Finding a Quartic Polynomial Equation

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Homework Help Overview

The discussion revolves around finding a quartic polynomial given specific zeros and a condition for its value at zero. The original poster states that the polynomial has zeros at 2 + i, 2 - i, 3 - 2i, and 3 + 2i, and seeks to determine the polynomial such that P(0) = 13.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of the polynomial's zeros and the condition P(0) = 13. Some suggest writing the polynomial in factored form and multiplying the factors, while others question the correctness of the constant term derived from their calculations.

Discussion Status

There is ongoing exploration of the polynomial's structure and the relationship between its coefficients and roots. Some participants have provided calculations and suggestions for scaling the polynomial, while others express confusion about the implications of their findings.

Contextual Notes

Participants are navigating the constraints of the problem, including the requirement for real coefficients and the specific condition for P(0). Additionally, there are multiple questions being raised regarding a separate problem involving projectile motion, indicating a broader context of homework assistance.

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Homework Statement



A quartic polynomial P(x) with real coefficients has zeros 2 + i and 3 - 2i. The other zeros are obviously 2 - i and 3 + 2i. If P(0) = 13, find a rule for P(x).

Homework Equations



-b/a = sum of all roots ; e/a = product of all roots

ax^4 + bx^3 + cx^2 + dx + e = 0

The Attempt at a Solution



I'm pretty sure e has to equal -13 because that would be the only way P(0) = 13, right?
If someone could kindly tell me how to solve this problem, it would be great.
 
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I Need This Asap Please!
 
Write the polynomial as a*(x-(2+i))*(x-(2-i))*(x-(3+2i))*(x-(3-2i)). You could just multiply it out and find a.
 
Simply multiply together the linear factors, and scale the coefficients to fit:

[tex](x-(2+i))(x-(2-i))(x-(3-2i))(x-(3+2i)) = x^4 - 10x^3 + 42x^2 - 82 x + 65[/tex]

Then scale by 5 to get the constant coefficient to be 13.
 
1/5x^4 - 2x^3 + 42/5x^2 - 82/5x + 13? that wouldn't work because the constant coefficient has to be negative 13 in order for P(0) to equal 13... would i then make everything negative? it already looks messy enough.
 
P(0) IS 13. What are you talking about?
 
oops, my bad. thanks, Dick and genneth.

One more thing I'm having trouble with...

From a platform 35m above the ground, a ball is thrown upward with an initial speed of 30 m/s. The approximate height of the ball above the ground t seconds later is given by h(t) = 35 + 30t - 5t^2. After how many seconds does the ball hit the ground? What is the domain and range of h? After how many seconds does the ball reach its maximum height above the ground?

Thanks again!
 
jeez, you ppl on this forum are smart as hell. ^_^
 
Also need this one ASAP.
 
  • #10
Please!
 
  • #11
Need your input first. It's not that hard. Start the problem out, how do you find when it hits the ground?
 
  • #12
35 + 30t - 5t^2...

divide everything by 5?

7 + 6t - t^2?

rewrite it = t^2 - 6t - 7?

factor? (x - 7) (x + 1) ?

cant be negative, so 7 is the answer? after 7 seconds, the height is 0? i did that, but it seems waaaay too simple.
 
  • #13
and also how would u find its maximum height in air?

>_<

how do u graph a quintic? should i just plug it in my calculator and copy it or is there a better way?

please, everyone, i need answers quickly! aahhh.
 
  • #14
omg precal said:
35 + 30t - 5t^2...
The problem was to use that height formula to determine when it hit the ground: h= 0. You want to solve the equation 35+ 30t- 5t^2= 0.

divide everything by 5?

7 + 6t - t^2?
Now that it is an equation you can simplify the equation by dividing both sides by 5: 7+ 6t- t^2= 0.

rewrite it = t^2 - 6t - 7?
Well, multiply the entire equation by -1: t^2- 6t- 7= 0.

factor? (x - 7) (x + 1) ?

cant be negative, so 7 is the answer? after 7 seconds, the height is 0? i did that, but it seems waaaay too simple.
Yes, you factor the right side of the equation and have (x-7)(x-1)= 0. Would you prefer it were harder?
Please remember that you have to deal with equations not just expressions.

omg precal said:
and also how would u find its maximum height in air?
h(t)= 35+ 6t- t^2 is a quadratic. Complete the square to get h(t)= H- (t-a)^2 (you find H and a). When t= a, h(t)= H. For t any other value, h(t)= H minus something and is lower. the maximum height is H and that happens when t= a.

>_<

how do u graph a quintic? should i just plug it in my calculator and copy it or is there a better way?

please, everyone, i need answers quickly! aahhh.
Quintic? A fifth degree polynomial? Is this yet a third question in the same thread? I would recommend using a calculator.
 
  • #15
HallsofIvy said:
The problem was to use that height formula to determine when it hit the ground: h= 0. You want to solve the equation 35+ 30t- 5t^2= 0.

Thanks.

Now that it is an equation you can simplify the equation by dividing both sides by 5: 7+ 6t- t^2= 0.

Thanks again.

Well, multiply the entire equation by -1: t^2- 6t- 7= 0.

Once more, thank you.

Yes, you factor the right side of the equation and have (x-7)(x-1)= 0. Would you prefer it were harder? Not really... :-p
Please remember that you have to deal with equations not just expressions.


h(t)= 35+ 6t- t^2 is a quadratic. Complete the square to get h(t)= H- (t-a)^2 (you find H and a). When t= a, h(t)= H. For t any other value, h(t)= H minus something and is lower. the maximum height is H and that happens when t= a.

Can you explain this once more? I don't get it.

Quintic? A fifth degree polynomial? Is this yet a third question in the same thread? I would recommend using a calculator.

Ok, thanks.

Look at Bold.
 
Last edited:

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