^2 * (x^2 + 9) = (x+3)*[(x+3)^2*(x-3)^2)]*(x^2+9)

 

which is (x+3) * [(x^2 - 9)(x^2 - 9)]* (x^2+9)

 

re-grouping:  (x+3) * (x^2 - 9) * [(x^2 - 9)* (x^2 + 9)]

 

(x+3)(x^2 - 9) * (x^4  - 81) which I leave to you to multiply out to get a 7th degree polynomial..unless they are ok with it being factored in which again, use the form ** above. Hope this helps! Let me know if you have any questions.