Solve Polynomial Problem: Find k (3x^2+kx-8) w/ 2 as Zero

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Homework Help Overview

The discussion revolves around polynomial problems, specifically finding coefficients and understanding polynomial properties. One problem involves determining the value of k in the polynomial P(x)=3x^2+kx-8 given that 2 is a zero. Another problem seeks to construct a cubic polynomial with specific zeros.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the implications of having 2 as a zero in the polynomial and attempt to substitute values into the equation. Questions arise about the correctness of polynomial forms and the nature of coefficients. Some participants express confusion over the steps taken in the calculations.

Discussion Status

There is ongoing exploration of the problems, with participants questioning the validity of presented solutions and the clarity of expressions. Some guidance has been offered regarding the presentation of solutions, but no consensus has been reached on the correctness of the approaches taken.

Contextual Notes

Participants note that the problems are part of Algebra 2 homework, and there are discussions about the nature of the coefficients and the requirements for integral values in polynomial expressions.

mustang
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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.
 
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Is this right?

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!


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

What are your thoughts on this?
 
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?
 
3(2)^2+k(2)-8=28+k^2=sqrt(-28)=2isqrt(7).

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


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?

Again, what are your thoughts?


(P.S. what kind of homework is this?)
 
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
 
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:
mustang said:
Your step 2: 36+k(2)-8
Your step 3: 28+k^2
I think we found your problem. And don't forget to thank Hurkyl for the extremely helpful suggestion in presentation.




mustang said:
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?
It depends on the size of the barn (whether it is greater than some critical value, hint, hint).
 
Last edited:
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?
 
Sorry!

I made a mistake, SORRY!
 
  • #10
Let's go back to problem 16:
Problem 16.
"Find a cubic polynomial with integral coefficients having 3i and -6 as zeros."

You gave, as your solution "x^2-(6-3i)x+(-18i)" and then Hurkyl asked several
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
THanks!

Thanks a lot for everything.!
 

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