College Algebra

[lola09]

Need help on this problem: a polynomial function has zeros of -5 and 1+3i. Determine the function.

 

[mmm4444bot]

Hi. As you've shown no work nor asked questions, am I to guess that you're stuck at the beginning? Otherwise, please show your efforts, and explain why you're stuck.

In order to understand this exercise, you need to have already learned some stuff. :cool:



Know definition of polynomial and meaning of "polynomial's degree".



Know that polynomials may be factored in terms of their roots (zeros), AND the number of factors matches the polynomial's degree:

(x - zero1) (x - zero2) (x - zero3) ... (x - zeroN), where N is that polynomial's degree



Know how to multiply polynomials with other polynomials



Know definition of Complex numbers and conjugate pairs



Fact: All polynomial functions have the same number of Complex zeros as their degree



Fact: Complex zeros with an imaginary part always come in conjugate pairs



These topics are sufficient for you to write the factored form of the polynomial in your exercise.

f(x) = your factored polynomial

Although, I suspect that your instructor wants you to multiply all of the factors together, to write your answer with the polynomial in standard form.

Now, please tell us over which parts of this post you have questions, and we'll go from there. :cool:

 

[Bob Brown MSEE]

Polynomials are the simplest expressions in Algebra

Need help on this problem: a polynomial function has zeros of -5 and 1+3i. Determine the function.

We are used to seeing polynomials of x, written is standard form.
However if you ...
1) Only have one unknown, x
2) Only use add, subtract, multiply and known numbers.
3) The known numbers can be complex. (involve i as a known number)
Then no matter how you mix these operations and numbers, we WILL have a polynomial expression!

So don't worry about how it looks.
Let's use what we know to make P(x).
We need P(x) = 0 for two conditions,
P(x) = P1(x) * P2(x) is a good choice. The product of two polynomials will make a new polynomial.
If P1(-5) = 0 and P2(1+3i) = 0 then we are done. The following do just that...
P1= x+5
P2= x-1-3i
ANSWER: P(x) = (x+5)(x-1-3i) and we are DONE.

The problem did not state that you could not have i in the answer, but we can multiply by one more factor and fix that without screwing up the zeros.
Pretty ANSWER: P(x) = (x+5)(x-1-3i) (x-1+3i)

For extra credit: multiply out the polynomials and get the standard form, remember i*i=-1.

 

[Bob Brown MSEE]

P(x)=xxx+3xx+50 is a guess

 

[Deleted member 4993]

P(x)=xxx+3xx+50 is a guess

I do not think that the original problem defined the sought after polynomial as polynomial with real coefficients.

 

[Deleted member 4993]

Fact: All polynomial functions have the same number of Complex zeros as their degree - True if real numbers (with zero imaginary component) are subset of complex number.



Fact: Complex zeros always come in conjugate pairs - - False
if real numbers are subset of complex number (i.e. 3+0i and 3-0i need not come as pairs)
or
the polynomial has complex coefficients (with non-zero imaginary component).

:cool:

.

 

[JeffM]

Need help on this problem: a polynomial function has zeros of -5 and 1+3i. Determine the function.

There are an infinite number of polynomials that have these particular zeroes. You probably have not given the complete problem. Please give it exactly and completely.

 

[srmichael]

There are an infinite number of polynomials that have these particular zeroes. You probably have not given the complete problem. Please give it exactly and completely.

To expand on JeffM's comment, lola, we don't have enough information to determine the leading coefficient. To determine this we need an ininitial value. For example, f(0) = 3 or something similar.

Also, we don't know if any of the roots may have a multiplicity other than 1.

These two missing pieces lead to an infinite number of possible polynomials. However, knowing these two pieces gets us a unique polynomial.

 

[JeffM]

To expand on JeffM's comment, lola, we don't have enough information to determine the leading coefficient. To determine this we need an ininitial value. For example, f(0) = 3 or something similar.

Also, we don't know if any of the roots may have a multiplicity other than 1.

These two missing pieces lead to an infinite number of possible polynomials. However, knowing these two pieces gets us a unique polynomial.

To expand on Sir Michael's comment, if we knew that the polynomial was of degree 3 and had real coefficients, we would also have a unique polynomial. The point is that there are various pieces of information that would permit finding a unique polynomial if you shared them with us.

 

[mmm4444bot]

P(x)=xxx+3xx+50 is a guess

I'm with you, Bob. :cool:

I interpreted original instruct as "Determine a function".

My gut tells me that the coefficients are Real, in this student's course.

 

[mmm4444bot]

if real numbers are subset of complex number (i.e. 3+0i and 3-0i need not come as pairs)
or
the polynomial has complex coefficients (with non-zero imaginary component).

I have my doubts that this student's class is using polynomials with imaginary coefficients, but anything is possible.

I edited my post to clarify that I was speaking of Complex zeros with imaginary part always coming in pairs.