Pre-Calc Polynomial Help

[knpoe03]

Hello!

I will be linking a screenshot of the assignment. I am so confused about the second question. How do I go about finding h and k?

 

[pka]

Do you know about the factor theorem and that complex roots occur in conjugate pairs?
If you do look at this link. The result is a polynomial has integer coefficients but one real root.

 

[JeffM]

What method did you use to answer the first question?

 

[knpoe03]

What method did you use to answer the first question?

Well, I tried to multiply something like (x^2+?x+?)(x+?)(x+?)(x-?) because a friend told me that's how she picked out numbers with three real positive zeros.

 

[Dr.Peterson]

I am so confused about the second question. How do I go about finding h and k?
View attachment 22668

The problem confuses me a little. First, is the red part your own answer and explanation, or a supplied explanation?

Part 1 looks like it is meant to be solved by picking some random integers using Descartes' Rule of Signs. But it doesn't look like you did that.

Part 2 is, at least, poorly stated. Are we to assume the coefficients are those chosen in part 1? Without part 1, f(x) is not fully defined. But if part 1 was filled in with random numbers with appropriate signs, then it would in general be hard or impossible to answer part 2. So I have to guess that they want you to answer part 1 with part 2 already in view, so that you can be able to answer part 2; pka showed one way to do that.

Well, I tried to multiply something like (x^2+?x+?)(x+?)(x+?)(x-?) because a friend told me that's how she picked out numbers with three real positive zeros.

This would be an alternative way to answer part 1; you could pick almost any positive numbers for the ?'s, except that it would have one positive and two negative real zeros, so it's not quite right. Did your friend tell you why this would work, or are you just taking her word for it?

In any case, if you did this, do you see how you can use it to find the complex zeros?

 

[pka]

The problem confuses me a little. First, is the red part your own answer and explanation, or a supplied explanation?
Part 1 looks like it is meant to be solved by picking some random integers using Descartes' Rule of Signs. But it doesn't look like you did that.
Part 2 is, at least, poorly stated. Are we to assume the coefficients are those chosen in part 1? Without part 1, f(x) is not fully defined. But if part 1 was filled in with random numbers with appropriate signs, then it would in general be hard or impossible to answer part 2. So I have to guess that they want you to answer part 1 with part 2 already in view, so that you can be able to answer part 2; pka showed one way to do that.
This would be an alternative way to answer part 1; you could pick almost any positive numbers for the ?'s, except that it would have one positive and two negative real zeros, so it's not quite right. Did your friend tell you why this would work, or are you just taking her word for it?
In any case, if you did this, do you see how you can use it to find the complex zeros?

I must say that I do not see this as confusing. If \(f(x)=ax^5+bx^4+cx^3+dx^2+ex+g\) the question asks us to pick integral coefficients of \(f(x)\) such that \(f(x)\) has one or three real zeros. I offered this link which gives a polynomial with integer coefficients but having only \(4\) as a real zero.
Note that is a fifth degree which factors as \((x-4)(x-\{1-i\})(x-\{1+i\})(x-\{2-i\})(x-\{2+i\}) \)
It is clear from the factored form that there is only one real root and four complex roots which appear on conjugate pairs.

 

[knpoe03]

The problem confuses me a little. First, is the red part your own answer and explanation, or a supplied explanation?

Part 1 looks like it is meant to be solved by picking some random integers using Descartes' Rule of Signs. But it doesn't look like you did that.

Part 2 is, at least, poorly stated. Are we to assume the coefficients are those chosen in part 1? Without part 1, f(x) is not fully defined. But if part 1 was filled in with random numbers with appropriate signs, then it would in general be hard or impossible to answer part 2. So I have to guess that they want you to answer part 1 with part 2 already in view, so that you can be able to answer part 2; pka showed one way to do that.


This would be an alternative way to answer part 1; you could pick almost any positive numbers for the ?'s, except that it would have one positive and two negative real zeros, so it's not quite right. Did your friend tell you why this would work, or are you just taking her word for it?

In any case, if you did this, do you see how you can use it to find the complex zeros?

I found the numbers in red from performing (x^2-2x+6)(x-8)(x+12)(x+7). She explained to me why this would work (her teacher provided her with hints, while mine did not), and from her explanation it made sense. She read the explanation from an email her teacher sent to her, and I've summarized it below.

