SOLUTION: A polynomial function f(x) with real coefficients has the given degree, zeros, and solution point. Degree 3 Zeros -3,3+3square root3i Solution Point f(â

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Question 1183353: A polynomial function f(x) with real coefficients has the given degree, zeros, and solution point.
Degree 3

Zeros -3,3+3square root3i

Solution Point

f(−1) = −172
(a) Write the function in completely factored form.
f(x) =


(b) Write the function in polynomial form.
f(x) =

Found 2 solutions by Solver92311, Edwin McCravy:
Answer by Solver92311(821)   (Show Source): You can put this solution on YOUR website!


You are given one integer zero and one complex zero. Complex zeros of the form always appear in conjugate pairs. If is a zero of the function, then is also a zero of the function.

If is a zero of a polynomial function then is a factor of the polynomial.

Your factors are:







However, if you simply multiply these factors, you produce one element of a set of polynomial functions with infinite elements that differ from each other by the value of an as yet undetermined constant lead coefficient. You need to determine the value of the lead coefficient based on the initial value of .

In general, the expression:



represents the entire set of functions where


Multiply the three binomial factors, substitute for which calculation will allow you to solve for



John

My calculator said it, I believe it, that settles it

From
I > Ø
Answer by Edwin McCravy(20059)   (Show Source): You can put this solution on YOUR website!
Instead of doing yours for you, I'll do one step-by-step exactly like yours.
You can use it as a model to solve yours by.

A polynomial function f(x) with real coefficients has the given degree, zeros, and solution point.
Degree 3
Zeros -4,4+(4square root5)i
Solution Point
f(−2) = −696
(a) Write the function in completely factored form.
f(x) =
(b) Write the function in polynomial form.
f(x) =
 
Solution step-by-step:




[That will be the completely factored form, once we find k and substitute



The last two factors are conjugates, so we form their product which
is the difference of the squares of the terms:



































Completely factored form:





Edwin
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