# Homework Help: Polynomial problems

1. Apr 8, 2004

### mustang

Problem 16.
Find a cubic polynomial with integral coefficients having 3i and -6 as zeros.
This is what I have done:
(x-r1)(x-r2)
(x-3i)(x+6)
x^2-6x-3ix+(-18i)
x^2-(6-3i)x+(-18i)
Is this right????

Problem 18.
If 2 is a zero of the polynomial P(x)=3x^2+kx-8, find k.

2. Apr 8, 2004

### Hurkyl

Staff Emeritus
Is it cubic? Is it a polynomial? Does it have integral coefficients? Does it have 3i and -6 as zeroes? If so, then it's right!

What are your thoughts on this?

3. Apr 8, 2004

### mustang

My thoughts on problem 18.

ON problem 18 my thoughts are that since 2 is a zero of the polynomial
P(x)=3x^2+kx-8, find k. Then one of the x values is 2...so 3(2)^2+k(2)-8=28+k^2=sqrt(-28)=2isqrt(7).

I also have a question on problem 15.
A rectangular enclosure is to be made using a barn as one side and 80m of fencing to form the other three sides. What is the maximum area of such an enclosure?

4. Apr 8, 2004

### Hurkyl

Staff Emeritus
I find this confusing (and you probably do too); why don't you do it one step at a time?

(P.S. what kind of homework is this?)

5. Apr 9, 2004

### mustang

Problem 18.
3(2)^2+k(2)-8 ----I inserted the value of x equaling 2.
36+k(2)-8
28+k^2
k^2=-28
k=sqrt(-28)
k=2isqrt(7)

For the rectangle problem I believe that the problem is in the ax^2+bx+c=0.

This is Algebra 2 homework

6. Apr 9, 2004

### Muzza

The attempted solution to the second problem is pretty poorly presented. First you have some sort of expression that you simplify, then you magically move the 28 to the previously non-existant right hand side of the "equation"! Yeah, we can figure out what you mean, but still. Also, remember that k(2) does not equal k^2...

Last edited: Apr 9, 2004
7. Apr 9, 2004

### turin

I think we found your problem. And don't forget to thank Hurkyl for the extremely helpful suggestion in presentation.

It depends on the size of the barn (whether it is greater than some critical value, hint, hint).

Last edited: Apr 9, 2004
8. Apr 10, 2004

### mustang

Regarding problem 22 & 14.

Problem 14.
This is what i have done:
(sqrt[x-4]+10)^2=(x+4)^2
x-4+20sqrt(x-4)+100=x+4
20sqrt(x-4)=92
400(x-4) = 8,464 --I squared both sides.
400x-1600=8464
400x=10,064
x=25.16

Is this right??

Problem 22.
This is what I have done:
x^3+bx^2+cx-10= (x-a)(x-b)(x-c)
Would i input the values -2+i and -2-i for a and b. TO find the value for c???

9. Apr 10, 2004

### mustang

Sorry!!

10. Apr 10, 2004

### HallsofIvy

Let's go back to problem 16:
Problem 16.
"Find a cubic polynomial with integral coefficients having 3i and -6 as zeros."

questions, in particular, "is this a cubic equation?" which you never answered.
(In case you were wondering: no, x^2 is not cubic!)

Also, "-18i", being an imaginary number, is not what we would normally consider an integer (it is a "complex integer" or "Gaussian integer"). In order to have integer (in particular "real") solutions, you would have to have -3i as a root too. What does
(x-3i)(x+3i)(x+6) give you?

11. Apr 10, 2004

### mustang

THanks!!

Thanks alot for everything.!!