She said: "In a problem on page 3 of the packet, it will ask you to prove why h and k are zeros of f(x) by using long division. If you use something like (x^2+?x+?)(x+?)(x+?)(x-?), plug in whatever numbers you choose, and change some of the signs so that it will result in a fifth degree polynomial, the (x^2+?x+?) will then tell you what you need to divide the polynomial by (instead of guessing and checking to see if it results in 0). On top of this, the (x^2+?x+?) will be the solution to solving the complex conjugate of z1=h+ki and z1=h-ki. To do this, just do [x-(h-ki)][x-(h+ki)]. To find my h and k, I just plugged in some random numbers into this equation until it resulted in (x^2+?x+?). So to recap: use (x^2+?x+?)(x+?)(x+?)(x-?) and it will show you what you need to divide your fifth degree polynomial by, the solution of the complex conjugates z1=h+ki and z1=h-ki, and will give you an idea of what numbers you need to plug in for h and k."

However, I'm still unsure about the whole h and k part. Would it be easier if I just plugged in smaller numbers and re-started?

 

[knpoe03]

I must say that I do not see this as confusing. If \(f(x)=ax^5+bx^4+cx^3+dx^2+ex+g\) the question asks us to pick integral coefficients of \(f(x)\) such that \(f(x)\) has one or three real zeros. I offered this link which gives a polynomial with integer coefficients but having only \(4\) as a real zero.
Note that is a fifth degree which factors as \((x-4)(x-\{1-i\})(x-\{1+i\})(x-\{2-i\})(x-\{2+i\}) \)
It is clear from the factored form that there is only one real root and four complex roots which appear on conjugate pairs.

I agree, this is what I believe it is supposed to be. Sorry if I lacked explanation, I was worried if I put too much info then I would be freeloading off of your help.

 

[pka]

I agree, this is what I believe it is supposed to be. Sorry if I lacked explanation, I was worried if I put too much info then I would be freeloading off of your help.

Not to worry. While it is the forum's policy not to supply answers without evidence of effort. Professionals can usually tell the difference.
That is to say, in this particular case it is clear that we have someone who has clearly struggled with the heart of the question.

 

[Dr.Peterson]

I found the numbers in red from performing (x^2-2x+6)(x-8)(x+12)(x+7). She explained to me why this would work (her teacher provided her with hints, while mine did not), and from her explanation it made sense. She read the explanation from an email her teacher sent to her, and I've summarized it below.

She said: "In a problem on page 3 of the packet, it will ask you to prove why h and k are zeros of f(x) by using long division. If you use something like (x^2+?x+?)(x+?)(x+?)(x-?), plug in whatever numbers you choose, and change some of the signs so that it will result in a fifth degree polynomial, the (x^2+?x+?) will then tell you what you need to divide the polynomial by (instead of guessing and checking to see if it results in 0). On top of this, the (x^2+?x+?) will be the solution to solving the complex conjugate of z1=h+ki and z1=h-ki. To do this, just do [x-(h-ki)][x-(h+ki)]. To find my h and k, I just plugged in some random numbers into this equation until it resulted in (x^2+?x+?). So to recap: use (x^2+?x+?)(x+?)(x+?)(x-?) and it will show you what you need to divide your fifth degree polynomial by, the solution of the complex conjugates z1=h+ki and z1=h-ki, and will give you an idea of what numbers you need to plug in for h and k."

However, I'm still unsure about the whole h and k part. Would it be easier if I just plugged in smaller numbers and re-started?

So is all the red what you typed in? Or just the numbers in the blanks? And have you learned Descartes' Rule of Signs, which is what the "three or one" part suggests? These are the things I am confused about.

The funny thing in the hint is that if you obtained the polynomial as described, there is no need to do any long division! You already know all the factors.

The h and k part means that rather than randomly choosing (x^2-2x+6), you can choose h and k and expand [x-(h-ki)][x-(h+ki)] in order to obtain a quadratic with those zeros. If you instead picked the quadratic first, you would solve it to find the two zeros.

So, what did you use for h and k? (Note that you didn't need any trial and error; you could use any values you wanted. You were explicitly told that the signs did not have to be as indicated.)

It does seem that this teacher expects you to be looking ahead in the problem (even beyond what you have shown us) to decide how to solve it; that, if intended, is why I said the problem was not well written. My suspicion is that this teacher's ideas may not be what the author of the problem intended for you to do.